diff --git a/.gitignore b/.gitignore index 445dfa66bcafe01e09dacf4c02ed2994b4810bbc..3f0d61d2d4279f23d986d7008bc0da455c8c6a11 100644 --- a/.gitignore +++ b/.gitignore @@ -16,4 +16,5 @@ test.html ./src/__pycache__/*.pyc leetcode_class.md fix_bug.py -tools.py \ No newline at end of file +tools.py +lanqiao.cpp \ No newline at end of file diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/1.\347\214\234\345\271\264\351\276\204/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/1.\347\214\234\345\271\264\351\276\204/solution.md" index da6064cb87d09e88c71b12491176b7f88a21567b..b669c5140a17c45017c9178241a8575ecdff404c 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/1.\347\214\234\345\271\264\351\276\204/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/1.\347\214\234\345\271\264\351\276\204/solution.md" @@ -4,15 +4,7 @@ 一次，他参加某个重要会议，年轻的脸孔引人注目。于是有人询问他的年龄，他回答说：“我年龄的立方是个4位数。我年龄的4次方是个6位数。这10个数字正好包含了从0到9这10个数字，每个都恰好出现1次。” 请你推算一下，他当时到底有多年轻。 - -## aop - -### before - -```cpp - -``` -### after +以下代码的输出中包含了该数学家的真实年龄，请你从下面四个选项中选择正确答案。 ```cpp #include @@ -30,6 +22,20 @@ int main() } ``` + +## aop + +### before + +```cpp + +``` +### after + +```cpp + +``` + ## 答案 ```cpp diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/10.\345\244\247\346\225\260\344\271\230\346\263\225/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/10.\345\244\247\346\225\260\344\271\230\346\263\225/solution.md" index 12d3e64121c1d46af330f76e26c63613825ca918..e24c9eaa8e7926a3f26eeeb7cdb16f7c3158e536 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/10.\345\244\247\346\225\260\344\271\230\346\263\225/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/10.\345\244\247\346\225\260\344\271\230\346\263\225/solution.md" @@ -5,32 +5,12 @@ ![](https://img-blog.csdn.net/20160125091111485) 上图表示了分块乘法的原理。可以把大数分成多段（此处为2段）小数，然后用小数的多次运算组合表示一个大数。可以根据int的承载能力规定小块的大小，比如要把int分成2段，则小块可取10000为上限值。注意，小块在进行纵向累加后，需要进行进位校正。 - -## aop - -### before - -```cpp -#include -``` -### after +请从以下四个选项中选择正确答案，填补在空白处，使得代码能运行后能产生正确的结果。 ```cpp -int main(int argc, char *argv[]) -{ - int x[] = {0, 0, 0, 0}; +#include +using namespace std; - bigmul(87654321, 12345678, x); - - printf("%d%d%d%d\n", x[0], x[1], x[2], x[3]); - - return 0; -} -``` - -## 答案 - -```cpp void bigmul(int x, int y, int r[]) { int base = 10000; @@ -46,98 +26,62 @@ void bigmul(int x, int y, int r[]) r[3] = n1 % base; r[2] = n1 / base + n2 % base + n3 % base; - r[1] = n2 / base + n3 / base + n4 % base; + r[1] = __________________; r[0] = n4 / base; - r[1] += r[2] / base; + r[1] += r[2] / base; r[2] = r[2] % base; r[0] += r[1] / base; r[1] = r[1] % base; } + +int main(int argc, char *argv[]) +{ + int x[] = {0, 0, 0, 0}; + + bigmul(87654321, 12345678, x); + + printf("%d%d%d%d\n", x[0], x[1], x[2], x[3]); + + return 0; +} ``` -## 选项 +## aop -### A +### before ```cpp -void bigmul(int x, int y, int r[]) -{ - int base = 10000; - int x2 = x / base; - int x1 = x % base; - int y2 = y / base; - int y1 = y % base; - int n1 = x1 * y1; - int n2 = x1 * y2; - int n3 = x2 * y1; - int n4 = x2 * y2; +``` +### after - r[3] = n1 % base; - r[2] = n1 / base + n2 % base + n3 % base; - r[1] = n2 / base + n3 / base + n4 % base; - r[0] = n4 / base; +```cpp - r[1] += r[2] % base; - r[2] = r[2] / base; - r[0] += r[1] / base; - r[1] = r[1] % base; -} ``` -### B +## 答案 ```cpp -void bigmul(int x, int y, int r[]) -{ - int base = 10000; - int x2 = x / base; - int x1 = x % base; - int y2 = y / base; - int y1 = y % base; +n2 / base + n3 / base + n4 % base +``` +## 选项 - int n1 = x1 * y1; - int n2 = x1 * y2; - int n3 = x2 * y1; - int n4 = x2 * y2; - r[3] = n1 % base; - r[2] = n1 / base + n2 % base + n3 % base; - r[1] = n2 / base + n3 / base + n4 / base; - r[0] = n4 / base; +### A - r[1] += r[2] / base; - r[2] = r[2] % base; - r[0] += r[1] / base; - r[1] = r[1] % base; -} +```cpp +n2 % base + n3 / base + n4 / base ``` -### C +### B ```cpp -void bigmul(int x, int y, int r[]) -{ - int base = 10000; - int x2 = x / base; - int x1 = x % base; - int y2 = y / base; - int y1 = y % base; - - int n1 = x1 * y1; - int n2 = x1 * y2; - int n3 = x2 * y1; - int n4 = x2 * y2; +n2 / base + n3 % base + n4 / base +``` - r[3] = n1 % base; - r[2] = n1 / base + n2 % base + n3 % base; - r[1] = n2 / base + n3 % base + n4 / base; - r[0] = n4 / base; +### C - r[1] += r[2] / base; - r[2] = r[2] % base; - r[0] += r[1] / base; - r[1] = r[1] % base; -} +```cpp +n2 / base + n3 / base + n4 / base ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/11.9\346\225\260\347\256\227\345\274\217/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/11.9\346\225\260\347\256\227\345\274\217/solution.md" index 18510b397c9f181d994353ee9e5e6f6e7759e3c9..946ecea53f57252f7b3b87ddeb8c3c4d8cce3b46 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/11.9\346\225\260\347\256\227\345\274\217/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/11.9\346\225\260\347\256\227\345\274\217/solution.md" @@ -14,6 +14,8 @@ 1. 总数目包含题目给出的那个示例。 2. 乘数和被乘数交换后作为同一方案来看待。 +以下错误的是？ + ## aop ### before @@ -21,8 +23,6 @@ ```cpp #include using namespace std; -int bei[10]; -map mp; ``` ### after @@ -33,15 +33,18 @@ map mp; ## 答案 ```cpp +int bei[10]; +map mp; + int main() { int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int res = 0; - do + while (next_permutation(a, a + 9)) { for (int i = 1; i < 9; i++) { - memset(bei, 0, sizeof(bei)); + memset(bei, 0, sizeof(bei)); long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0; for (int j = 0; j <= i; j++) { @@ -52,7 +55,6 @@ int main() for (int k = i + 1; k < 9; k++) { right = right * 10 + a[k]; - x = x * 10 + a[k]; } y = right; y = y * 10; @@ -61,7 +63,7 @@ int main() ans = left * right; long long int ff = ans; - while (ans > 0) + while (ans >= 0) { int x = ans % 10; ans = ans / 10; @@ -71,14 +73,14 @@ int main() t++; } } - if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0) + if (mp.count(x) == 0 && mp.count(y) == 0) { res++; mp[x] = 1; mp[y] = 1; } } - } while (next_permutation(a, a + 9)); + }; cout << res << endl; return 0; } @@ -89,52 +91,55 @@ int main() ### A ```cpp +const int N = 1e3 + 5; +int a[15] = {9, 2, 1, 3, 8, 5, 6, 7, 4}, res = 0, vis[12], temp1; +bool fun(int x) +{ + int t1 = 0, t2 = 0; + for (int i = x, j = 1; i >= 0; i--, j *= 10) + { + t1 = a[i] * j + t1; + } + for (int i = 8, j = 1; i > x; i--, j *= 10) + { + t2 = a[i] * j + t2; + } + + temp1 = t1 * t2; + int temp = temp1; + memset(vis, 0, sizeof(vis)); + while (temp) + { + vis[temp % 10] = 1; + temp /= 10; + } + for (int i = 1; i <= 9; i++) + { + if (!vis[i]) + return false; + } + return true; +} int main() { - int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; - int res = 0; - while (next_permutation(a, a + 9)) + + for (int i = 0; i < 9; i++) + a[i] = i + 1; + + set st; + do { - for (int i = 1; i < 9; i++) + for (int i = 0; i < 8; i++) { - memset(bei, 0, sizeof(bei)); - long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0; - for (int j = 0; j <= i; j++) - { - left = left * 10 + a[j]; - } - x = left; - x = x * 10; - for (int k = i + 1; k < 9; k++) - { - right = right * 10 + a[k]; - } - y = right; - y = y * 10; - for (int j = 0; j <= i; j++) - y = y * 10 + a[j]; - - ans = left * right; - long long int ff = ans; - while (ans >= 0) - { - int x = ans % 10; - ans = ans / 10; - if (bei[x] == 0 && x != 0) - { - bei[x] = 1; - t++; - } - } - if (mp.count(x) == 0 && mp.count(y) == 0) + if (fun(i)) { + st.insert(temp1); res++; - mp[x] = 1; - mp[y] = 1; } } - }; - cout << res << endl; + } while (next_permutation(a, a + 