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skill_tree_algorithm
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data/1.算法初阶/1.蓝桥杯/不同子串/solution.md
浏览文件 @ 62f75b25
......@@ -10,6 +10,9 @@
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after
```cpp
......@@ -18,21 +21,61 @@
## 答案
```cpp
int main()
{
set<string> s;
string str;
cin >> str;
for (int i = 0; i < str.size(); i++)
for (int j = i; j < str.size(); j++)
s.insert(str.substr(i, j - i + 1));
cout << s.size();
return 0;
}
```
## 选项
### A
```cpp
int main()
{
set<string> s;
string str;
cin >> str;
for (int i = 0; i < str.size(); i++)
for (int j = i; j < str.size(); j++)
s.insert(str.substr(i, j + i - 1));
cout << s.size();
return 0;
}
```
### B
```cpp
int main()
{
set<string> s;
string str;
cin >> str;
for (int i = 0; i < str.size(); i++)
for (int j = i; j < str.size(); j++)
s.insert(str.substr(i, j + 1));
cout << s.size();
return 0;
}
```
### C
```cpp
int main()
{
set<string> s;
string str;
cin >> str;
for (int i = 0; i < str.size(); i++)
for (int j = i; j < str.size(); j++)
s.insert(str.substr(i, j - i));
cout << s.size();
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/乘积尾零/solution.cpp
浏览文件 @ 62f75b25
......@@ -10,12 +10,12 @@ 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;
......
data/1.算法初阶/1.蓝桥杯/乘积尾零/solution.md
浏览文件 @ 62f75b25
......@@ -16,7 +16,8 @@
## aop
### before
```cpp
#include <iostream>
using namespace std;
```
### after
```cpp
......@@ -25,21 +26,106 @@
## 答案
```cpp
int main()
{
int count2 = 0, count5 = 0;
int num;
for (int i = 0; i < 100; i++)
{
cin >> num;
while (num % 5 == 0)
{
count5++;
num /= 5;
}
while (num % 2 == 0)
{
count2++;
num /= 2;
}
}
int ans = count2 < count5 ? count2 : count5;
cout << ans;
return 0;
}
```
## 选项
### A
```cpp
int main()
{
int count2 = 0, count5 = 0;
int num;
for (int i = 0; i < 100; i++)
{
cin >> num;
while (num % 5 == 0)
{
count5++;
num /= 5;
}
while (num % 2 == 0)
{
count2++;
num /= 2;
}
}
int ans = count2 < count5 ? count5 : count2;
cout << ans;
return 0;
}
```
### B
```cpp
int main()
{
int count2 = 0, count5 = 0;
int num;
for (int i = 0; i < 100; i++)
{
cin >> num;
while (num % 5 == 0)
{
count5++;
num %= 5;
}
while (num % 2 == 0)
{
count2++;
num %= 2;
}
}
int ans = count2 < count5 ? count2 : count5;
cout << ans;
return 0;
}
```
### C
```cpp
int main()
{
int count2 = 0, count5 = 0;
int num;
for (int i = 0; i < 100; i++)
{
cin >> num;
while (num % 5 == 0)
{
count5++;
num += 5;
}
while (num % 2 == 0)
{
count2++;
num += 2;
}
}
int ans = count2 < count5 ? count2 : count5;
cout << ans;
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/乘积最大/solution.md
浏览文件 @ 62f75b25
......@@ -47,7 +47,10 @@
## aop
### before
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
```
### after
```cpp
......@@ -56,21 +59,174 @@
## 答案
```cpp
int main()
{
int n, k;
long long ans;
cin >> n >> k;
vector<long long> 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)
{
ans = num.back();
k = k - 1;
num.pop_back();
}
else
ans = 1;
while (k > 0)
{
if ((num[0] * num[1]) > num.at(num.size() - 1) * num.at(num.size() - 2))
{
ans = ans * num[0] * num[1] % 1000000009;
num.erase(num.begin(), num.begin() + 1);
}
else
{
ans = ans * num.at(num.size() - 1) * num.at(num.size() - 2) % 1000000009;
num.pop_back();
num.pop_back();
}
k -= 2;
}
cout << ans;
return 0;
}
```
## 选项
### A
```cpp
int main()
{
int n, k;
long long ans;
cin >> n >> k;
vector<long long> 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)
{
ans = num.back();
k = k - 1;
num.pop_back();
}
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);
}
else
{
ans = ans * num.at(num.size()) * num.at(num.size() - 1) % 1000000009;
num.pop_back();
num.pop_back();
}
k -= 2;
}
cout << ans;
return 0;
}
```
### B
```cpp
int main()
{
int n, k;
long long ans;
cin >> n >> k;
vector<long long> 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)
{
ans = num.back();
k = k - 1;
num.pop_back();
}
else
ans = 1;
while (k > 0)
{
if ((num[0] * num[1]) > num.at(num.size() - 1) * num.at(num.size() - 1))
{
ans = ans * num[0] * num[1] % 1000000009;
num.erase(num.begin(), num.begin() + 1);
}
else
{
ans = ans * num.at(num.size() - 1) % 1000000009;
num.pop_back();
num.pop_back();
}
k -= 2;
}
cout << ans;
