小树学信奥

T1#include <bits/stdc++.h>using namespace std;bool isPerfectSquare(int n){ int root = sqrt(n); return root * root == n;}int main(){ int n; cin >> n; if(isPerfectSquare(n)) { cout << n; ret

题目https://iai.sh.cn/contest/indexT1#include <iostream>using namespace std;int main(){ int n, cnt = 0; while(cin >> n) { cnt++; } if(3 == cnt) { if(6 == n) { cout << 1000

题目https://iai.sh.cn/contest/7T1#include <iostream>using namespace std;int main(){ int a, x, y; cin >> a >> x >> y; cout << y + x << ' ' << a * (y + x) - x; return 0;}T2分析最后四个测试点，数据很大。不能

解法一一、分析采用二维数组的前缀和来算。二、时间复杂度O(n^4)三、代码#include <iostream>using namespace std;const int maxN = 800 + 5;int n, m, k, sum[maxN][maxN];char a[maxN][maxN];void input(){ cin >> n >> m >> k; for(int r=1; r<=n; r++)

T1#include <iostream>using namespace std;int main(){ int n; cin >> n; int a = 1;//初始化，0刀是一块 for(int i = 1; i <= n; i++) { //切第i刀时会多i块 a += i; } cout << a; return 0;}T2#include

T1#include <iostream>using namespace std;int main(){ int day = 3, remain = 10; int n; cin >> n; while(remain != n) { day++; remain --; if(day % 7 == 1 || day % 7 == 2) { rema