9)); + cout << res / 2 << '\n'; + return 0; } ``` @@ -142,52 +147,66 @@ int main() ### B ```cpp -int main() +int st[10], res[10], book[10], cnt; + +int cal(int l, int r) { - int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; - int res = 0; - do + int t = 0; + for (int i = l; i <= r; i++) { - for (int i = 1; i < 9; i++) - { - memset(bei, 0, sizeof(bei)); - long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0; - for (int j = 0; j <= i; j++) - { - left = left * 10 + a[j]; - } - x = left; - x = x * 10; - for (int k = i + 1; k < 9; k++) - { - right = right * 10 + a[k]; - x = x * 10 + a[k]; - } - y = right; - y = y * 10; - for (int j = 0; j <= i; j++) - y = y * 10 + a[j]; + t = t * 10 + res[i]; + } + return t; +} - long long int ff = ans; - while (ans > 0) - { - int x = ans % 10; - ans = ans / 10; - if (bei[x] == 0 && x != 0) - { - bei[x] = 1; - t++; - } - } - if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0) +bool check(int k) +{ + for (int i = 0; i < 10; i++) + book[i] = 0; + while (k) + { + book[k % 10]++; + k /= 10; + } + bool flag = true; + for (int i = 1; i <= 9; i++) + { + if (book[i] != 1) + flag = false; + } + return flag; +} + +void dfs(int k) +{ + if (k == 9) + { + for (int i = 0; i < 8; i++) + { + int a = cal(0, i); + int b = cal(i + 1, 8); + if (check(a * b)) { - res++; - mp[x] = 1; - mp[y] = 1; + cnt++; } } - } while (next_permutation(a, a + 9)); - cout << res << endl; + return; + } + for (int i = 1; i <= 9; i++) + { + if (!st[i]) + { + st[i] = 1; + res[k] = i; + dfs(k + 1); + st[i] = 0; + } + } +} +int main() +{ + dfs(0); + cout << cnt / 2; return 0; } ``` @@ -195,52 +214,87 @@ int main() ### C ```cpp -int main() +typedef long long LL; +const int N = 15; +bool vis[N]; +int ans; + +bool check_2(LL sumb) { - int a[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; - int res = 0; - while (next_permutation(a, a + 9)) + bool st[N]; + for (int i = 1; i <= 9; i++) { - for (int i = 1; i < 9; i++) + st[i] = 0; + } + while (sumb) + { + st[sumb % 10] = true; + sumb = sumb / 10; + } + bool flag = false; + for (int i = 1; i <= 9; i++) + { + if (!st[i]) { - memset(bei, 0, sizeof(bei)); - long long int ans, left = 0, right = 0, t = 0, x = 0, y = 0; - for (int j = 0; j <= i; j++) - { - left = left * 10 + a[j]; - } - x = left; - x = x * 10; - for (int k = i + 1; k < 9; k++) - { - right = right * 10 + a[k]; - } - y = right; - y = y * 10; - for (int j = 0; j <= i; j++) - y = y * 10 + a[j]; + return false; + } + } + return true; +} - ans = left * right; - long long int ff = ans; - while (ans > 0) - { - int x = ans % 10; - ans = ans / 10; - if (bei[x] == 0 && x != 0) - { - bei[x] = 1; - t++; - } - } - if (t == 9 && mp.count(x) == 0 && mp.count(y) == 0) - { - res++; - mp[x] = 1; - mp[y] = 1; - } +bool check_1(int suma, int sumc) +{ + LL sumb = (LL)suma * sumc; + if (check_2(sumb)) + { + return true; + } + return false; +} + +void dfs_c(int u, int suma, int sumc) +{ + if (u > 9) + { + return; + } + + if (check_1(suma, sumc)) + { + ans++; + return; + } + + for (int i = 1; i <= 9; i++) + { + if (!vis[i]) + { + vis[i] = true; + dfs_c(u + 1, suma, sumc * 10 + i); + vis[i] = false; } - }; - cout << res << endl; + } +} + +void dfs_a(int u, int suma) +{ + dfs_c(u, suma, 0); + + for (int i = 1; i <= 9; i++) + { + if (!vis[i]) + { + vis[i] = true; + dfs_a(u + 1, suma * 10 + i); + vis[i] = false; + } + } +} + +int main() +{ + dfs_a(0, 0); + cout << ans / 2 << endl; return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/12.\346\257\224\351\205\222\351\207\217/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/12.\346\257\224\351\205\222\351\207\217/solution.md" index f61d38922a5321c7d2eef11a7e494e1fbc7c3274..04eff5de1ce9e49edff3ea924e33c32cc24d5b77 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/12.\346\257\224\351\205\222\351\207\217/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/12.\346\257\224\351\205\222\351\207\217/solution.md" @@ -15,47 +15,27 @@ 例如,有一种可能是：```20,5,4,2,0``` - +以下错误的一项是？ ## aop ### before ```cpp -#include -#include +#include using namespace std; -int d, a1[4]; ``` ### after ```cpp -int main() -{ - for (int sum = 20; sum >= 1; sum--) - { - for (int a = 1; a <= 20; a++) - { - for (int b = 1; b <= 20; b++) - { - for (int c = 1; c <= 20; c++) - { - a1[0] = sum, a1[1] = a, a1[2] = b, a1[3] = c; - if (getS(a1) == a1[0] && a > b && b > c && sum > a) - { - printf("%d, %d, %d, %d, 0\n", sum, a, b, c); - } - } - } - } - } - return 0; -} + ``` ## 答案 ```cpp +int d, a1[4]; + int d1(int *a1) { int sum = a1[0]; @@ -74,12 +54,31 @@ int getS(int *a1) int index = d1(a1); a1[0] = a1[0] * a1[index]; ss = a1[index]; + sum += ss; } - for (int i = 1; i < 4; i++) + return sum; +} + +int main() +{ + for (int sum = 20; sum >= 1; sum--) { - sum += (a1[0] / a1[i]); + for (int a = 1; a <= 20; a++) + { + for (int b = 1; b <= 20; b++) + { + for (int c = 1; c <= 20; c++) + { + a1[0] = sum, a1[1] = a, a1[2] = b, a1[3] = c; + if (getS(a1) == a1[0] && a > b && b > c && sum > a) + { + printf("%d, %d, %d, %d, 0\n", sum, a, b, c); + } + } + } + } } - return sum + ss; + return 0; } ``` ## 选项 @@ -88,86 +87,99 @@ int getS(int *a1) ### A ```cpp -int d1(int *a1) +int main() { - int sum = a1[0]; - for (int i = 1; i < 4; i++) + float n; + float a, b, c; + float s1, s2, s3; + for (n = 5; n <= 20; n++) { - if (sum % a1[i] != 0) - return i; + for (a = 1; a <= n; a++) + for (b = 1; b <= n; b++) + for (c = 1; c <= n; c++) + { + s1 = n - a; + s2 = n - a - b; + s3 = n - a - b - c; + if (1 / n + 1 / s1 + 1 / s2 + 1 / s3 == 1 && s1 > 0 && s2 > 0 && s3 > 0) + cout << n << "," << s1 << "," << s2 << "," << s3 << "," + << "0" << endl; + } } return 0; } -int getS(int *a1) -{ - int sum = 0, ss = 1; - while (d1(a1) != 0) - { - int index = d1(a1); - a1[0] = a1[0] * a1[index]; - ss = a1[index]; - } - for (int i = 1; i <= 4; i++) - { - sum += (a1[0] / a1[i]); - } - return sum + ss; -} ``` ### B ```cpp -int d1(int *a1) +int cm(int a, int b) { - int sum = a1[0]; - for (int i = 1; i <= 4; i++) + int temp; + int mul; + mul = a * b; + if (a % b == 0) + return a; + else { - if (sum % a1[i] != 0) - return i; + while (a % b != 0) + { + temp = a % b; + a = b; + b = temp; + } + return mul * b; } - return 0; } -int getS(int *a1) +int main() { - int sum = 0, ss = 1; - while (d1(a1) != 0) - { - int index = d1(a1); - a1[0] = a1[0] * a1[index]; - ss = a1[index]; - } - for (int i = 1; i <= 4; i++) - { - sum += (a1[0] / a1[i]); - } - return sum + ss; + int a, b, c, d; + int temp; + int ta, tb, tc, td; + for (a = 20; a >= 5; a--) + for (b = a - 1; b >= 4; b--) + for (c = b - 1; c >= 3; c--) + for (d = c - 1; d >= 2; d--) + { + temp = cm(a, b); + temp = cm(temp, c); + temp = cm(temp, d); + ta = temp / a; + tb = temp / b; + tc = temp / c; + td = temp / d; + if (ta + tb + tc + td == temp) + cout << a << "," << b << "," << c << "," << d << "," + << "0" << endl; + } + return 0; } ``` ### C ```cpp -int d1(int *a1) +int main(void) { - int sum = a1[0]; - for (int i = 1; i < 4; i++) - { - if (sum % a1[i] != 0) - return i; - } - return 0; -} -int getS(int *a1) -{ - int sum = 0, ss = 1; - while (d1(a1) != 0) + float a, b, c, d; + float x, y, z, m; + for (x = 20; x >= 5; x--) { - int index = d1(a1); - a1[0] = a1[0] * a1[index]; - ss = a1[index]; - sum += ss; + for (y = x - 1; y >= 4; y--) + { + for (z = y - 1; z >= 3; z--) + { + for (m = y - 1; m >= 2; m--) + { + if (y * z * m + x * z * m + x * y * m + x * y * z == x * y * z * m) + + { + cout << x << "," << y << "," << z << "," << m << "," + << "0" << endl; + } + } + } + } } - return sum; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/13.