return 0;
}
```
### C
```cpp
int main()
{
int n, k;
long long ans;
cin >> n >> k;
vector<long long> 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)
{
ans = num.back();
k = k - 1;
num.pop_back();
}
else
ans = 1;
while (k > 0)
{
if ((num[0] * num[1]) > num.at(num.size() - 1) * num.at(num.size() - 2))
{
ans = ans * num[0] * num[1] % 1000000009;
num.erase(num.begin(), num.begin() + 1);
}
else
{
ans = ans * num.at(num.size() - 1) % 1000000009;
num.pop_back();
num.pop_back();
}
k -= 2;
}
cout << ans;
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/凑算式/solution.md
浏览文件 @ 62f75b25
......@@ -16,30 +16,124 @@ A + —- + ——–- = 10
## aop
### before
```cpp
#include <stdio.h>
#include <stdbool.h>
int ans = 0;
int nums[10];
bool visited[10];
void judge()
{
int i;
if (nums[0] + (double)nums[1] / nums[2] + (double)(nums[3] * 100 + nums[4] * 10 + nums[5]) / (nums[6] * 100 + nums[7] * 10 + nums[8]) == 10)
{
printf("%d + %d/%d + %d%d%d/%d%d%d", nums[0], nums[1], nums[2], nums[3], nums[4], nums[5], nums[6], nums[7], nums[8]);
printf("\n");
ans++;
}
}
```
### after
```cpp
int main()
{
dfs(0);
printf("%d\n", ans);
return 0;
}
```
## 答案
```cpp
void dfs(int index)
{
if (index >= 9)
{
judge();
return;
}
int i;
for (i = 1; i < 10; i++)
{
if (visited[i] == false)
{
visited[i] = true;
nums[index] = i;
dfs(index + 1);
visited[i] = false;
}
}
}
```
## 选项
### A
```cpp
void dfs(int index)
{
if (index >= 9)
{
judge();
return;
}
int i;
for (i = 1; i < 10; i++)
{
if (visited[i] == false)
{
visited[i] = true;
nums[index] = i;
dfs(index);
visited[i] = false;
}
}
}
```
### B
```cpp
void dfs(int index)
{
if (index >= 9)
{
judge();
return;
}
int i;
for (i = 1; i < 10; i++)
{
if (visited[i] == true)
{
visited[i] = false;
nums[index] = i;
dfs(index);
visited[i] = true;
}
}
}
```
### C
```cpp
void dfs(int index)
{
if (index >= 9)
{
judge();
return;
}
int i;
for (i = 1; i < 10; i++)
{
if (visited[i] == true)
{
visited[i] = false;
nums[index] = i;
dfs(index + 1);
visited[i] = true;
}
}
}
```
data/1.算法初阶/1.蓝桥杯/大数乘法/solution.md
浏览文件 @ 62f75b25
# 大数乘法
对于32位字长的机器，大约超过20亿，用int类型就无法表示了，我们可以选择int64类型，但无论怎样扩展，固定的整数类型总是有表达的极限！如果对超级大整数进行精确运算呢？一个简单的办法是：仅仅使用现有类型，但是把大整数的运算化解为若干小整数的运算，即所谓：“分块法”。
![](https://img-blog.csdn.net/20160125091111485)
![](https:
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 <stdio.h>
```
### after
```cpp
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;
}
```
## 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;
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;
r[1] += r[2] / base;
r[2] = r[2] % base;
r[0] += r[1] / base;
r[1] = r[1] % base;
}
```
## 选项
### A
```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;
r[3] = n1 % base;
r[2] = n1 / base + n2 % base + n3 % base;
r[1] = n2 / base + n3 / base + n4 % base;
r[0] = n4 / base;
r[1] += r[2] % base;
r[2] = r[2] / base;
r[0] += r[1] / base;
r[1] = r[1] % base;
}
```
### 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;
r[3] = n1 % base;
r[2] = n1 / base + n2 % base + n3 % base;
r[1] = n2 / base + n3 / base + n4 / base;
r[0] = n4 / base;
r[1] += r[2] / base;
r[2] = r[2] % base;
r[0] += r[1] / base;
r[1] = r[1] % base;
}
```
### C
```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;
r[3] = n1 % base;
r[2] = n1 / base + n2 % base + n3 % base;
r[1] = n2 / base + n3 % base + n4 / base;
r[0] = n4 / base;
r[1] += r[2] / base;
r[2] = r[2] % base;
r[0] += r[1] / base;
r[1] = r[1] % base;
}
```
data/1.算法初阶/1.蓝桥杯/大臣的旅费/solution.md
浏览文件 @ 62f75b25
......@@ -42,30 +42,165 @@ J大臣想知道：他从某一个城市出发，中间不休息，到达另一
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
#define ll long long
const double eps = 3e-8;
const int mod = 10;
const int maxn = 10005;
vector<int> ed[maxn];
ll edge[maxn][maxn];
ll dis[maxn];
ll sum = 0;
```
### after
```cpp
int main()
{
int n, a, b;
ll c;
scanf("%d", &n);
for (int i = 1; i < n; i++)
{
scanf("%d%d%lld", &a, &b, &c);
ed[a].push_back(b);
ed[b].push_back(a);
edge[a][b] = c;
edge[b][a] = c;
}
int starta = 1;
int endnode, startnode;
sum = 0;
endnode = bfs(starta);
sum = 0;
startnode = bfs(endnode);
double ans = sum * (sum + 1.0) / 2 + 10.0 * sum;
printf("%.0f\n", ans);
}
```
## 答案
```cpp
int bfs(int node)
{
mem(dis, -1);
queue<int> que;
que.push(node);
