Merge remote-tracking branch 'origin/master'

# Conflicts:
#	exercises/wumf/简单/ShuiXianHua.java
#	exercises/wumf/简单/exercies.md

.gitignore

@@ -6,5 +6,4 @@ release/
 debug/
 *.o
 *.obj
-*.a
-*.out
+*.a
\ No newline at end of file
\ No newline at end of file

exercises/wcg/中等/大整数替换数位/exercises.md

+# 大整数替换数位
+
+以字符串的形式给你一个长度为 M 的整数 N，请你计算出对这个数进行一次操作后模 9 的值为 1 的所有可能的不同操作方式。
+
+在一次操作中， 我们可以选择 N 的一个数位 N[i]，并把它替换成另一个不同的 0 到 9 范围之内的数 B，两种操作方式不同当且仅当它们选择的 i 或 B 不同。
+
+## 输入描述
+第一行包含一个整数 M (1 <= M <= 100000)。
+
+第二行包含一个大整数 N。
+
+## 输出描述
+
+第一行输出一个整数， 代表对 N 执行一次操作后使 N 模 9 的值为 1 的所有可能的不同操作方式数量。
+
+## 输入用例
+
+4
+
+2345
+
+## 输出用例
+
+5
+
+## 提示
+
+一共有 5 种不同的操作方式使操作后的数模 9 = 1。
+
+前两种是:
+
+i = 0, B = 7
+
+7345 % 9 = 1
+
+i = 1, B = 8
+
+2845 % 9 = 1

exercises/wcg/中等/大整数替换数位/solution.cpp

+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define out(x) cout << #x << '=' << x << endl
+#define out2(x, y) cout << #x << '=' << x << ',' << #y << '=' << y << endl 
+#define no cout << "No" << endl; return
+#define yes cout << "Yes" << endl; return
+#define outvec(a) for (int v : a) { cout << v << ' '; } cout << endl
+#define lowbit(x) (x & -x)
+#define gcd __gcd 
+#define inf 0x3f3f3f3f3f3f3f3fLL
+#define infi 0x3f3f3f3f
+
+using ll = long long;
+using pii = pair<int, int>;
+
+
+void solve() {
+	int n;
+	string s;
+	cin >> n >> s;
+	int sum = 0;
+	for (int i = 0; i < n; i++) {
+		sum += s[i] - '0';
+	}
+	int ans = 0;
+	for (int i = 0; i < n; i++) {
+		int k = s[i] - '0';
+		for (int j = 0; j < 10; j++) {
+			if (j == k) continue;
+			if ((sum - k + j) % 9 == 1) {
+				ans++;
+			}
+		}
+	}
+	cout << ans << endl;
+}
+
+int main(void) {
+    ios::sync_with_stdio(false);
+    cin.tie(0); cout.tie(0);
+    int t = 1;
+    //cin >> t;
+    
+    while (t--) {
+    	solve();
+	}
+}

exercises/wcg/中等/大整数替换数位/test_cases/1.out

+51786

exercises/wcg/中等/大整数替换数位/test_cases/10.in

+10004
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exercises/wcg/中等/大整数替换数位/test_cases/10.out

