#include<bits/stdc++.h> using namespace std; int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; bool isLeapYear(int y){ retu...

#include<bits/stdc++.h> using namespace std; int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; bool isLeapYear(int y){ retu...

#include<bits/stdc++.h> using namespace std; int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; bool isLeapYear(int y){ retu...

#include<bits/stdc++.h> using namespace std; int main(){ int m,d; cin >> m >> d; int a[13] = {0, 31, 28, 31, 30, ...

#include<bits/stdc++.h> using namespace std; void merge(int *b){ int front = b[0]; int flag = 0; // front是否为合并后的结果 for(int i = 1;...

矩阵旋转题目，把握好顺时针旋转90度的逻辑、和纵向对折交换即可 tip：建议采用vector容器，便于数组处理 #include<bits/stdc++.h> using namespace std; // 矩阵顺时针旋转90度 vector<vec...

关键： 1.根据题目要求进行模拟，d用于控制填充方向  2.注意输出格式 #include<bits/stdc++.h> using namespace std; int main(){ int n; cin >> n; ...

 关键点： 1.找到旋转90度的规律 2.如何比较数组相等 tip：采用vector容器代替数组 #include<bits/stdc++.h> using namespace std; // 顺时针旋转90度 vector<v...

#include<bits/stdc++.h> using namespace std; int main(){ int n; while(cin >> n){ int a[10][10]; int b[10][10]; ...

#include<bits/stdc++.h> using namespace std; int main(){ int n; cin >> n; int a[105][105] = {0}; a[1][1] = 1; a[2][1] ...

#include<bits/stdc++.h> using namespace std; int main(){ int num; while(scanf("%d", &num) != EOF){ char out[105]; int ...

#include<bits/stdc++.h> using namespace std; int main(){ int num; while(scanf("%d", &num) != EOF){ char out[105]; int ...

#include<bits/stdc++.h> using namespace std; int main(){ char s[105]; while(scanf("%s",s) != EOF){ int res = 0; int len = ...

#include<bits/stdc++.h> using namespace std; int main(){ for(int i = 1000; i <= (9999/9); i++) { int res = i*9; ...

#include<bits/stdc++.h> using namespace std; int w[] = {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2}; int trans[] = {1,0,'X'-'0',9,8,7,6,5,4,3...