int ans = node;
dis[node] = 0;
while (!que.empty())
{
int now = que.front();
que.pop();
for (int i = 0; i < ed[now].size(); i++)
{
int temp = ed[now][i];
if (dis[temp] < 0)
{
dis[temp] = dis[now] + edge[now][temp];
if (dis[temp] > sum)
{
ans = temp;
sum = dis[temp];
}
que.push(temp);
}
}
}
return ans;
}
```
## 选项
### A
```cpp
int bfs(int node)
{
mem(dis, -1);
queue<int> que;
que.push(node);
int ans = node;
dis[node] = 0;
while (!que.empty())
{
int now = que.front();
que.pop();
for (int i = 0; i < ed[now].size(); i++)
{
int temp = ed[now][i];
if (dis[temp] < 0)
{
dis[temp] = dis[now] + edge[now][temp];
ans = temp;
sum = dis[temp];
que.push(temp);
}
}
}
return ans;
}
```
### B
```cpp
int bfs(int node)
{
mem(dis, -1);
queue<int> que;
que.push(node);
int ans = node;
dis[node] = 0;
while (!que.empty())
{
int now = que.front();
que.pop();
for (int i = 0; i < ed[now].size(); i++)
{
int temp = ed[now][i];
dis[temp] = dis[now] + edge[now][temp];
ans = temp;
sum = dis[temp];
que.push(temp);
}
}
return ans;
}
```
### C
```cpp
int bfs(int node)
{
mem(dis, -1);
queue<int> que;
que.push(node);
int ans = node;
dis[node] = 0;
while (!que.empty())
{
int now = que.front();
que.pop();
for (int i = 0; i < ed[now].size(); i++)
{
int temp = ed[now][i];
if (dis[temp] < 0)
{
dis[temp] = dis[now] + edge[now][temp];
if (dis[temp] < sum)
{
ans = temp;
sum = dis[temp];
}
que.push(temp);
}
}
}
return ans;
}
```
data/1.算法初阶/1.蓝桥杯/成绩分析/desc.md
浏览文件 @ 62f75b25
#### 问题描述
小蓝给学生们组织了一场考试，卷面总分为 分，每个学生的得分都是一个 到 的整数。
小蓝给学生们组织了一场考试，卷面总分为100分，每个学生的得分都是一个0到100的整数。
请计算这次考试的最高分、最低分和平均分。
#### 输入格式
输入的第一行包含一个整数 ，表示考试人数。
输入的第一行包含一个整数n，表示考试人数。
接下来 行，每行包含一个 至 的整数，表示一个学生的得分。
接下来n行，每行包含一个0至100的整数，表示一个学生的得分。
#### 输出格式
输出三行。
......@@ -20,6 +20,7 @@
#### 样例输入
```
7
80
92
56
......
data/1.算法初阶/1.蓝桥杯/成绩分析/solution.md
浏览文件 @ 62f75b25
# 成绩分析
#### 问题描述
小蓝给学生们组织了一场考试，卷面总分为 分，每个学生的得分都是一个 到 的整数。
小蓝给学生们组织了一场考试，卷面总分为100分，每个学生的得分都是一个0到100的整数。
请计算这次考试的最高分、最低分和平均分。
#### 输入格式
输入的第一行包含一个整数 ，表示考试人数。
输入的第一行包含一个整数n，表示考试人数。
接下来 行，每行包含一个 至 的整数，表示一个学生的得分。
接下来n行，每行包含一个0至100的整数，表示一个学生的得分。
#### 输出格式
输出三行。
......@@ -21,6 +21,7 @@
#### 样例输入
```
7
80
92
56
......@@ -39,7 +40,10 @@
## aop
### before
```cpp
#include <stdio.h>
#include <iostream>
using namespace std;
```
### after
```cpp
......@@ -48,21 +52,102 @@
## 答案
```cpp
int main()
{
int n;
cin >> n;
int sum = 0;
int top = 0;
int low = 100;
int score;
for (int i = 0; i < n; i++)
{
cin >> score;
if (score > top)
top = score;
if (score < low)
low = score;
sum += score;
}
printf("%d\n%d\n%.2lf", top, low, (sum * 1.0 / n));
return 0;
}
```
## 选项
### A
```cpp
int main()
{
int n;
cin >> n;
int sum = 0;
int top = 0;
int low = 100;
int score;
for (int i = 0; i < n; i++)
{
cin >> score;
if (score > top)
top = score;
if (score < low)
low = score;
sum += score;
}
printf("%d\n%d\n%.f", top, low, (sum * 1.0 / n));
return 0;
}
```
### B
```cpp
int main()
{
int n;
cin >> n;
int sum = 0;
int top = 0;
int low = 100;
int score;
for (int i = 0; i < n; i++)
{
cin >> score;
if (score > top)
score = top;
if (score < low)
score = low;
sum += score;
}
printf("%d\n%d\n%.2lf", top, low, (sum * 1.0 / n));
return 0;
}
```
### C
```cpp
int main()
{
int n;
cin >> n;
int sum = 0;
int top = 0;
int low = 100;
int score;
for (int i = 0; i < n; i++)
{
cin >> score;
if (score < top)
top = score;
if (score > low)
low = score;
sum += score;
}
printf("%d\n%d\n%d", top, low, (sum / n));
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/成绩统计/solution.md
浏览文件 @ 62f75b25
......@@ -20,7 +20,8 @@ yx1
## aop
### before
```cpp
#include <iostream>
using namespace std;
```
### after
```cpp
......@@ -29,21 +30,154 @@ yx1
## 答案
```cpp
int main()
{
struct student
{
string xm;
int xh;
double yy;
double sx;
double cpp;
};
student a[1000];
int n;
double sum = 0, min = 301, max = 0;
string mins, maxs;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i].xm >> a[i].xh >> a[i].yy >> a[i].sx >> a[i].cpp;
sum = a[i].yy + a[i].sx + a[i].cpp;
if (min > sum)
{
min = sum;
mins = a[i].xm;
}
if (max < sum)
{
max = sum;
maxs = a[i].xm;
}
}
cout << maxs << endl
<< mins;
return 0;
}
```
## 选项
### A
```cpp
int main()
{
struct student
{
string xm;
int xh;
double yy;
double sx;
double cpp;
};
student a[1000];
int n;