+10860

exercises/wcg/中等/大整数替换数位/test_cases/2.out

+57947

exercises/wcg/中等/大整数替换数位/test_cases/3.out

+54572

exercises/wcg/中等/大整数替换数位/test_cases/4.out

+61898

exercises/wcg/中等/大整数替换数位/test_cases/5.out

+82200

exercises/wcg/中等/大整数替换数位/test_cases/6.out

+31646

exercises/wcg/中等/大整数替换数位/test_cases/7.out

+37675

exercises/wcg/中等/大整数替换数位/test_cases/8.out

+70657

exercises/wcg/中等/大整数替换数位/test_cases/9.out

+37707

exercises/wcg/中等/近视的小张/exercises.md

+# 近视的小张
+
+小张和他的 M 个朋友来到了一个十分神奇的地方，在这里有 N 个 柱子， 对于每个 1 <= i <= N， 第 i 个柱子都有两个属性 : H[i], P[i]。
+
+H[i] 表示柱子 i 的高度， 而 P[i] 则表示柱子 i 当前所处的位置，题目保证同一个位置不会有多个柱子。
+
+在一个柱子 i 在另一个不比他低的柱子 j 的后面时 (P[i] > P[j] && H[i] <= H[j])， 这个柱子会被遮挡住， 也就不再能被清晰的看到。
+小张和他的朋友们在位置 0 休息时， 发现似乎朋友们能清晰看到的柱子数量并不相同，在他反复思考后， 他认为这可能是近视度数导致的，于是他询问了每一个朋友的近视度数 A[i]。
+
+为了方便计算， 我们认为对于朋友 j 来说，对每一个柱子 i， 如果有 P[i] > A[j]， 那么第 i 个柱子无法被清晰看见。
+
+请你计算出每个小张的朋友能清晰看到的最远一个柱子的位置, 如果那个朋友一个柱子都没有清晰看到， 请输出 -1。
+
+## 输入描述
+
+第一行包含两个整数 M (1 <= M <= 100000), N(1 <= N <= 100000)。
+
+第二行包含 N 个整数， 其中第 i 个整数代表第 i 个柱子的高度 H[i] (1 <= H[i] <= 1000000000)。
+
+第三行包含 N 个整数， 其中第 i 个整数代表第 i 个柱子的位置 P[i] (1 <= P[i] <= 1000000000)。
+
+第四行包含 M 个整数， 其中第 i 个整数代表第 i 个张浩扬的朋友的近视度数 A[i] (1 <= A[i] <= 1000000000)。
+
+## 输出描述
+
+按顺序打印小张每一个同学能清晰看到的最远的柱子，每个答案占一行。
+
+## 输入用例
+
+4 3
+
+1 3 2
+
+2 3 4
+
+1 2 3 4
+
+## 输出用例
+
+-1
+
+2
+
+3
+
+3
+
+## 提示
+
+对于最后一个朋友来说， 因为最后一个柱子被第 2 个柱子遮挡住了， 所以他只能看到第 2 个柱子， 而这个柱子的位置为 3。

exercises/wcg/中等/近视的小张/solution.cpp

+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define out(x) cout << #x << '=' << x << endl
+#define out2(x, y) cout << #x << '=' << x << ',' << #y << '=' << y << endl 
+#define no cout << "No" << endl; return
+#define yes cout << "Yes" << endl; return
+#define outvec(a) for (int v : a) { cout << v << ' '; } cout << endl
+#define lowbit(x) (x & -x)
+#define gcd __gcd 
+#define inf 0x3f3f3f3f3f3f3f3fLL
+#define infi 0x3f3f3f3f
+
+using ll = long long;
+using pii = pair<int, int>;
+
+
+void solve() {
+	int m, n;
+	cin >> m >> n;
+	vector<int> h(n);
+	vector<int> p(n);
+	vector<int> id(n); 
+	for (int i = 0; i < n; i++) {
+		cin >> h[i];
+		id[i] = i;
+	}
+	for (int i = 0; i < n; i++) {
+		cin >> p[i];
+	}
+	sort(id.begin(), id.end(), [&](int a, int b) -> int {
+		return p[a] < p[b];
+	});
+	vector<int> stk;
+	for (int i = 0; i < n; i++) {
+		if (stk.empty() || h[stk.back()] < h[id[i]]) {
+			stk.push_back(id[i]);
+		}
+	}
+	for (int i = 0; i < m; i++) {
+		int a;
+		cin >> a;
+		int left = 0;
+		int right = (int) stk.size() - 1;
+		int ans = -1;
+		while (left <= right) {
+			int mid = left + (right - left) / 2;
+			if (p[stk[mid]] > a) {
+				right = mid - 1;
+			} else {
+				left = mid + 1;
+				ans = p[stk[mid]];
+			}
+		}
+		cout << ans << endl;
+	}
+}
+
+int main(void) {
+    ios::sync_with_stdio(false);
+    cin.tie(0); cout.tie(0);
+    int t = 1;
+    //cin >> t;
+    
+    while (t--) {
+    	solve();
+	}
+}

exercises/wcg/困难/小张的手速大比拼/exercises.md

+# 小张的手速大比拼
+
+在很久很久以前，小张找到了一颗有 N 个节点的有根树 T 。 树上的节点编号在 1 到 N 范围内。
+
+他很快发现树上的每个节点 i 都有一个对应的整数值 V[i]。
+
+一个老爷爷对他说，给你一个整数 X， 如果你能回答我的 M 个问题，他就给小张购买一些零食。
+
+对于每个问题 Q[i], 请你找到 在 T 中以节点 Q[i] 为根的子树中的所有节点(包括 Q[i])中， 有没有两个节点 A, B (A != B) 的值 V[A] ^ V[B] 的异或和为 X。
+
+如果有这样的两个节点， 请你输出 YES。
+
+否则你需要输出 NO 表示没有节点符合上面的条件。
+
+## 输入描述
+
+第一行包含三个整数 N (2 <= N <= 100000), X (0 <= X <= 1000000000)，M（1 <= M <= N) , 表示树上节点的数量 和 老爷爷给的数 X 和 老爷爷的问题数量。
+
+第二行包含 N 个数， 第 1 个数代表节点 1 的父节点， 第 2 个数代表节点 2 的父节点， 以此类推。
+
+题目保证输入是一颗合法的有根树， 如果第 i 个数为 -1 表示 节点 i 没有父节点。
+
+第三行包含 N 个数， 第 1 个数代表节点 1 的值 V[1]， 第 2 个数代表节点 2 的值 V[2]， 以此类推。
+
+题目保证对于每一个下标1 <= i <= N 都有 0 <= V[i] <= 1000000000。
+
+接下来有 M 行， 每一行包括一个整数Q[i] (1 <= Q[i] <= N)， 代表老爷爷问小张的第 i 个问题。
+
+## 输出描述
+
+按顺序输出问题 Q[1], Q[2] ... Q[M] 的答案， 每个答案占一行。
+
+## 输入用例
+
+4 2 3
+
+-1 1 1 2
+
+4 3 2 1
+
+1
+
+3
+
+4
+
+## 输出用例
+
+YES
+
+NO
+
+NO
+
+## 提示
+
+第一个用例给出的树是
+
+                1(4)
+          2(3)         3(2)
+    4(1)
+
+括号左边的数为节点编号， 括号内的数为节点值。
+
+对于第一个测试用例的第一个问题， 显然有节点 4 与 节点 2 的值 (1 ^ 3) == 2。
+
+对于第一个测试用例的第二个问题， 以节点 3 为根的子树里只有 3 一个节点 并没有一对节点满足 A != B 且 (V[A] ^ V[B]) == 2。
+