\345\210\206\346\225\260/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/13.\345\210\206\346\225\260/solution.md" index 0474194f922d92318eb6c509318d7b5a16cf518f..23ac9c67d7a7c0f0ea34e27b6c9ffeb09ea1e451 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/13.\345\210\206\346\225\260/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/13.\345\210\206\346\225\260/solution.md" @@ -4,6 +4,7 @@ 类似：3/2 当然，这只是加了前2项而已。分子分母要求互质。 +下面哪一项是错误的？ ## aop ### before @@ -15,17 +16,7 @@ using namespace std; ### after ```cpp -int gcd(long a, long b) -{ - if (b == 0) - return a; - return gcd(b, a % b); -} -int main() -{ - cout << gcd(pow_2(20) - 1, pow_2(19)) << endl; - cout << pow_2(20) - 1 << "/" << pow_2(19) << endl; -} + ``` ## 答案 @@ -44,6 +35,17 @@ long pow_2(int b) } return res; } + +int gcd(long a, long b) +{ + if (b == 0) + return a; + return gcd(b, a % b); +} +int main() +{ + cout << pow_2(20) << "/" << pow_2(19) << endl; +} ``` ## 选项 @@ -59,45 +61,63 @@ long pow_2(int b) { if (b & 1) res *= x; - b <<= 1; + b >>= 1; x = x * x; } return res; } + +int gcd(long a, long b) +{ + if (b == 0) + return a; + return gcd(b, a % b); +} +int main() +{ + cout << pow_2(20) - 1 << "/" << pow_2(19) << endl; +} ``` ### B ```cpp -long pow_2(int b) +int gcd(int a, int b) { - long x = 2; - long res = 1; - while (b > 0) - { - if (b && 1) - res *= x; - b <<= 1; - x = x * x; - } - return res; + return a % b ? gcd(b, a % b) : b; +} + +int main() +{ + int a = 1048575, b = 524288; + int t = gcd(a, b); + cout << a / t << "/" << b / t << endl; + return 0; } ``` ### C ```cpp -long pow_2(int b) +int gcd(int a, int b) { - long x = 2; - long res = 1; while (b > 0) { - if (b & 1) - res = x; - b >>= 1; - x = x * x; + int c; + c = a % b; + a = b; + b = c; } - return res; + return a; +} +int main() +{ + int a = 1048575; + int b = 524288; + int c = gcd(a, b); + + cout << a / c << "/" << b / c << endl; + + return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/14.\346\210\220\347\273\251\345\210\206\346\236\220/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/14.\346\210\220\347\273\251\345\210\206\346\236\220/solution.md" index a362b07a9c11d716b6a48c1cb7bc6212bca6e688..fce4b0400f0a58b1fc4506cc30c3490bdfca400c 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/14.\346\210\220\347\273\251\345\210\206\346\236\220/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/14.\346\210\220\347\273\251\345\210\206\346\236\220/solution.md" @@ -43,14 +43,14 @@ 71.29 ``` +以下选项错误的是？ + ## aop ### before ```cpp -#include -#include - +#include using namespace std; ``` ### after @@ -66,24 +66,22 @@ int main() { int n; cin >> n; - int sum = 0; - int top = 0; - int low = 100; - int score; + int maxv = 0, minv = 100, sum = 0; for (int i = 0; i < n; i++) { - cin >> score; - if (score > top) - top = score; - if (score < low) - low = score; - sum += score; + int x; + cin >> x; + minv = min(minv, x); + maxv = max(maxv, x); + sum += x; } - printf("%d\n%d\n%.2lf", top, low, (sum * 1.0 / n)); + + cout << maxv << endl; + cout << minv << endl; + printf("%.2f", sum / n); return 0; } - ``` ## 选项 @@ -93,23 +91,32 @@ int main() ```cpp int main() { - int n; + int n, max = 0, min = 0; + double sum = 0, average = 0; cin >> n; - int sum = 0; - int top = 0; - int low = 100; - - int score; - for (int i = 0; i < n; i++) + int a[10000]; + cin >> a[0]; + sum = a[0]; + max = a[0]; + min = a[0]; + for (int i = 1; i < n; i++) { - cin >> score; - if (score > top) - top = score; - if (score < low) - low = score; - sum += score; + cin >> a[i]; + if (max < a[i]) + { + max = a[i]; + } + if (min > a[i]) + { + min = a[i]; + } + sum += a[i]; } - printf("%d\n%d\n%.f", top, low, (sum * 1.0 / n)); + average = sum / n; + cout << max << endl + << min << endl + << setiosflags(ios::fixed) << setprecision(2) << average; + return 0; } ``` @@ -117,25 +124,56 @@ int main() ### B ```cpp +int n; + +int Max(int m[10005]) +{ + int max = m[0]; + for (int i = 1; i < n; i++) + { + if (max < m[i]) + { + max = m[i]; + } + } + return max; +} + +int Min(int m[10005]) +{ + int min = m[0]; + for (int i = 1; i < n; i++) + { + if (min > m[i]) + { + min = m[i]; + } + } + return min; +} + int main() { - int n; cin >> n; - int sum = 0; - int top = 0; - int low = 100; + int m[10005]; + for (int i = 0; i < n; i++) + { + cin >> m[i]; + } + + int ans = 0; - int score; for (int i = 0; i < n; i++) { - cin >> score; - if (score > top) - score = top; - if (score < low) - score = low; - sum += score; + ans += m[i]; } - printf("%d\n%d\n%.2lf", top, low, (sum * 1.0 / n)); + + int a = Max(m); + int b = Min(m); + + cout << "最高分 = " << a << endl; + cout << "最低分 = " << b << endl; + printf("%.2lf", 1.0 * ans / n); return 0; } ``` @@ -155,13 +193,13 @@ int main() for (int i = 0; i < n; i++) { cin >> score; - if (score < top) + if (score > top) top = score; - if (score > low) + if (score < low) low = score; sum += score; } - printf("%d\n%d\n%d", top, low, (sum / n)); + printf("%d\n%d\n%.2lf", top, low, (sum * 1.0 / n)); return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/15.\347\254\254\345\207\240\344\270\252\345\271\270\350\277\220\346\225\260/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/15.\347\254\254\345\207\240\344\270\252\345\271\270\350\277\220\346\225\260/solution.md" index 9095308a21b6b346507457ccfe21ee6c7b53a7e3..35ead1ac1bc858c5978011418652604086912f5e 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/15.\347\254\254\345\207\240\344\270\252\345\271\270\350\277\220\346\225\260/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/15.\347\254\254\345\207\240\344\270\252\345\271\270\350\277\220\346\225\260/solution.md" @@ -12,17 +12,14 @@ x星的国王有个怪癖，他只喜欢数字3,5和7。 请你帮小明计算一下，59084709587505是第几个幸运数字。 - +以下选项错误的是？ ## aop ### before ```cpp -#include -#include -#include -#include +#include using namespace std; ``` ### after @@ -34,34 +31,27 @@ using namespace std; ## 答案 ```cpp +typedef long long LL; +const LL MAX = 59084709587505; int main() { - set st; - priority_queue, greater> pq; - const int ok[3] = {3, 5, 7}; - st.insert(1); - pq.push(1); - int times = 0; + int a[3] = {3, 5, 7}; + LL tou = 1; + set s; while (true) { - long long lucky = pq.top(); - pq.pop(); - if (lucky == 59084709587505) - { //49 - cout << times << endl; - return 0; - } - times++; + for (int i = 0; i < 3; i++) { - long long b = lucky * ok[i]; - if (!st.count(b)) - { - st.insert(b); - pq.push(b); - } + LL tt = tou * a[i]; + if (tt <= MAX) + s.insert(tt); } + tou = s.upper_bound(tou); + if (tou >= MAX) + break; } + cout << s.size() << endl; return 0; } ``` @@ -71,34 +61,32 @@ int main() ### A ```cpp +#define LL long long +LL maxs = 59084709587505; +set q; int main() { - set st; - priority_queue, greater> pq; - const int ok[3] = {3, 5, 7}; - st.insert(1); - pq.push(1); - int times = 0; - while (true) + q.insert(3); + q.insert(5); + q.insert(7); + set::iterator it; + it = q.begin(); + LL mid; + while (*it <= maxs) { - long long lucky = pq.top(); - pq.pop(); - if (lucky == 59084709587505) - { //49 - cout << times << endl; - return 0; - } - times++; - for (int i = 0; i < 3; i++) - { - long long b = lucky * ok[i + 1]; - if (!st.count(b)) - { - st.insert(b); - pq.push(b); - } - } + mid = *it; + q.insert(mid * 3); + q.insert(mid * 5); + q.insert(mid * 7); + it++; } + int num = 0; + for (it = q.begin(); it != q.end(); it++) + { + if (*it <= maxs) + num++; + } + cout << num; return 0; } ``` @@ -106,37 +94,35 @@ int main() ### B ```cpp -int main() +typedef long long LL; +const LL Max = 59084709587505; +int a[3] = {3, 5, 7}; + +void Find(LL Max) { - set st; - priority_queue, greater> pq; - const int ok[3] = {3, 5, 7}; - st.insert(1); - pq.push(1); - int times = 0; - while (true) + set se; + LL t = 1; + while (1) { - long long lucky = pq.top(); - pq.pop(); - if (lucky == 59084709587505) - { //49 - cout << times << endl; - return 0; - } - times++; - for (int i = 0; i < 3; i++) + for (int i = 0; i < 3; ++i) { - long long b = lucky * ok[i] + 1; - if (!st.count(b)) - { - st.insert(b); - pq.push(b); - } + LL tt = t * a[i]; + if (tt <= Max) + se.insert(tt); } + t = *se.upper_bound(t); + if (t == Max) + break; } - return 0; + cout << se.size() << endl; } +int main(void) +{ + Find(Max); + + return 0; +} ``` ### C @@ -155,14 +141,14 @@ int main() long long lucky = pq.top(); pq.pop(); if (lucky == 59084709587505) - { //49 + { cout << times << endl; return 0; } times++; for (int i = 0; i < 3; i++) { - long long b = lucky + ok[i]; + long long b = lucky * ok[i]; if (!st.count(b)) { st.insert(b); @@ -172,5 +158,4 @@ int main() } return 0; } - ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/16.\347\255\211\345\267\256\347\264\240\346\225\260\345\210\227/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/16.\347\255\211\345\267\256\347\264\240\346\225\260\345\210\227/solution.md" index 693604f5503e5fe96f60661aa494d19b3c6aa3e7..9f3f45ab943edffb1acb0d1da30bf72f389b4e4e 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/16.\347\255\211\345\267\256\347\264\240\346\225\260\345\210\227/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/16.\347\255\211\345\267\256\347\264\240\346\225\260\345\210\227/solution.md" @@ -12,84 +12,60 @@ 长度为10的等差素数列，其公差最小值是多少？ +以下选项错误的是？ ## aop ### before ```cpp -#include -#include -#include +#include using namespace std; +``` +### after -typedef long long LL; -set all; +```cpp -bool isPrimt(LL t) -{ - for (int i = 2; i < t / 2; ++i) - { - if (t % i == 0) - return false; - } - return true; -} ``` -### after + +## 答案 ```cpp -const int N = 5000; -int main() -{ - LL a[N]; +const int N = 10010; - a[0] = 2; - a[1] = 3; - all.insert(2); - all.insert(3); +int k; +int prime[N]; +bool st[N], check[N]; - int index = 2; - LL t = 5; - while (index < N) - { - if (isPrimt(t)) +void init() +{ + for (int i = 2; i < N; i++) + if (!st[i]) { - a[index++] = t; - all.insert(t); + prime[k++] = i; + check[i] = true; + for (int j = i + i; j < N; j += i) + st[j] = true; } - t++; - } - - printf("%d\n", f(a, N)); - return 0; } -``` - -## 答案 -```cpp -int f(LL a[], int n) -{ - for (int i = 0; i < n; ++i) - { - LL first = a[i]; - for (int delta = 1; delta < a[n - 1] - first; ++delta) - { - int m = first; - for (int j = 1; j < 10; ++j) - { - m += delta; - if (all.find(m) == all.end()) - break; - if (m > a[n - 1]) - break; - if (j == 9) - return delta; +int main() +{ + init(); + for (int i = 1; i < N; i++) + for (int j = 0; j < k; j++) + { + int k = prime[j], cnt = 1; + while (check[k + i]) + { + k += i; + if (cnt++ == 10) + { + cout << i << endl; + return 0; + } } } - } - return -1; } ``` ## 选项 @@ -98,80 +74,171 @@ int f(LL a[], int n) ### A ```cpp -int f(LL a[], int n) -{ - for (int i = 0; i < n; ++i) - { - LL first = a[i]; - for (int delta = 1; delta < a[n] - first; ++delta) - { - int m = first; - for (int j = 1; j < 10; ++j) - { - m += delta; - if (all.find(m) == all.end()) - break; - if (m > a[n - 1]) - break; - if (j == 9) - return delta; +typedef long long ll; +const ll maxn = 1e6 + 50; +ll a[maxn]; +bool ok(ll n, ll cha) +{ + for (ll i = 0; i < 10; i++) + { + if (!a[n + i * cha]) + return 0; + } + return 1; +} + +int main() +{ + a[1] = 0; + a[2] = 1; + a[3] = 1; + for (ll i = 4; i <= 1000000; i++) + { + bool flag = 0; + for (ll j = 2; j * j <= i; j++) + { + if (i % j == 0) + { + flag = 1; + break; + } + } + if (flag) + a[i] = 0; + else + a[i] = 1; + } + + for (ll cha = 1;; cha++) + { + for (ll i = 2; i < 1000000; i++) + { + if (a[i] && ok(i, cha)) + { + printf("%lld\n", cha); + return 0; } } } - return -1; } ``` ### B ```cpp -int f(LL a[], int n) -{ - for (int i = 0; i < n; ++i) - { - LL first = a[i]; - for (int delta = 1; delta < a[n + 1] - first; ++delta) - { - int m = first; - for (int j = 1; j < 10; ++j) - { - m += delta; - if (all.find(m) == all.end()) - break; - if (m > a[n - 1]) +const int N = 10001; +int prime[N]{0}; + +int main() +{ + + for (int i = 2; i != N; ++i) + prime[i] = 1; + + for (int i = 2; i < N; ++i) + { + for (int j = i + 1; j < N; ++j) + { + if (j % i == 0) + { + if (prime[j] == 1) + prime[j] = 0; + } + } + } + + for (int d = 1; d <= 300; ++d) + { + for (int i = 2; i < N; ++i) + { + int a1 = i, flag = 1, len = 1; + while (len < 10) + { + if (prime[a1 + len * d] == 0) + { + flag = 0; break; - if (j == 9) - return delta; + } + else + { + + ++len; + } + } + + if (flag) + { + cout << d << endl; + break; } } } - return -1; + + return 0; } ``` ### C ```cpp +typedef long long LL; +set all; + +bool isPrimt(LL t) +{ + for (int i = 2; i < t / 2; ++i) + { + if (t % i == 0) + return false; + } + return true; +} + int f(LL a[], int n) -{ +{ for (int i = 0; i < n; ++i) - { + { LL first = a[i]; - for (int delta = 1; delta < a[n + 1] + first; ++delta) - { + for (int delta = 1; delta < a[n] - first; ++delta) + { int m = first; for (int j = 1; j < 10; ++j) - { + { m += delta; if (all.find(m) == all.end()) - break; - if (m > a[n - 1]) break; - if (j == 9) + if (j == 9) return delta; } } } return -1; } + +const int N = 5000; +int main() +{ + LL a[N]; + + a[0] = 2; + a[1] = 3; + all.insert(2); + all.insert(3); + + int index = 2; + LL t = 5; + while (index < N) + { + if (isPrimt(t)) + { + a[index++] = t; + all.insert(t); + } + t++; + } + + printf("%d\n", f(a, N)); + return 0; +} + ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/17.\344\271\230\347\247\257\346\234\200\345\244\247/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/17.\344\271\230\347\247\257\346\234\200\345\244\247/solution.md" index cb8c4e058c3553a3fa0df4d3b4132ab9bd99d576..d202fc7e8f4472199354388450f2f0431e347573 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/17.\344\271\230\347\247\257\346\234\200\345\244\247/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/17.\344\271\230\347\247\257\346\234\200\345\244\247/solution.md" @@ -54,14 +54,14 @@ -999999829 ``` +以下选项错误的是？ + ## aop ### before ```cpp -#include -#include -#include +#include using namespace std; ``` ### after @@ -73,43 +73,48 @@ using namespace std; ## 答案 ```cpp +#define ll long long +#define inf 0x7fffffff +using namespace std; +const int N = 1e5 + 10; +const int mod = 1e9 + 9; +int n, k; +int a[N]; +ll ans = 1; int main() { - int n, k; - long long ans; cin >> n >> k; - vector num; - for (int i = 0; i < n; i++) - { - int temp; - cin >> temp; - num.push_back(temp); - } - sort(num.begin(), num.end()); - if (k % 2 != 0) + for (int i = 1; i <= n; i++) + scanf("%d", &a[i]); + + sort(a + 1, a + n + 1); + + int sign = 1; + if (k % 2) { - ans = num.back(); - k = k - 1; - num.pop_back(); + ans = a[n]; + n--; + k--; } - else - ans = 1; - while (k > 0) + + int i = 0, j = n; + while (k) { - if ((num[0] * num[1]) > num.at(num.size() - 1) * num.at(num.size() - 2)) + ll tmp1 = (ll)a[i] * a[i + 1], tmp2 = (ll)a[j] * a[j - 1]; + if (tmp1 * sign > tmp2 * sign) { - ans = ans * num[0] * num[1] % 1000000009; - num.erase(num.begin(), num.begin() + 1); + ans = (tmp1 % mod * ans) % mod; + i += 2; } else { - ans = ans * num.at(num.size() - 1) * num.at(num.size() - 2) % 