double sum = 0, min = 301, max = 0;
string mins, maxs;
cin >> n;
for (int i = 1; i < n; i++)
{
cin >> a[i].xm >> a[i].xh >> a[i].yy >> a[i].sx >> a[i].cpp;
sum = a[i].yy + a[i].sx + a[i].cpp;
if (min > sum)
{
min = sum;
mins = a[i].xm;
}
if (max < sum)
{
max = sum;
maxs = a[i].xm;
}
}
cout << maxs << endl
<< mins;
return 0;
}
```
### B
```cpp
int main()
{
struct student
{
string xm;
int xh;
double yy;
double sx;
double cpp;
};
student a[1000];
int n;
double sum = 0, min = 301, max = 0;
string mins, maxs;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i].xm >> a[i].xh >> a[i].yy >> a[i].sx >> a[i].cpp;
sum = a[i].yy + a[i].sx + a[i].cpp;
if (min > sum)
{
min = sum;
mins = a[i]->xm;
}
if (max < sum)
{
max = sum;
maxs = a[i]->xm;
}
cout << maxs << endl
<< mins;
return 0;
}
}
```
### C
```cpp
int main()
{
struct student
{
string xm;
int xh;
double yy;
double sx;
double cpp;
};
student a[1000];
int n;
double sum = 0, min = 301, max = 0;
string mins, maxs;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i].xm >> a[i].xh >> a[i].yy >> a[i].sx >> a[i].cpp;
sum = a[i].yy + a[i].sx + a[i].cpp;
if (min > sum)
{
min = sum;
mins = a[i].xm;
}
if (max < sum)
{
max = sum;
maxs = a[i].xm;
}
}
cout << mins << endl
<< maxs;
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/打印图形/solution.md
浏览文件 @ 62f75b25
......@@ -63,30 +63,111 @@ ran=6
## aop
### before
```cpp
#include <cstdio>
#define N 70
```
### after
```cpp
int main()
{
char a[N][N];
int i, j;
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
a[i][j] = ' ';
f(a, 4, 0, 0);
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
printf("%c", a[i][j]);
printf("\n");
}
return 0;
}
```
## 答案
```cpp
void f(char a[][N], int rank, int row, int col)
{
if (rank == 1)
{
a[row][col] = '*';
return;
}
int w = 1;
int i;
for (i = 0; i < rank - 1; i++)
w *= 2;
f(a, rank - 1, row, col + w / 2);
f(a, rank - 1, row + w / 2, col);
f(a, rank - 1, row + w / 2, col + w);
}
```
## 选项
### A
```cpp
void f(char a[][N], int rank, int row, int col)
{
if (rank == 1)
{
a[row][col] = '*';
return;
}
int w = 1;
int i;
for (i = 0; i < rank; i++)
w *= 2;
f(a, rank - 1, row, col + w / 2);
f(a, rank - 1, row + w / 2, col);
f(a, rank - 1, row + w / 2, col + w);
}
```
### B
```cpp
void f(char a[][N], int rank, int row, int col)
{
if (rank == 1)
{
a[row][col] = '*';
return;
}
int w = 1;
int i;
for (i = 0; i < rank - 1; i++)
w *= 2;
f(a, rank + 1, row, col + w / 2);
f(a, rank + 1, row + w / 2, col);
f(a, rank + 1, row + w / 2, col + w);
}
```
### C
```cpp
void f(char a[][N], int rank, int row, int col)
{
if (rank == 1)
{
a[row][col] = '*';
return;
}
int w = 0;
int i;
for (i = 0; i < rank - 1; i++)
w *= 2;
f(a, rank - 1, row, col + w / 2);
f(a, rank - 1, row + w / 2, col);
f(a, rank - 1, row + w / 2, col + w);
}
```
data/1.算法初阶/1.蓝桥杯/打印菱形/config.json 已删除 100644 → 0
浏览文件 @ 7ddf3532
{
"node_id": "569d5e11c4fc5de7844053d9a733c5e8",
"keywords": [],
"children": [],
"export": []
}
\ No newline at end of file
data/1.算法初阶/1.蓝桥杯/打印菱形/desc.md 已删除 100644 → 0
浏览文件 @ 7ddf3532
给出菱形的边长，在控制台上打印出一个菱形来。
为了便于比对空格，我们把空格用句点代替。
当边长为8时，菱形为：
```
*
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
*
```
\ No newline at end of file
data/1.算法初阶/1.蓝桥杯/打印菱形/solution.java 已删除 100644 → 0
浏览文件 @ 7ddf3532
public class Main {
public static void main(String[] args) {
f(8);
}
public static void f(int n) {
String s = "*";
for (int i = 0; i < 2 * n - 3; i++)
s += ".";
s += "*";
String s1 = s + "\n";
String s2 = "";
for (int i = 0; i < n - 1; i++) {
s = "." + s.substring(0, s.length() - 3) + "*"; // 填空
s1 = s + "\n" + s1;
s2 += s + "\n";
}
System.out.println(s1 + s2);
}
}
\ No newline at end of file
data/1.算法初阶/1.蓝桥杯/打印菱形/solution.md 已删除 100644 → 0
浏览文件 @ 7ddf3532
# 打印菱形
给出菱形的边长，在控制台上打印出一个菱形来。
为了便于比对空格，我们把空格用句点代替。
当边长为8时，菱形为：
```
*
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
* *
*
```
## aop
### before
```cpp
```
### after
```cpp
```
## 答案
```cpp
```
## 选项
### A
```cpp
```
### B
```cpp
```
### C
```cpp
```
data/1.算法初阶/1.蓝桥杯/承压计算/solution.cpp
浏览文件 @ 62f75b25
......@@ -21,7 +21,7 @@ int main()
for (ll j = 1; j <= i; j++)
{
a[i + 1][j] += (0.5 * a[i][j]);
a[i + 1][j + 1] += (0.5 * a[i][j]);
a[i][j + 1] += (0.5 * a[i][j]);
}
}
double minn = 1000000000, maxx = -1;
......
