摘要: 1.从1遍历到40; 2.若与40的最小公因数为1则输出 #include<iostream> using namespace std; int g(int m,int n){if(m<m){swap(m,n);}int t=1;while(t){t=m%n;m=n;n=t;}return m;} 阅读全文
posted @ 2023-05-22 19:47 The-rich 阅读(13) 评论(0) 推荐(0)