1000000009; - num.pop_back(); - num.pop_back(); + ans = (tmp2 % mod * ans) % mod; + j -= 2; } k -= 2; } - cout << ans; + + printf("%lld\n", ans); return 0; } @@ -120,43 +125,164 @@ int main() ### A ```cpp +#define mem(a, b) memset(a, b, sizeof(a)) +#define mod 1000000009 + +typedef long long ll; +const int maxn = 1e6 + 5; +const double esp = 1e-7; +const int ff = 0x3f3f3f3f; +map::iterator it; + +struct node +{ + ll x; + int f; +} a[maxn]; + +int n, k; + +bool cmp(node x, node y) +{ + return x.x > y.x; +} + +ll solve(int o) +{ + ll ans = 1; + int cnt = 0; + if (o == 0) + { + for (int i = 1; i <= k; i++) + { + ans = (ans * a[i].x) % mod; + if (a[i].f == 1) + cnt++; + } + } + else + { + for (int i = n; i > n - k; i--) + { + ans = (ans * a[i].x) % mod; + if (a[i].f == 1) + cnt++; + } + } + + if (cnt & 1) + return ans * (-1); + return ans; +} + int main() { - int n, k; - long long ans; cin >> n >> k; - vector num; - for (int i = 0; i < n; i++) + + int flag = 0; + int cnt = 0; + for (int i = 1; i <= n; i++) { - int temp; - cin >> temp; - num.push_back(temp); + scanf("%lld", &a[i].x); + if (a[i].x < 0) + { + a[i].f = 1; + a[i].x = -a[i].x; + cnt++; + } + else if (a[i].x > 0) + a[i].f = 0; + else + { + i--; + n--; + flag = 1; + } } - sort(num.begin(), num.end()); - if (k % 2 != 0) + + sort(a + 1, a + n + 1, cmp); + + ll ans = 0; + + if (n < k) + ans = 0; + else if (cnt == n) { - ans = num.back(); - k = k - 1; - num.pop_back(); + if (k & 1) + ans = solve(1); + else + ans = solve(0); } + else if (cnt == 0) + ans = solve(0); else - ans = 1; - while (k > 0) { - if ((num[0] * num[1]) > num.at(num.size()) * num.at(num.size() - 1)) - { - ans = ans * num[0] * num[1] % 1000000009; - num.erase(num.begin(), num.begin() + 1); - } + int tmp = 0; + for (int i = 1; i <= k; i++) + if (a[i].f == 1) + tmp++; + + if (tmp % 2 == 0) + ans = solve(0); else { - ans = ans * num.at(num.size()) * num.at(num.size() - 1) % 1000000009; - num.pop_back(); - num.pop_back(); + ans = -1; + + int p = -1, q = -1; + for (int i = k + 1; i <= n; i++) + if (a[i].f == 0) + { + q = i; + break; + } + for (int i = k; i >= 1; i--) + if (a[i].f == 1) + { + p = i; + break; + } + + if (p != -1 && q != -1) + { + swap(a[p], a[q]); + ans = solve(0); + swap(a[p], a[q]); + } + + p = -1, q = -1; + for (int i = k + 1; i <= n; i++) + if (a[i].f == 1) + { + q = i; + break; + } + for (int i = k; i >= 1; i--) + if (a[i].f == 0) + { + p = i; + break; + } + + if (p != -1 && q != -1) + { + swap(a[p], a[q]); + ans = max(ans, solve(0)); + swap(a[p], a[q]); + } + + if (ans < 0) + ans = solve(1); } - k -= 2; } - cout << ans; + + if (ans < 0) + if (flag) + { + cout << 0 << endl; + return 0; + } + cout << ans << endl; + return 0; } ``` @@ -164,43 +290,72 @@ int main() ### B ```cpp +#define mod 1000000009 +const int N = 100010; +typedef long long LL; +int a[N]; +int n, k; int main() { - int n, k; - long long ans; cin >> n >> k; - vector num; for (int i = 0; i < n; i++) { - int temp; - cin >> temp; - num.push_back(temp); + scanf("%d", &a[i]); } - sort(num.begin(), num.end()); - if (k % 2 != 0) + sort(a, a + n); + LL res = 1; + if (n == k) { - ans = num.back(); - k = k - 1; - num.pop_back(); + for (int i = 0; i < n; i++) + { + res = (res * a[i]) % mod; + } + cout << res % mod; + return 0; } - else - ans = 1; - while (k > 0) + + if (a[n - 1] < 0) { - if ((num[0] * num[1]) > num.at(num.size() - 1) * num.at(num.size() - 1)) + if (k % 2 == 1) { - ans = ans * num[0] * num[1] % 1000000009; - num.erase(num.begin(), num.begin() + 1); + for (int i = n - 1; i >= n - k; i--) + { + res = (res * a[i]) % mod; + } } else { - ans = ans * num.at(num.size() - 1) % 1000000009; - num.pop_back(); - num.pop_back(); + for (int i = 0; i < k; i++) + { + res = (res * a[i]) % mod; + } + } + cout << res << endl; + return 0; + } + + int l = 0, r = n - 1; + if (k % 2 == 1) + { + res = a[r--]; + k--; + } + while (k) + { + LL x = (LL)a[l] * a[l + 1], y = (LL)a[r] * a[r - 1]; + if (x > y) + { + res = x % mod * res % mod; + l += 2; + } + else + { + res = y % mod * res % mod; + r -= 2; } k -= 2; } - cout << ans; + cout << res << endl; return 0; } ``` @@ -238,7 +393,7 @@ int main() } else { - ans = ans * num.at(num.size() - 1) % 1000000009; + ans = ans * num.at(num.size() - 1) * num.at(num.size() - 2) % 1000000009; num.pop_back(); num.pop_back(); } diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/2.\344\271\230\347\247\257\345\260\276\351\233\266/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/2.\344\271\230\347\247\257\345\260\276\351\233\266/solution.md" index e4d93213e9b39d797b01d172ae19e034ad90ed7f..b66768cc2d9e2ecee2343cf42c8c90bb4fabb6d2 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/2.\344\271\230\347\247\257\345\260\276\351\233\266/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/2.\344\271\230\347\247\257\345\260\276\351\233\266/solution.md" @@ -14,12 +14,14 @@ 689 5510 8243 6114 337 4096 8199 7313 3685 211 ``` +以下错误的选项是？ + ## aop ### before ```cpp -#include +#include using namespace std; ``` ### after @@ -41,12 +43,12 @@ int main() while (num % 5 == 0) { count5++; - num /= 5; + num %= 5; } while (num % 2 == 0) { count2++; - num /= 2; + num %= 2; } } int ans = count2 < count5 ? count2 : count5; @@ -62,24 +64,20 @@ int main() ```cpp int main() { - int count2 = 0, count5 = 0; - int num; + + long long int a, sum = 1, aum = 0; for (int i = 0; i < 100; i++) { - cin >> num; - while (num % 5 == 0) + cin >> a; + sum = sum * a; + while ((sum % 10) == 0) { - count5++; - num /= 5; - } - while (num % 2 == 0) - { - count2++; - num /= 2; + sum = sum / 10; + aum++; } + sum = sum % 10000; } - int ans = count2 < count5 ? count5 : count2; - cout << ans; + printf("%lld", aum); return 0; } ``` @@ -87,29 +85,30 @@ int main() ### B ```cpp +int a[] = {5650, 4542, 3554, 473, 946, 4114, 3871, 9073, 90, 4329, + 2758, 7949, 6113, 5659, 5245, 7432, 3051, 4434, 6704, 3594, + 9937, 1173, 6866, 3397, 4759, 7557, 3070, 2287, 1453, 9899, + 1486, 5722, 3135, 1170, 4014, 5510, 5120, 729, 2880, 9019, + 2049, 698, 4582, 4346, 4427, 646, 9742, 7340, 1230, 7683, + 5693, 7015, 6887, 7381, 4172, 4341, 2909, 2027, 7355, 5649, + 6701, 6645, 1671, 5978, 2704, 9926, 295, 3125, 3878, 6785, + 2066, 4247, 4800, 1578, 6652, 4616, 1113, 6205, 3264, 2915, + 3966, 5291, 2904, 1285, 2193, 1428, 2265, 8730, 9436, 7074, + 689, 5510, 8243, 6114, 337, 4096, 8199, 7313, 3685, 211}; int main() { - int count2 = 0, count5 = 0; - int num; + int k1 = 0, k2 = 0; for (int i = 0; i < 100; i++) { - cin >> num; - while (num % 5 == 0) - { - count5++; - num %= 5; - } - while (num % 2 == 0) - { - count2++; - num %= 2; - } + while (!(a[i] % 2)) + a[i] /= 2, k1++; + while (!(a[i] % 5)) + a[i] /= 5, k2++; } - int ans = count2 < count5 ? count2 : count5; - cout << ans; + cout << min(k1, k2); + return 0; } - ``` ### C @@ -117,24 +116,25 @@ int main() ```cpp int main() { - int count2 = 0, count5 = 0; - int num; + int t_cnt, f_cnt, data = 0; + t_cnt = f_cnt = 0; + for (int i = 0; i < 100; i++) { - cin >> num; - while (num % 5 == 0) + cin >> data; + while (data % 5 == 0) { - count5++; - num += 5; + f_cnt++; + data /= 5; } - while (num % 2 == 0) + while (data % 2 == 0) { - count2++; - num += 2; + t_cnt++; + data /= 2; } } - int ans = count2 < count5 ? count2 : count5; - cout << ans; + cout << min(t_cnt, f_cnt) << endl; + return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/3.\345\244\247\350\241\215\346\225\260\345\210\227/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/3.