data/1.算法初阶/1.蓝桥杯/承压计算/solution.md
浏览文件 @ 62f75b25
......@@ -54,6 +54,14 @@ X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X
## aop
### before
```cpp
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 1e6 + 50;
double a[1050][1050];
```
### after
......@@ -63,21 +71,123 @@ X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X
## 答案
```cpp
int main()
{
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
scanf("%lf", &a[i][j]);
}
}
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
a[i + 1][j] += (0.5 * a[i][j]);
a[i + 1][j + 1] += (0.5 * a[i][j]);
}
}
double minn = 1000000000, maxx = -1;
for (ll i = 1; i <= 30; i++)
{
maxx = max(maxx, a[30][i]);
minn = min(minn, a[30][i]);
printf("%lf\n", a[30][i]);
}
printf("%lf\n", maxx * 2086458231 / minn);
}
```
## 选项
### A
```cpp
int main()
{
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
scanf("%lf", &a[i][j]);
}
}
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
a[i + 1][j] += (0.5 * a[i][j]);
a[i + 1][j + 1] += (0.5 * a[i][j]);
}
}
double minn = 1000000000, maxx = -1;
for (ll i = 1; i <= 29; i++)
{
maxx = max(maxx, a[30][i]);
minn = min(minn, a[30][i]);
printf("%lf\n", a[30][i]);
}
printf("%lf\n", maxx * 2086458231 / minn);
}
```
### B
```cpp
int main()
{
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
scanf("%lf", &a[i][j]);
}
}
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
a[i + 1][j] = (0.5 * a[i][j]);
a[i + 1][j + 1] = (0.5 * a[i][j]);
}
}
double minn = 1000000000, maxx = -1;
for (ll i = 1; i <= 29; i++)
{
maxx = max(maxx, a[30][i]);
minn = min(minn, a[30][i]);
printf("%lf\n", a[30][i]);
}
printf("%lf\n", maxx * 2086458231 / minn);
}
```
### C
```cpp
int main()
{
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
scanf("%lf", &a[i][j]);
}
}
for (ll i = 1; i <= 29; i++)
{
for (ll j = 1; j <= i; j++)
{
a[i + 1][j] += (0.5 * a[i][j]);
a[i][j + 1] += (0.5 * a[i][j]);
}
}
double minn = 1000000000, maxx = -1;
for (ll i = 1; i <= 30; i++)
{
maxx = max(maxx, a[30][i]);
minn = min(minn, a[30][i]);
printf("%lf\n", a[30][i]);
}
printf("%lf\n", maxx * 2086458231 / minn);
}
```
data/1.算法初阶/1.蓝桥杯/抽签/solution.md
浏览文件 @ 62f75b25
......@@ -36,30 +36,96 @@ C国最多可以派出2人。
## aop
### before
```cpp
#include <iostream>
using namespace std;
int a[10000], N, i, ans = 0;
```
### after
```cpp
int main()
{
cin >> N;
for (i = 0; i < N; i++)
{
cin >> a[i];
}
findAns(a, 0, 0);
cout << ans << endl;
return 0;
}
```
## 答案
```cpp
void findAns(int a[], int start, int An)
{
if (start == N)
{
if (An == 5)
ans++;
return;
}
for (i = 0; i <= a[start]; i++)
{
findAns(a, start + 1, An + i);
}
}
```
## 选项
### A
```cpp
void findAns(int a[], int start, int An)
{
if (start == N)
{
if (An == 5)
ans++;
return;
}
for (i = 0; i <= a[start]; i++)
{
findAns(a, start + 1, An);
}
}
```
### B
```cpp
void findAns(int a[], int start, int An)
{
if (start == N)
{
if (An == 5)
ans++;
return;
}
for (i = 0; i <= a[start]; i++)
{
findAns(a, start, An + i);
}
}
```
### C
```cpp
void findAns(int a[], int start, int An)
{
if (start == N)
{
if (An == 5)
ans++;
return;
}
for (i = 0; i <= a[start]; i++)
{
findAns(a, start, An + 1);
}
}
```
data/1.算法初阶/1.蓝桥杯/搭积木/solution.md
浏览文件 @ 62f75b25
......@@ -25,30 +25,160 @@
## aop
### before
```cpp
#include <bits/stdc++.h>
using namespace std;
int row = 4, col;
int seq[5][8];
int tmp[11];
int cnt = 0;
bool used[10];
void change()
{
int cur = 1;
for (int i = 1; i <= row; ++i)
{
for (int j = 1; j <= i; ++j)
{
seq[i][j] = tmp[cur++];
}
}
}
bool judge()
{
for (int i = 1; i <= row - 1; ++i)
{
for (int j = 1; j <= i; ++j)
{
if (seq[i][j] > seq[i + 1][j] || seq[i][j] > seq[i + 1][j + 1])
return false;
}
}
return true;
}
int ans = 0;
```
### after
```cpp
int main()
{
memset(used, false, sizeof(used));
dfs(1);
cout << cnt << endl;
return 0;
}
```
## 答案
```cpp
void dfs(int idx)
{
if (idx == 11)
{
change();
if (judge())
{
cnt++;
return;
}
return;
}
for (int i = 0; i <= 9; ++i)
{
if (!used[i])
{
tmp[idx] = i;
used[i] = true;
dfs(idx + 1);
used[i] = false;
}
}
}
```
## 选项
### A
```cpp
void dfs(int idx)
{
if (idx == 11)
{
change();
