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#include “pch.h” #include using namespace std; int main() { int s; cin >> s; if(s>3&&s % 2 == 0) cout << “YES”; else cout << “NO”; }

#include “pch.h” #include using namespace std; int main() { char a[3]; while (cin >> a) { char*p = a; for (int i =0 ; i < 2; ++i) { for (int t = i; t < 3; ++t) { if (p[i] > ...

#include “pch.h” #include using namespace std; int main() { int A, B; while(cin >> A >> B) { if ((A + B) % 86 == 0) cout << “yes” << endl; else cout &

#include “pch.h” #include #include using namespace std; int main() { double r, PI=3.1415927,V; cin >> r; V = 4.0 / 3.0PIrrr; cout << setprecision(3)<<fixed<&lt

#include <iostream> using namespace std; int main() { int x = 9999; char a; int n; char *A=NULL; A = new char[x]; int *N = NULL; N = new int[x]; int k; for (k=0;;++k) { cin >&g...

G - 7 POJ - 3414 BFS算法 #include <iostream> #include <queue> using namespace std; int *color; int *pi; int *d; int *operation; //保存操作 int BFS(int A, int B, int C) { if (C == 0) return ...

```#include <iostream> using namespace std; int J(int a); int main() { int m, n; while (cin >> m >> n && m <= n) { for (int i = m,k=0 ; i <= n ; ++i) { ...

#include <iostream> using namespace std; void sort(int**M, int**N, int**p, int t); int main() { int n, m; int**p = new int*; int **M=new int*; int **N = new int*; int k; //记录数据组数 for (...

#include <iostream> using namespace std; int main() //二位指针数组 { int T; //1<=T<=13 cin >> T; // int n, m; //n为验题人数量，m为AC队伍数 (2≤n≤20,0≤m≤500) int**N = new int*[T]; //二维数组n储存T道题...

H - 8 POJ - 3126 BFS 在这里插入#include <iostream> #include <cmath> #include <stdio.h> #include <string> #include <cstring> #include <map> #include <set> #includ

在这里插入代码片 #include <iostream> using namespace std; int j(double a, double b, double c); int main() { int M; cin >> M; double a, b, c; double*p = nullptr; p = new double; for (int i =...

G - 7 HDU - 1213 并查集——套模板-递归思想 #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; int father[1001]; int find(int x) //模板（并查集...

F - 6 HDU - 1016 深搜问题DFS #include <iostream> #include <cstdio> #include <cmath> using nam

G - 7 HDU - 1241 八连块问题——dfs算法 参考// 算法竞赛入门经典（第二版）P163 #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn = 105; char pic[maxn][maxn]; in

A - Overturned Numbers URAL - 2031 思路：先列举1-99所有符合题意的数，翻转后排序，之后判断是否有符合n的连续序列，符合的话翻转（按翻转前从大到小的顺序！！！）输出。 #include <iostream> #include <algorithm> #include <cstring> using namespace std;...

H - 8 UVA - 11389 贪心问题 方法：先对早、晚班车时长排序（一个升序，一个降序，之后对应项相加与d比较，计算出超时总量即可） #include <iostream> #include <cstdio> #include <cstdlib> #include

G - 7 HDU - 1010 //DFS+奇偶剪枝 #include <iostream> #include <cstring> using namespace std; const int maxn = 10; const int maxT = 60; char map[maxn][maxn]; int sx, sy, ex, ey; //起点坐标、终点坐标 in

I - 9 HDU - 1576 #include <iostream> using namespace std; int main() { long long x, B, n,T; cin >> T; int *q = nullptr; q = new int[T]; for (int i = 0; i < T; ++i) { cin &am