摘要:
点击查看代码 #include<bits/stdc++.h> using namespace std; int m; int main() { scanf("%d",&m); for(int k1=sqrt(2*m);k1>1;k1--){ if(2*m%k1==0&&(k1+2*m/k1)%2){ 阅读全文
posted @ 2025-11-21 10:31
AnoSky
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点击查看代码 #include <iostream> using namespace std; const int N = 1005; int n, pre[N][N], ans; int main() { cin >> n; for (int i = 1; i <= n; i++) { for ( 阅读全文