if (judge())
{
cnt++;
return;
}
return;
}
for (int i = 0; i <= 9; ++i)
{
if (!used[i])
{
tmp[idx] = i;
used[i] = true;
dfs(idx);
used[i] = false;
}
}
}
```
### B
```cpp
void dfs(int idx)
{
if (idx == 11)
{
change();
if (judge())
{
cnt++;
return;
}
return;
}
for (int i = 0; i <= 9; ++i)
{
if (!used[i])
{
tmp[idx] = i;
used[i] = false;
dfs(idx);
used[i] = true;
}
}
}
```
### C
```cpp
void dfs(int idx)
{
if (idx == 11)
{
change();
if (judge())
{
cnt++;
return;
}
return;
}
for (int i = 0; i <= 9; ++i)
{
if (!used[i])
{
tmp[idx] = i;
used[i] = false;
dfs(idx + 1);
used[i] = true;
}
}
}
```
data/1.算法初阶/1.蓝桥杯/次数差/solution.md
浏览文件 @ 62f75b25
......@@ -37,7 +37,8 @@ bbccccddaaaacccc
## aop
### before
```cpp
#include <stdio.h>
#include <string.h>
```
### after
```cpp
......@@ -46,21 +47,117 @@ bbccccddaaaacccc
## 答案
```cpp
int main()
{
char s[1000];
scanf("%s", &s);
int len = strlen(s);
int helper[256] = {0};
int max = 0, min = len;
for (int i = 0; i < len; i++)
{
helper[s[i]]++;
}
for (int i = 0; i < 256; i++)
{
if (helper[i] > max)
{
max = helper[i];
}
if (helper[i] < min && helper[i] != 0)
{
min = helper[i];
}
}
printf("%d\n", max - min);
return 0;
}
```
## 选项
### A
```cpp
int main()
{
char s[1000];
scanf("%s", &s);
int len = strlen(s);
int helper[256] = {0};
int max = 0, min = len;
for (int i = 0; i < len; i++)
{
helper[s[i]]++;
}
for (int i = 0; i < 256; i++)
{
if (helper[i] > max)
{
max = helper[i];
}
if (helper[i] < min)
{
min = helper[i];
}
}
printf("%d\n", max - min);
return 0;
}
```
### B
```cpp
int main()
{
char s[1000];
scanf("%s", &s);
int len = strlen(s);
int helper[256] = {0};
int max = 0, min = len;
for (int i = 0; i < len; i++)
{
helper[s[i]]++;
}
for (int i = 0; i < 256; i++)
{
if (helper[i] > max)
{
max = helper[i];
}
if (helper[i] < min && helper[i] == 0)
{
min = helper[i];
}
}
printf("%d\n", max - min);
return 0;
}
```
### C
```cpp
int main()
{
char s[1000];
scanf("%s", &s);
int len = strlen(s);
int helper[256] = {0};
int max = 0, min = len;
for (int i = 0; i < len; i++)
{
helper[s[i]]++;
}
for (int i = 0; i < 256; i++)
{
if (helper[i] < max)
{
max = helper[i];
}
if (helper[i] > min)
{
min = helper[i];
}
}
printf("%d\n", max - min);
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/波动数列/solution.md
浏览文件 @ 62f75b25
......@@ -25,7 +25,17 @@
## aop
### before
```cpp
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int N = 1e3 + 5;
const int mod = 100000007;
int n, s, a, b, up;
ll v;
int dp[2][N * (N + 1) / 2], now;
int ans;
```
### after
```cpp
......@@ -34,21 +44,111 @@
## 答案
```cpp
int main()
{
scanf("%d%d%d%d", &n, &s, &a, &b);
dp[now][0] = 1;
for (int i = 1; i < n; ++i)
{
now = !now;
up = i * (i + 1) / 2;
for (int j = 0; j <= up; ++j)
{
dp[now][j] = dp[!now][j];
if (j >= i)
dp[now][j] = (dp[now][j] + dp[!now][j - i]) % mod;
}
}
for (int i = 0; i <= up; ++i)
{
v = 1ll * s - 1ll * i * a + 1ll * (up - i) * b;
if (v % n == 0)
ans = (ans + dp[now][i]) % mod;
}
printf("%d\n", ans);
return 0;
}
```
## 选项
### A
```cpp
int main()
{
scanf("%d%d%d%d", &n, &s, &a, &b);
dp[now][0] = 1;
for (int i = 1; i < n; ++i)
{
now = !now;
up = i * (i + 1) / 2;
for (int j = 0; j <= up; ++j)
{
dp[now][j] = dp[now][j];
if (j >= i)
dp[now][j] = (dp[now][j] + dp[!now][j - i]) % mod;
}
}
for (int i = 0; i <= up; ++i)
{
v = 1ll * s - 1ll * i * a + 1ll * (up - i) * b;
if (v % n == 0)
ans = (ans + dp[now][i]) % mod;
}
printf("%d\n", ans);
return 0;
}
```
### B
```cpp
int main()
{
scanf("%d%d%d%d", &n, &s, &a, &b);
dp[now][0] = 1;
for (int i = 1; i < n; ++i)
{
up = i * (i + 1) / 2;
for (int j = 0; j <= up; ++j)
{
dp[now][j] = dp[!now][j];
if (j >= i)
dp[now][j] = (dp[now][j] + dp[now][j - i]) % mod;
}
}
for (int i = 0; i <= up; ++i)
{
v = 1ll * s - 1ll * i * a + 1ll * (up - i) * b;
if (v % n == 0)
ans = (ans + dp[now][i]) % mod;
}
printf("%d\n", ans);
return 0;
}
```
### C
```cpp
int main()
{
scanf("%d%d%d%d", &n, &s, &a, &b);
dp[now][0] = 1;
for (int i = 1; i < n; ++i)
{
up = i * (i + 1) / 2;
for (int j = 0; j <= up; ++j)
{
dp[now][j] = dp[!now][j];
if (j >= i)