\345\244\247\350\241\215\346\225\260\345\210\227/solution.md" index 45454e047cfc556d780c0a4c2f9575ed4166fd12..72b1c0df51216d382cc0b8930a025de8a29a7b83 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/3.\345\244\247\350\241\215\346\225\260\345\210\227/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/3.\345\244\247\350\241\215\346\225\260\345\210\227/solution.md" @@ -8,12 +8,30 @@ 以下的代码打印出了大衍数列的前 100 项。 +请填补空白处的内容。 + +```cpp +#include +int main() +{ + int i; + for (i = 1; i <= 100; i++) + { + if (__________________) + printf("%d ", i * i / 2); + else + printf("%d ", (i * i - 1) / 2); + } + printf("\n"); +} +``` + ## aop ### before ```cpp -#include + ``` ### after @@ -24,18 +42,7 @@ ## 答案 ```cpp -int main() -{ - int i; - for (i = 1; i <= 100; i++) - { - if (i % 2 == 0) - printf("%d ", i * i / 2); - else - printf("%d ", (i * i - 1) / 2); - } - printf("\n"); -} +i % 2 == 0 ``` ## 选项 @@ -43,50 +50,17 @@ int main() ### A ```cpp -int main() -{ - int i; - for (i = 1; i < 100; i++) - { - if (i % 2 == 0) - printf("%d ", i * i / 2); - else - printf("%d ", (i * i - 1) / 2); - } - printf("\n"); -} +i % 2 == 1 ``` ### B ```cpp -int main() -{ - int i; - for (i = 1; i <= 100; i++) - { - if (i / 2 == 0) - printf("%d ", i * i / 2); - else - printf("%d ", (i * i - 1) / 2); - } - printf("\n"); -} +i / 2 == 0 ``` ### C ```cpp -int main() -{ - int i; - for (i = 1; i <= 100; i++) - { - if (i % 2 == 0) - printf("%d ", i * i % 2); - else - printf("%d ", (i * i - 1) / 2); - } - printf("\n"); -} +i / 2 == 1 ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/4.\346\226\220\346\263\242\351\202\243\345\245\221/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/4.\346\226\220\346\263\242\351\202\243\345\245\221/solution.md" index 4999d656750f55ad75b5ddc3a1ba740c43f9e060..b624bb376541c47c47c09858864ed376ba1e1644 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/4.\346\226\220\346\263\242\351\202\243\345\245\221/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/4.\346\226\220\346\263\242\351\202\243\345\245\221/solution.md" @@ -36,12 +36,15 @@ Fibonacci数列的递推公式为：Fn=Fn-1+Fn-2，其中F1=F2=1。 1 <= n <= 1,000,000。 ``` +下列哪一个选项会超时？ + ## aop ### before ```cpp -#include +#include +using namespace std; ``` ### after @@ -52,22 +55,18 @@ Fibonacci数列的递推公式为：Fn=Fn-1+Fn-2，其中F1=F2=1。 ## 答案 ```cpp -int main() +int fib(int n) { - int n, b; - scanf("%d", &n); - int a[n]; - a[0] = a[1] = 1; - - for (int i = 2; i < n; i++) - { - a[i] = (a[i - 1] + a[i - 2]) % 10007; - b = a[i]; - } - if (n > 2) - printf("%d", b); + if (n < 3) + return 1; else - printf("1"); + return fib(n - 1) + fib(n - 2); +} +int main() +{ + int n; + cin >> n; + cout << fib(n) % 10007 << endl; return 0; } ``` @@ -77,22 +76,28 @@ int main() ### A ```cpp -int main() +int fib(int N) { - int n, b; - scanf("%d", &n); - int a[n]; - a[0] = a[1] = 1; - - for (int i = 2; i < n; i++) + if (N <= 0) + return 0; + if (N == 1 || N == 2) + return 1; + int a = 1, b = 1; + for (int i = 3; i <= N; i++) { - a[i] = (a[i - 1] + a[i - 2]) % 10007; - b = a[i] + 1; + int c = (a + b) % 10007; + a = b; + b = c; } - if (n > 2) - printf("%d", b); - else - printf("1"); + return b; +} +int main() +{ + int n; + cin >> n; + + cout << fib(n) << endl; + return 0; } ``` @@ -110,7 +115,7 @@ int main() for (int i = 2; i < n; i++) { a[i] = (a[i - 1] + a[i - 2]) % 10007; - b = a[i - 1]; + b = a[i]; } if (n > 2) printf("%d", b); @@ -123,22 +128,24 @@ int main() ### C ```cpp +int f[1000000 + 1]; + +void meter(int n) +{ + f[1] = 1; + f[2] = 1; + if (n >= 3) + for (int i = 3; i <= n; i++) + { + f[i] = (f[i - 1] + f[i - 2]) % 10007; + } +} int main() { - int n, b; - scanf("%d", &n); - int a[n]; - a[0] = a[1] = 1; - - for (int i = 2; i < n; i++) - { - a[i] = (a[i - 1] + a[i - 2]) % 10007; - b = a[i - 2]; - } - if (n > 2) - printf("%d", b); - else - printf("1"); + meter(1000000); + int n; + cin >> n; + cout << f[n]; return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/5.\351\200\222\345\242\236\344\270\211\345\205\203\347\273\204/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/5.\351\200\222\345\242\236\344\270\211\345\205\203\347\273\204/solution.md" index 6509956a573113da10228b585b8db84019d3f857..ee0c880cc6921fa9c884cf3793d868ea0cdd66df 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/5.\351\200\222\345\242\236\344\270\211\345\205\203\347\273\204/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/5.\351\200\222\345\242\236\344\270\211\345\205\203\347\273\204/solution.md" @@ -37,22 +37,15 @@ C = [C1, C2, ... CN]， 27 ``` +以下错误的一项是？ ## aop ### before ```cpp -#include -#include - +#include using namespace std; - -typedef long long LL; - -const int N = 1e5 + 10; -int a[N], b[N], c[N], sa[N], sc[N], s[N]; - ``` ### after @@ -66,34 +59,47 @@ int a[N], b[N], c[N], sa[N], sc[N], s[N]; int main() { int n; + long long ans = 0; cin >> n; - for (int i = 0; i < n; i++) - cin >> a[i], a[i]++; - for (int i = 0; i < n; i++) - cin >> b[i], b[i]++; - for (int i = 0; i < n; i++) - cin >> c[i], c[i]++; - - for (int i = 0; i < n; i++) - s[a[i]]++; - for (int i = 1; i < N; i++) - s[i] += s[i - 1]; - for (int i = 0; i < n; i++) - sa[i] = s[b[i] - 1]; - - memset(s, 0, sizeof s); - - for (int i = 0; i < n; i++) - s[c[i]]++; - for (int i = 1; i < N; i++) - s[i] += s[i - 1]; - for (int i = 0; i < n; i++) - sc[i] = s[N - 1] - s[b[i]]; - - LL res = 0; - for (int i = 0; i <= n; i++) - res += (LL)sa[i] * sc[i]; - cout << res; + int I[n], J[n], K[n]; + for (int i = 0; i < n; i++) + { + cin >> I[i]; + } + for (int i = 0; i < n; i++) + { + cin >> J[i]; + } + for (int i = 0; i < n; i++) + { + cin >> K[i]; + } + sort(I, I + n); + sort(J, J + n); + sort(K, K + n); + int index_min_i = 0, index_min_j = 0; + for (int i = 0; i < n; i++) + { + int count1 = 0, count2 = 0; + for (int j = index_min_i; j < n; j++) + { + if (I[i] < J[j]) + { + index_min_i = j; + count1 = n - j; + for (int k = index_min_j; k < n; k++) + { + + count2 = n - k; + index_min_j = k; + ans += (long long)(count1 * count2); + } + break; + } + } + } + cout << ans << endl; + return 0; } ``` ## 选项 @@ -102,80 +108,115 @@ int main() ### A ```cpp +const int maxn = 1e5 + 10; +typedef long long ll; +int a[maxn]; +int b[maxn]; +int c[maxn]; int main() { int n; + ll ans = 0; cin >> n; for (int i = 0; i < n; i++) - cin >> a[i], a[i]++; - for (int i = 0; i < n; i++) - cin >> b[i], b[i]++; + cin >> a[i]; for (int i = 0; i < n; i++) - cin >> c[i], c[i]++; - + cin >> b[i]; for (int i = 0; i < n; i++) - s[a[i]]++; - for (int i = 1; i < N; i++) - s[i] += s[i - 1]; + cin >> c[i]; + sort(a, a + n); + sort(b, b + n); + sort(c, c + n); + ll cnt1 = 0, cnt2 = 0; for (int i = 0; i < n; i++) - sa[i] = s[b[i] - 1]; + { + while (cnt1 < n && a[cnt1] < b[i]) + cnt1++; + while (cnt2 < n && c[cnt2] <= b[i]) + cnt2++; + ans += cnt1 * (n - cnt2); + } - memset(s, 0, sizeof s); - - for (int i = 0; i < n; i++) - s[c[i]]++; - for (int i = 1; i < N; i++) - s[i] += s[i - 1]; - for (int i = 0; i < n; i++) - sc[i] = s[N] - s[b[i]]; - - LL res = 0; - for (int i = 0; i <= n; i++) - res += (LL)sa[i] * sc[i]; - cout << res; + cout << ans << endl; } ``` ### B ```cpp -int main() -{ - int n; - cin >> n; - for (int i = 0; i < n; i++) - cin >> a[i], a[i]++; - for (int i = 0; i < n; i++) - cin >> b[i], b[i]++; - for (int i = 0; i < n; i++) - cin >> c[i], c[i]++; +#define ll long long +using namespace std; - for (int i = 0; i < n; i++) - s[a[i]]++; - for (int i = 1; i < N; i++) - s[i] += s[i - 1]; - for (int i = 0; i < n; i++) - sa[i] = s[b[i] - 1]; +const int N = 100000 + 10; - memset(s, 0, sizeof s); +int a[N], b[N], c[N]; +int