题解 | #小美的因子查询#
小美的因子查询
https://www.nowcoder.com/practice/1870e68256794c6aa727c8bb71fd9737
#include <stdio.h> int main() { int t,x; scanf("%d",&t); for(int i=0;i<t;i++){ scanf("%d",&x); if (x%2==0) { printf("YES\n"); }else printf("NO\n"); } return 0; }