dp[now][j] = (dp[!now][j] + dp[!now][j - i]) % mod;
}
}
for (int i = 0; i <= up; ++i)
{
v = 1ll * s - 1ll * i * a + 1ll * (up - i) * b;
if (v % n == 0)
ans = (ans + dp[now][i]) % mod;
}
printf("%d\n", ans);
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/测试次数/solution.md
浏览文件 @ 62f75b25
......@@ -16,30 +16,96 @@ x星球有很多高耸入云的高塔，刚好可以用来做耐摔测试。塔
## aop
### before
```cpp
#include <iostream>
using namespace std;
int num[5][1010] = {0};
```
### after
```cpp
int main()
{
cout << dp(3, 1000) << endl;
return 0;
}
```
## 答案
```cpp
int dp(int k, int n)
{
int res = 10000;
if (n == 0)
return 0;
if (k == 1)
return n;
if (num[k][n])
return num[k][n];
for (int i = 1; i <= n; i++)
{
res = min(res, max(dp(k - 1, i - 1), dp(k, n - i)) + 1);
}
num[k][n] = res;
return res;
}
```
## 选项
### A
```cpp
int dp(int k, int n)
{
int res = 10000;
if (n == 0)
return 0;
if (k == 1)
return n;
if (num[k][n])
return num[k][n];
for (int i = 1; i <= n; i++)
{
res = max(res, min(dp(k - 1, i - 1), dp(k, n - 1)) + 1);
}
num[k][n] = res;
return res;
}
```
### B
```cpp
int dp(int k, int n)
{
int res = 10000;
if (n == 0)
return 0;
if (k == 1)
return n;
if (num[k][n])
return num[k][n];
for (int i = 1; i <= n; i++)
{
res = min(res, max(dp(k + 1, i + 1), dp(k, n - i)) + 1);
}
num[k][n] = res;
return res;
}
```
### C
```cpp
int dp(int k, int n)
{
int res = 10000;
if (n == 0)
return 0;
if (k == 1)
return n;
if (num[k][n])
return num[k][n];
for (int i = 1; i <= n; i++)
{
res = min(res, max(dp(k - 1, i), dp(k, n - i)) + 1);
}
num[k][n] = res;
return res;
}
```
data/1.算法初阶/1.蓝桥杯/猜字母/solution.md
浏览文件 @ 62f75b25
......@@ -6,7 +6,9 @@
## aop
### before
```cpp
#include <iostream>
#include <vector>
using namespace std;
```
### after
```cpp
......@@ -15,21 +17,89 @@
## 答案
```cpp
int main()
{
vector<char> vc1;
int i;
for (i = 0; i < 2014; i++)
vc1.push_back('a' + (i % 19));
while (vc1.size() != 1)
{
vector<char> vc2;
cout << vc1.size() << endl;
for (i = 1; i < vc1.size(); i += 2)
vc2.push_back(vc1[i]);
vc1.assign(vc2.begin(), vc2.end());
cout << vc1.size() << endl;
}
cout << vc1[0] << endl;
return 0;
}
```
## 选项
### A
```cpp
int main()
{
vector<char> vc1;
int i;
for (i = 0; i < 2014; i++)
vc1.push_back('a' + (i % 19));
while (vc1.size() != 1)
{
vector<char> vc2;
cout << vc1.size() << endl;
for (i = 0; i < vc1.size(); i++)
vc2.push_back(vc1[i]);
vc1.assign(vc2.begin(), vc2.end());
cout << vc1.size() << endl;
}
cout << vc1[0] << endl;
return 0;
}
```
### B
```cpp
int main()
{
vector<char> vc1;
int i;
for (i = 0; i < 2014; i++)
vc1.push_back('a' + (i % 19));
while (vc1.size() != 1)
{
vector<char> vc2;
cout << vc1.size() << endl;
for (i = 0; i < vc1.size(); i += 1)
vc2.push_back(vc1[i]);
vc1.assign(vc2.begin(), vc2.end());
cout << vc1.size() << endl;
}
cout << vc1[0] << endl;
return 0;
}
```
### C
```cpp
int main()
{
vector<char> vc1;
int i;
for (i = 0; i < 2014; i++)
vc1.push_back('a' + (i % 19));
while (vc1.size() != 1)
{
vector<char> vc2;
cout << vc1.size() << endl;
for (i = 0; i < vc1.size(); i += 2)
vc2.push_back(vc1[i]);
vc1.assign(vc2.begin(), vc2.end());
cout << vc1.size() << endl;
}
cout << vc1[0] << endl;
return 0;
}
```
data/1.算法初阶/1.蓝桥杯/猜年龄/solution.md
浏览文件 @ 62f75b25
......@@ -11,26 +11,38 @@
```
### after
```cpp
#include <iostream>
using namespace std;
int main()
{
for (int i = 11; i < 35; i++)
{
int i1 = i * i * i;
int i2 = i * i1;
if ((i1 >= 1000 && i1 < 10000) && (i2 >= 100000 && i2 < 1000000))
cout << i << " " << i1 << " " << i2 << endl;
}
return 0;
}
```
## 答案
```cpp
18
```
## 选项
### A
```cpp
19
```
### B
```cpp
20
```
### C
```cpp
21
```
data/1.算法初阶/1.蓝桥杯/超级胶水/solution.cpp
浏览文件 @ 62f75b25
#include<bits/stdc++.h>
#include <bits/stdc++.h>
using namespace std;
const int maxn=100005;
const int maxn = 100005;
int numv[maxn];
int v[maxn];
int dfs(int idx){
if(idx==0)
return 0;
if(idx==1)
return v[0]*v[1];
int max_,a=0x3f3f3f,b=0x3f3f3f3f,c=0;
if(idx>=2)
a=(v[idx]*v[idx-1]+dfs(idx-2)+numv[idx-2]*(v[idx]+v[idx-1])); //与左边粘
b=v[idx]*numv[idx-1]+dfs(idx-1); //与左边不粘
c=min(a,b);
int dfs(int idx)
{
if (idx == 0)
return 0;
if (idx == 1)
return v[0] * v[1];
