A. Serval and Mocha‘s Array codeforces 1789A

Problem - A - Codeforces

题目大意：有一长为n的数组，可以随意改变数组中元素的顺序，问能否使任意的前缀数组都满足所有数的gcd<=该数组长度

2<=n<=5e4

思路：求多个数的gcd时，随着数字增多，gcd不会变大，所以我们只要满足前两个数的gcd<=2即可，那么我们枚举每两个数，只要找到两个数gcd<=2，答案就是YES

//#include<__msvc_all_public_headers.hpp>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int gcd(int a, int b)
{//求两个数的最大公因数return b ? gcd(b, a % b) : a;
}
const int N = 105;
int a[N];
int main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);int t;cin >> t;while (t--){int n;cin >> n;int now;int gc;for (int i = 1; i <= n; i++){cin >> a[i];}bool flag = 0;for (int i = 1; i <= n; i++){for (int j = i + 1; j <= n; j++){if (gcd(a[i], a[j]) <= 2){flag = 1;break;}}}if (flag){cout << "YES" << endl;}else{cout << "NO" << endl;}}return 0;
}

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