题解 | #小美的因子查询#
小美的因子查询
https://www.nowcoder.com/practice/1870e68256794c6aa727c8bb71fd9737
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;
}
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