int max_, a = 0x3f3f3f, b = 0x3f3f3f3f, c = 0;
if (idx >= 2)
a = (v[idx] * v[idx - 1] + dfs(idx - 2) + numv[idx - 2] * (v[idx] + v[idx - 1])); //与左边粘
b = v[idx] * numv[idx - 1] + dfs(idx - 1); //与左边不粘
c = min(a, b);
return c;
}
int main(){
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&v[i]);
if(i==0) numv[i]=v[i];
numv[i]=numv[i-1]+v[i];
}
printf("%d",dfs(n-1));
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d", &v[i]);
if (i == 0)
numv[i] = v[i];
numv[i] = numv[i - 1] + v[i];
}
printf("%d", dfs(n - 1));
}
\ No newline at end of file
data/1.算法初阶/1.蓝桥杯/超级胶水/solution.md
浏览文件 @ 62f75b25
......@@ -21,30 +21,107 @@
## aop
### before
```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
int numv[maxn];
int v[maxn];
```
### after
```cpp
int main(){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&v[i]);
if(i==0) numv[i]=v[i];
numv[i]=numv[i-1]+v[i];
}
printf("%d",dfs(n-1));
}
```
## 答案
```cpp
int dfs(int idx){
if(idx==0)
return 0;
if(idx==1)
return v[0]*v[1];
int max_,a=0x3f3f3f,b=0x3f3f3f3f,c=0;
if(idx>=2)
a=(v[idx]*v[idx-1]+dfs(idx-2)+numv[idx-2]*(v[idx]+v[idx-1]));
b=v[idx]*numv[idx-1]+dfs(idx-1);
c=min(a,b);
return c;
}
```
## 选项
### A
```cpp
int dfs(int idx)
{
if (idx == 0)
return 0;
if (idx == 1)
return v[0] * v[1];
int max_, a = 0x3f3f3f, b = 0x3f3f3f3f, c = 0;
if (idx >= 2)
a = (v[idx] * v[idx] + dfs(idx - 2) + numv[idx - 2] * (v[idx] + v[idx - 1]));
b = v[idx] * numv[idx] + dfs(idx - 1);
c = min(a, b);
return c;
}
```
### B
```cpp
int dfs(int idx)
{
if (idx == 0)
return 0;
if (idx == 1)
return v[0] * v[1];
int max_, a = 0x3f3f3f, b = 0x3f3f3f3f, c = 0;
if (idx >= 2)
a = (v[idx - 2] * v[idx] + dfs(idx - 2) + numv[idx - 2] * (v[idx] + v[idx - 1]));
b = v[idx - 2] * numv[idx] + dfs(idx - 1);
c = min(a, b);
return c;
}
```
### C
```cpp
int dfs(int idx)
{
if (idx == 0)
return 0;
if (idx == 1)
return v[0] * v[1];
int max_, a = 0x3f3f3f, b = 0x3f3f3f3f, c = 0;
if (idx >= 2)
a = (v[idx] * v[idx] + dfs(idx - 2) + numv[idx - 2] * (v[idx] + v[idx - 1]));
b = v[idx - 1] * numv[idx] + dfs(idx - 1);
c = min(a, b);
return c;
}
```
data/1.算法初阶/1.蓝桥杯/错误票据/solution.md
浏览文件 @ 62f75b25
......@@ -52,7 +52,14 @@
## aop
### before
```cpp
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int ans[10005];
char str[100001];
```
### after
```cpp
......@@ -61,21 +68,145 @@
## 答案
```cpp
int main()
{
int n;
cin >> n;
int i, k = 0;
getchar();
for (i = 0; i < n; i++)
{
gets(str);
char *p;
p = strtok(str, " ");
ans[k++] = atoi(p);
while (p)
{
p = strtok(NULL, " ");
if (p)
ans[k++] = atoi(p);
}
}
sort(ans, ans + k);
int ans1, ans2;
for (i = 0; i < k - 1; i++)
{
if (ans[i + 1] - ans[i] == 2)
{
ans1 = ans[i] + 1;
}
if (ans[i + 1] == ans[i])
ans2 = ans[i];
}
cout << ans1 << " " << ans2;
}
```
## 选项
### A
```cpp
int main()
{
int n;
cin >> n;
int i, k = 0;
getchar();
for (i = 0; i < n; i++)
{
gets(str);
char *p;
p = strtok(str, " ");
ans[k++] = atoi(p);
while (p)
{
p = strtok(NULL, " ");
if (p)
ans[k] = atoi(p);
}
}
sort(ans, ans + k);
int ans1, ans2;
for (i = 0; i < k - 1; i++)
{
if (ans[i + 1] - ans[i] == 2)
{
ans1 = ans[i] + 1;
}
if (ans[i + 1] == ans[i])
ans2 = ans[i];
}
cout << ans1 << " " << ans2;
}
```
### B
```cpp
int main()
{
int n;
cin >> n;
int i, k = 0;
getchar();
for (i = 0; i < n; i++)
{
gets(str);
char *p;
p = strtok(str, " ");
ans[k++] = atoi(p);
while (p)
{
p = strtok(NULL, " ");
if (p)
ans[k++] = atoi(p);
}
}
sort(ans, ans + k);
int ans1, ans2;
for (i = 0; i < k - 1; i++)
{
if (ans[i + 1] - ans[i] == 2)
{
ans1 = ans[i + 1];
}
if (ans[i + 1] == ans[i])
ans2 = ans[i];
}
cout << ans1 << " " << ans2;
}
```
### C
```cpp
int main()
{
int n;
cin >> n;
int i, k = 0;
getchar();
for (i = 0; i < n; i++)
{
gets(str);
char *p;
p = strtok(str, " ");
ans[k++] = atoi(p);
while (p)
{
p = strtok(NULL, " ");
if (p)
ans[k++] = atoi(p);
}
}
sort(ans, ans + k);
int ans1, ans2;
for (i = 0; i < k - 1; i++)
{
if (ans[i + 1] - ans[i] == 1)
{
ans1 = ans[i] + 1;
}
if (ans[i + 1] == ans[i])
ans2 = ans[i];
}
cout << ans1 << " " << ans2;
}
```
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