exercises/wcg/困难/小张的手速大比拼/solution.cpp

+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define out(x) cout << #x << '=' << x << endl
+#define out2(x, y) cout << #x << '=' << x << ',' << #y << '=' << y << endl 
+#define no cout << "No" << endl; return
+#define yes cout << "Yes" << endl; return
+#define outvec(a) for (int v : a) { cout << v << ' '; } cout << endl
+#define lowbit(x) (x & -x)
+#define gcd __gcd 
+#define inf 0x3f3f3f3f3f3f3f3fLL
+#define infi 0x3f3f3f3f
+
+using ll = long long;
+using pii = pair<int, int>;
+
+
+void solve() {
+	int n, x, m;
+	cin >> n >> x >> m;
+	vector<vector<int>> graph(n + 1);
+	
+	vector<int> p(n + 1);
+	for (int i = 1; i <= n; i++) {
+		cin >> p[i];
+		if (p[i] != -1) graph[p[i]].push_back(i);
+	}
+	
+	vector<int> v(n + 1);
+	for (int i = 1; i <= n; i++) {
+		cin >> v[i];
+	}
+	int dfn = 0;
+	vector<int> dfns(n + 1);
+	vector<int> dfne(n + 1);
+	vector<int> idx(n + 1);
+	function<int(int)> dfs = [&](int node) -> int {
+		dfns[node] = dfne[node] = ++dfn;
+		idx[dfn] = node;
+		for (int next : graph[node]) {
+			dfne[node] = dfs(next);
+		}
+		return dfne[node];
+	};
+	for (int i = 1; i <= n; i++) {
+		if (p[i] == -1) {
+			dfs(i);
+			break;
+		}
+	}
+	map<int, int> mp;
+	vector<int> pre(n + 1);
+	for (int i = 1; i <= n; i++) {
+		pre[i] = pre[i - 1];
+		int back = v[idx[i]] ^ x;
+		if (mp.count(back)) {
+			pre[i] = max(pre[i], mp[back]);
+		}
+		mp[v[idx[i]]] = i;
+	}
+	for (int i = 1; i <= m; i++) {
+		int q;
+		cin >> q;
+		int l = dfns[q];
+		int r = dfne[q];
+		
+		if (pre[r] < l) {
+			cout << "NO" << endl;
+		} else {
+			cout << "YES" << endl;
+		}
+	}
+}
+
+int main(void) {
+    ios::sync_with_stdio(false);
+    cin.tie(0); cout.tie(0);
+    int t = 1;
+    //cin >> t;
+    
+    while (t--) {
+    	solve();
+	}
+}

exercises/wcg/简单/exercies.md

+# 异或和
+
+小张找到了一个整数 N，他想问问你从 1 到 N 的所有不同整数的异或和是多少， 请你回答他的问题。
+
+## 输入描述
+
+第一行包含一个整数 N (1 <= N <= 100000)。
+
+## 输出描述
+
+第一行输出一个整数， 代表从 1 到 N 的所有不同整数的异或和。
+
+## 输入用例
+
+5
+
+## 输出用例
+
+1
+
+## 提示
+
+1 ^ 2 ^ 3 ^ 4 ^ 5 = 3 ^ 3 ^ 4 ^ 5 = 4 ^ 5 = 1。