n; - for (int i = 0; i < n; i++) - s[c[i]]++; - for (int i = 1; i < N; i++) - s[i] = s[i - 1]; - for (int i = 0; i < n; i++) - sc[i] = s[N - 1] - s[b[i]]; +int find2(int x, int y[]) +{ + int l = 1, r = n; + while (l < r) + { + int mid = (l + r) >> 1; + if (y[mid] > x) + r = mid; + else + l = mid + 1; + } + if (y[r] <= x) + return 0; + return n - r + 1; +} - LL res = 0; - for (int i = 0; i <= n; i++) - res += (LL)sa[i] * sc[i]; - cout << res; +int find1(int x, int y[]) +{ + int l = 1, r = n; + while (l < r) + { + int mid = (l + r + 1) >> 1; + if (y[mid] < x) + l = mid; + else + r = mid - 1; + } + if (y[l] >= x) + return 0; + return l; +} + +int main() +{ + cin >> n; + for (int i = 1; i <= n; i++) + cin >> a[i]; + for (int i = 1; i <= n; i++) + cin >> b[i]; + for (int i = 1; i <= n; i++) + cin >> c[i]; + sort(a + 1, a + n + 1); + sort(b + 1, b + n + 1); + sort(c + 1, c + n + 1); + long long ans = 0; + for (int i = 1; i <= n; i++) + { + int x = find1(b[i], a); + int y = find2(b[i], c); + + ans += 1ll * x * y; + } + cout << ans; + return 0; } ``` ### C ```cpp +typedef long long LL; + +const int N = 1e5 + 10; +int a[N], b[N], c[N], sa[N], sc[N], s[N]; + int main() { int n; @@ -199,9 +240,9 @@ int main() for (int i = 0; i < n; i++) s[c[i]]++; for (int i = 1; i < N; i++) - s[i] = s[i - 1]; + s[i] += s[i - 1]; for (int i = 0; i < n; i++) - sc[i] = s[N - 1] - s[b[i] - 1]; + sc[i] = s[N - 1] - s[b[i]]; LL res = 0; for (int i = 0; i <= n; i++) diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/6.\345\200\215\346\225\260\351\227\256\351\242\230/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/6.\345\200\215\346\225\260\351\227\256\351\242\230/solution.md" index f5e7f5e49a2b6b46df6ff0e4d8fca873a37fe0ce..948e102bb9043faa69de7d53158aef3dd7c8bf2f 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/6.\345\200\215\346\225\260\351\227\256\351\242\230/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/6.\345\200\215\346\225\260\351\227\256\351\242\230/solution.md" @@ -34,66 +34,73 @@ 选择2、3、4。 ``` +以下错误的一项是？ + ## aop ### before ```cpp #include -#include -#include -#include -#include -#include -#include -#define MAX 1000000000 using namespace std; -int n, k, a[100010]; -int b[4]; -int flag = 0; ``` ### after ```cpp -int main() -{ - cin >> n >> k; - a[0] = MAX; - for (int i = 1; i <= n; i++) - cin >> a[i]; - sort(a + 1, a + n + 1); - reverse(a + 1, a + n + 1); - dfs(a, n, 1); - return 0; -} + ``` ## 答案 ```cpp -void dfs(int a[], int n, int s) +#define ll long long +using namespace std; + +const int maxx = 1e5 + 100; +const int maxm = 1e3 + 100; +int a[maxx]; +vector p[maxm]; +int n, k; + +bool cmp(int a, int b) { return a > b; } +int main() { - if (flag == 1) - return; - if (s == 4) - { - int sum = b[1] + b[2] + b[3]; - if (sum % k == 0) - { - flag = 1; - cout << sum << endl; - } - return; - } - for (int i = 1; i <= n; i++) - { - if (a[i] < a[s - 1]) - { - b[s] = a[i]; - dfs(a, n, s + 1); - } - } + scanf("%d%d", &n, &k); + for (int i = 1; i <= n; i++) + { + scanf("%d", &a[i]); + p[a[i] % k].push_back(a[i]); + } + for (int i = 0; i < k; i++) + sort(p[i].begin(), p[i].end(), cmp); + int _max = 0; + for (int i = 0; i < k; i++) + { + if (p[i].size() == 0) + continue; + if ((3 * i == k || i == 0) && p[i].size() >= 3) + _max = max(_max, p[i][0] + p[i][1] + p[i][2]); + for (int j = 0; j < k; j++) + { + if (!p[j].size()) + continue; + int z = k - (i + j) % k; + if (z < 0) + continue; + if (z == i && z == j) + continue; + else if ((i != j) && (z == i || z == j) && p[z].size() >= 2) + _max = max(_max, p[i][0] + p[j][0] + p[z][1]); + else if (i == j && p[i].size() >= 2 && p[z].size() >= 1) + _max = max(_max, p[i][0] + p[j][1] + p[z][0]); + else if (i != j && i != z && j != z && p[i].size() >= 1) + _max = max(_max, p[i][0] + p[j][0] + p[z][0]); + } + } + printf("%d\n", _max); + return 0; } + ``` ## 选项 @@ -101,83 +108,145 @@ void dfs(int a[], int n, int s) ### A ```cpp -void dfs(int a[], int n, int s) +#define FOR0(a, b) for (int i = a; i < b; ++i) +#define FORE(a, b) for (int i = a; i <= b; ++i) +typedef long long ll; +typedef pair pii; + +ll dp[2][4][1005]; +int n, k, a[100005], v[100005]; +int main() { - if (flag == 1) - return; - if (s == 4) - { - int sum = b[1] + b[2] + b[3]; - if (sum % k == 0) - { - flag = 1; - cout << sum << endl; - } - return; - } - for (int i = 1; i <= n; i++) - { - if (a[i] < a[s - 1]) - { - b[s] = a[i]; - dfs(a, n, s); - } - } + scanf("%d%d", &n, &k); + for (int i = 1; i <= n; ++i) + { + scanf("%d", &a[i]); + v[i] = a[i]; + v[i] %= k; + } + memset(dp, -0x3f3f3f3f, sizeof dp); + dp[0][0][0] = 0; + int d = 0; + for (int i = 1; i <= n; ++i) + { + for (int j = 0; j < k; ++j) + { + for (int p = 0; p <= 3; ++p) + { + if (i < p) + continue; + dp[d ^ 1][p][j] = max(dp[d ^ 1][p][j], dp[d][p][j]); + if (p > 0) + dp[d ^ 1][p][j] = max(dp[d ^ 1][p][j], dp[d][p - 1][((j - v[i]) % k + k) % k] + a[i]); + } + } + d ^= 1; + } + cout << dp[d][3][0] << endl; + return 0; } ``` ### B ```cpp -void dfs(int a[], int n, int s) +int n, k, ans = 0; +int temp[3], num[100005], vis[100005] = {0}; + +void dfs(int s) { - if (flag == 1) - return; - if (s == 4) - { - int sum = b[1] + b[2] + b[3]; - if (sum % k == 0) - { - flag = 1; - cout << sum << endl; - } - return; - } - for (int i = 1; i <= n; i++) - { - if (a[i] < a[s + 1]) - { - b[s] = a[i]; - dfs(a, n, s + 1); - } - } + if (s == 3) + { + int t = temp[0] + temp[1] + temp[2]; + if (t % k == 0 && t > ans) + ans = t; + return; + } + else + { + for (int i = 0; i < n; i++) + { + if (!vis[i]) + { + temp[s] = num[i]; + vis[i] = 1; + dfs(s + 1); + vis[i] = 0; + } + } + } +} + +int main() +{ + scanf("%d%d", &n, &k); + for (int i = 0; i < n; i++) + scanf("%d", &num[i]); + dfs(0); + printf("%d\n", ans); + return 0; } ``` ### C ```cpp -void dfs(int a[], int n, int s) +int N, k; +int main() { - if (flag == 1) - return; - if (s == 4) - { - int sum = b[1] + b[2] + b[3]; - if (sum % k == 0) - { - flag = 1; - cout << sum << endl; - } - return; - } - for (int i = 1; i <= n; i++) - { - if (a[i] < a[s + 1]) - { - b[s] = a[i]; - dfs(a, n, s); - } - } + ios::sync_with_stdio(0); + cin.tie(0); + cout.tie(0); + cin >> N >> k; + vector> vec(k, vector(3, -1)); + for (int i = 0; i < N; ++i) + { + int temp; + cin >> temp; + int y = temp % k; + if (temp > vec[y][0]) + { + vec[y][2] = vec[y][1]; + vec[y][1] = vec[y][0]; + vec[y][0] = temp; + } + else if (temp > vec[y][1]) + { + vec[y][2] = vec[y][1]; + vec[y][1] = temp; + } + else if (temp > vec[y][2]) + vec[y][2] = temp; + } + + vector ans(3); + int result = 0; + for (int i = 0; i < k; ++i) + for (int j = i; j < k; ++j) + { + int r = (k - i + k - j) % k; + ans[0] = vec[i][0]; + if (i == j) + { + ans[1] = vec[i][1]; + if (r == i) + ans[2] = vec[i][2]; + else + ans[2] = vec[r][0]; + } + else + { + ans[1] = vec[j][0]; + if (r == i) + ans[2] = vec[i][1]; + else if (r == j) + ans[2] = vec[j][1]; + else + ans[2] = vec[r][0]; + } + if (ans[0] != -1 && ans[1] != -1 && ans[2] != -1 && ans[1] + ans[2] + ans[0] > result) + result = ans[1] + ans[2] + ans[0]; + } + cout << result; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/7.\347\254\25439\347\272\247\345\217\260\351\230\266/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/7.\347\254\25439\347\272\247\345\217\260\351\230\266/solution.md" index ea8ca4d3cb31b5eaffb798f07f1b5e6d727fb297..668a945a7b113cb06e83cbb3c5afd89a7e56101c 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/7.\347\254\25439\347\272\247\345\217\260\351\230\266/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/7.