题解 | #小美的因子查询#

新手求拷打
#include <stdio.h>
int main() {
    int n,j;
    scanf("%d",&n);
    int a[n];
    for (int i=0;i<n;i++){
        scanf("%d",&j);
        if(j%2==0){a[i]=1;}
        else{a[i]=0;}
    }
    for(int i=0;i<n;i++){
        if(a[i]){printf("YES\n");}
        else{printf("NO\n");}
    }

    return 0;
}