exercises/wcg/简单/异或和/exercies.md

+# 异或和
+
+小张找到了一个整数 N，他想问问你从 1 到 N 的所有不同整数的异或和是多少， 请你回答他的问题。
+
+## 输入描述
+
+第一行包含一个整数 N (1 <= N <= 100000)。
+
+## 输出格式
+
+第一行输出一个整数， 代表从 1 到 N 的所有不同整数的异或和。
+
+## 输入用例
+
+5
+
+## 输出用例
+
+1
+
+## 提示
+
+1 ^ 2 ^ 3 ^ 4 ^ 5 = 3 ^ 3 ^ 4 ^ 5 = 4 ^ 5 = 1。

exercises/wcg/简单/异或和/solution.cpp

+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define out(x) cout << #x << '=' << x << endl
+#define out2(x, y) cout << #x << '=' << x << ',' << #y << '=' << y << endl 
+#define no cout << "No" << endl; return
+#define yes cout << "Yes" << endl; return
+#define outvec(a) for (int v : a) { cout << v << ' '; } cout << endl
+#define lowbit(x) (x & -x)
+#define gcd __gcd 
+#define inf 0x3f3f3f3f3f3f3f3fLL
+#define infi 0x3f3f3f3f
+
+using ll = long long;
+using pii = pair<int, int>;
+
+
+void solve() {
+	int n;
+	cin >> n;
+	int s = 0;
+	for (int i = 1; i <= n; i++) {
+		s ^= i;
+	}
+	cout << s << endl;
+}
+
+int main(void) {
+    ios::sync_with_stdio(false);
+    cin.tie(0); cout.tie(0);
+    int t = 1;
+    //cin >> t;
+    
+    while (t--) {
+    	solve();
+	}
+}

exercises/wcg/简单/异或和/test_cases/1.in

+56893

exercises/wcg/简单/异或和/test_cases/1.out

+1

exercises/wcg/简单/异或和/test_cases/10.in

+92830

exercises/wcg/简单/异或和/test_cases/10.out

+92831

exercises/wcg/简单/异或和/test_cases/2.in

+218

exercises/wcg/简单/异或和/test_cases/2.out

+219

exercises/wcg/简单/异或和/test_cases/3.in

+53049

exercises/wcg/简单/异或和/test_cases/3.out

+1

exercises/wcg/简单/异或和/test_cases/4.in

+25769

exercises/wcg/简单/异或和/test_cases/4.out

+1

exercises/wcg/简单/异或和/test_cases/5.in

+2085

exercises/wcg/简单/异或和/test_cases/5.out

+1

exercises/wcg/简单/异或和/test_cases/6.in

+89909

exercises/wcg/简单/异或和/test_cases/6.out

+1

exercises/wcg/简单/异或和/test_cases/7.in

+88297

exercises/wcg/简单/异或和/test_cases/7.out

+1

exercises/wcg/简单/异或和/test_cases/8.in

+17800

exercises/wcg/简单/异或和/test_cases/8.out

+17800

exercises/wcg/简单/异或和/test_cases/9.in

+30815

exercises/wcg/简单/异或和/test_cases/9.out

+0

exercises/wumf/简单/ShuiXianHua.java

-import java.util.Scanner;
-
-/**
- * @description:
- * @author: wmf
- * @create: 2022-10-13 14:21
- **/
-public class TestMain {
-    public static void main(String[] args) {
-
-        Scanner myscanner = new Scanner(System.in);
-        System.out.println("请输入数字");  // 提示用户输入
-        while(true){
-            int number = myscanner.nextInt(); // 接收用户输入
-
-            int number1 = number / 100;  // 取百位
-            int number2 = number % 100 / 10; // 取十位
-            int number3 = number % 10; // 取个位
-
-            if (number1 * number1 * number1 +
-                    number2 * number2 * number2 +
-                    number3 * number3 * number3 == number) {
-                System.out.println("是");
-            } else {
-                System.out.println("不是");
-            }
-        }
-
-    }
-}

exercises/wumf/简单/exercies.md

-# 水仙花数
-
-水仙花数，一个三位数，其各位数字得立方和等于本身，
-例如 153 = 1*1*1 + 5*5*5 + 3*3*3。
-输入一个大于100，小于1000的整数，输出是否是水仙花数。
-
-## 输入描述
-输入一个数字。
-
-## 输出描述
-是否是水仙花数。
-
-## 输入样例
-153
-
-## 输出样例
-是/不是
-
-## 提示
-无

exercises/wumf/简单/test_cases/1.in

-153
\ No newline at end of file

exercises/wumf/简单/test_cases/10.in

-408
\ No newline at end of file

exercises/wumf/简单/test_cases/2.in

-123
\ No newline at end of file

exercises/wumf/简单/test_cases/3.in

-234
\ No newline at end of file

exercises/wumf/简单/test_cases/4.in

-999
\ No newline at end of file

exercises/wumf/简单/test_cases/5.in

-879
\ No newline at end of file

exercises/wumf/简单/test_cases/6.in

-666
\ No newline at end of file

exercises/wumf/简单/test_cases/7.in

-587
\ No newline at end of file