\347\254\25439\347\272\247\345\217\260\351\230\266/solution.md" @@ -9,16 +9,15 @@ 请你利用计算机的优势，帮助小明寻找答案。 +以下哪一项不能得到正确答案？ + ## aop ### before ```cpp -#include -#define LEFT 0 -#define RIGHT 1 +#include using namespace std; -int stage[40][2]; ``` ### after @@ -29,19 +28,32 @@ int stage[40][2]; ## 答案 ```cpp -int main() +int ans = 0; +int sum = 0; +vector a(40, 0); + +void dfs(int steps) { - int i; - stage[1][LEFT] = 1; - stage[1][RIGHT] = 0; - stage[2][LEFT] = 1; - stage[2][RIGHT] = 1; - for (i = 3; i <= 39; i++) + if (sum >= 39) { - stage[i][LEFT] = stage[i - 1][RIGHT] + stage[i - 2][RIGHT]; - stage[i][RIGHT] = stage[i - 1][LEFT] + stage[i - 2][LEFT]; + if ((steps - 1) % 2 == 0 && sum == 39) + ans++; + return; } - cout << stage[39][RIGHT] << endl; + + for (int i = 1; i <= 2; i++) + { + a[steps] = i; + sum += i; + dfs(steps + 1); + sum -= i; + } +} + +int main() +{ + dfs(0); + cout << ans << endl; return 0; } ``` @@ -51,19 +63,24 @@ int main() ### A ```cpp -int main() +int ans = 0; +void dfs(int k, int n) { - int i; - stage[1][LEFT] = 1; - stage[1][RIGHT] = 0; - stage[2][LEFT] = 1; - stage[2][RIGHT] = 1; - for (i = 3; i <= 39; i++) + if (k == 39 && n % 2 == 0) { - stage[i][LEFT] = stage[i - 1][RIGHT] + stage[i - 2][RIGHT]; - stage[i][RIGHT] = stage[i - 1][LEFT] + stage[i - 2][LEFT]; + ans++; } - cout << stage[39][LEFT] << endl; + if (k > 39) + return; + dfs(k + 1, n + 1); + dfs(k + 2, n + 1); +} +int main() +{ + + dfs(0, 0); + printf("%d\n", ans); + return 0; } ``` @@ -71,19 +88,27 @@ int main() ### B ```cpp -int main() +int countt = 0; + +void f(int stair, int step) { - int i; - stage[1][LEFT] = 1; - stage[1][RIGHT] = 0; - stage[2][LEFT] = 1; - stage[2][RIGHT] = 1; - for (i = 3; i <= 39; i++) + if (stair < 0) + return; + if (step % 2 == 0 && stair == 0) { - stage[i][LEFT] = stage[i + 1][RIGHT] + stage[i - 1][RIGHT]; - stage[i][RIGHT] = stage[i + 1][LEFT] + stage[i - 1][LEFT]; + countt++; + return; } - cout << stage[39][RIGHT] << endl; + for (int i = 1; i <= 2; i++) + { + f(stair - i, step + 1); + } +} + +int main(void) +{ + f(39, 0); + cout << countt << endl; return 0; } ``` @@ -91,6 +116,11 @@ int main() ### C ```cpp +#define LEFT 0 +#define RIGHT 1 +using namespace std; +int stage[40][2]; + int main() { int i; @@ -100,10 +130,10 @@ int main() stage[2][RIGHT] = 1; for (i = 3; i <= 39; i++) { - stage[i][LEFT] = stage[i + 1][RIGHT] + stage[i - 1][RIGHT]; - stage[i][RIGHT] = stage[i + 1][LEFT] + stage[i - 1][LEFT]; + stage[i][LEFT] = stage[i - 1][RIGHT] + stage[i - 2][RIGHT]; + stage[i][RIGHT] = stage[i - 1][LEFT] + stage[i - 2][LEFT]; } - cout << stage[39][LEFT] << endl; + cout << stage[39][RIGHT] << endl; return 0; } ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/8.\347\254\254\345\207\240\345\244\251/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/8.\347\254\254\345\207\240\345\244\251/solution.md" index fc2f3e86750dc785e3608ff60ffadf3e50cc7c3c..b5036066d8d3207232e875b12047d8b6c033f7c7 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/8.\347\254\254\345\207\240\345\244\251/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/8.\347\254\254\345\207\240\345\244\251/solution.md" @@ -1,52 +1,25 @@ # 第几天 y年m月d日是哪一年的第几天。 -比如y年的1月1日是那一年的第一天，那么y年m月d日是哪一年的第几天。 - -**输入** - -``` -y m d -``` -**输出** - -输出一个整数 - -**样例** +比如y年的1月1日是那一年的第一天，那么2000年7月7日是那一年的第几天。 **输入** ``` 2000 7 7 ``` -**输出** -``` -189 -``` - - -## aop - -### before +请通过以下代码，选出正确答案。 ```cpp -#include +#include using namespace std; + bool is_leap(int year) { return (year % 4 == 0 && year % 100 != 0) || year % 400 == 0; } -``` -### after - -```cpp - -``` -## 答案 - -```cpp int main() { int y, m, d, ans = 0; @@ -54,9 +27,9 @@ int main() int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - if (is_leap(y)) + if (is_leap(y)) { - for (int i = 0; i < (m - 1); i++) + for (int i = 0; i < (m - 1); i++) { ans += L_m_d[i]; } @@ -73,7 +46,26 @@ int main() cout << ans << endl; return 0; } +``` + + +## aop + +### before + +```cpp +``` +### after + +```cpp + +``` + +## 答案 + +```cpp +189 ``` ## 选项 @@ -81,92 +73,17 @@ int main() ### A ```cpp -int main() -{ - int y, m, d, ans = 0; - cin >> y >> m >> d; - int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - - if (is_leap(y)) - { - for (int i = 0; i < m; i++) - { - ans += L_m_d[i]; - } - ans += d; - } - else - { - for (int i = 0; i < (m - 1); i++) - { - ans += nonL_m_d[i]; - } - ans += d; - } - cout << ans << endl; - return 0; -} +188 ``` ### B ```cpp -int main() -{ - int y, m, d, ans = 0; - cin >> y >> m >> d; - int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - - if (is_leap(y)) - { - for (int i = 0; i < (m - 1); i++) - { - ans += L_m_d[i]; - } - ans += d; - } - else - { - for (int i = 0; i < m; i++) - { - ans += nonL_m_d[i]; - } - ans += d; - } - cout << ans << endl; - return 0; -} +190 ``` ### C ```cpp -int main() -{ - int y, m, d, ans = 0; - cin >> y >> m >> d; - int L_m_d[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - int nonL_m_d[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; - - if (is_leap(y)) - { - for (int i = 0; i < m; i++) - { - ans += L_m_d[i]; - } - ans += d; - } - else - { - for (int i = 0; i < m; i++) - { - ans += nonL_m_d[i]; - } - ans += d; - } - cout << ans << endl; - return 0; -} +191 ``` diff --git "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/9.\346\226\271\351\230\265\350\275\254\347\275\256/solution.md" "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/9.\346\226\271\351\230\265\350\275\254\347\275\256/solution.md" index 414c835b90f5e41046c9b413dc1334c82d1e435e..03584518065d7693fd5bebed53840f97d548291b 100644 --- "a/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/9.\346\226\271\351\230\265\350\275\254\347\275\256/solution.md" +++ "b/data/1.\347\256\227\346\263\225\345\210\235\351\230\266/1.\350\223\235\346\241\245\346\235\257-\345\237\272\347\241\200/9.\346\226\271\351\230\265\350\275\254\347\275\256/solution.md" @@ -31,24 +31,13 @@ 7 9 ``` -## aop -### before +请从以下四个选项中选择正确的代码填补空白处，实现方阵转置功能。 ```cpp #include - using namespace std; -``` -### after - -```cpp - -``` - -## 答案 -```cpp int main() { int m, n; @@ -62,6 +51,27 @@ int main() cin >> a[j][i]; } } + __________________ + return 0; +} +``` + +## aop + +### before + +```cpp + +``` +### after + +```cpp + +``` + +## 答案 + +```cpp for (i = 0; i < n; i++) { for (j = 0; j < m; j++) @@ -70,9 +80,6 @@ int main() } cout << endl; } - return 0; -} - ``` ## 选项 @@ -80,83 +87,38 @@ int main() ### A ```cpp -int main() -{ - int m, n; - int a[20][20]; - int i, j; - cin >> m >> n; - for (i = 0; i < m; i++) - { - for (j = 0; j < n; j++) - { - cin >> a[j][i]; - } - } for (i = 0; i < n; i++) { - for (j = m; j > 0; j--) + for (j = 0; j < m; j++) { - cout << a[i][j] << " "; + cout << a[i-1][j] << " "; } cout << endl; } - return 0; -} ``` ### B ```cpp -int main() -{ - int m, n; - int a[20][20]; - int i, j; - cin >> m >> n; - for (i = 0; i < m; i++) - { - for (j = 0; j < n; j++) - { - cin >> a[j][i]; - } - } - for (i = n; i > 0; i--) + for (i = 0; i < n; i++) { for (j = 0; j < m; j++) { - cout << a[i][j] << " "; + cout << a[i][j-1] << " "; } cout << endl; } - return 0; -} ``` ### C ```cpp -int main() -{ - int m, n; - int a[20][20]; - int i, j; - cin >> m >> n; - for (i = 0; i < m; i++) - { - for (j = 0; j < n; j++) - { - cin >> a[j][i]; - } - } - for (i = n; i > 0; i--) + for (i = 0; i < n; i++) { - for (j = m; j > 0; j--) + for (j = 0; j < m; j++) { - cout << a[i][j] << " "; + cout << a[i-1][j-1] << " "; } cout << endl; } - return 0; -} ```