开始的时候一直执着于<abbr title="&#37;4&#61;&#61;0">这样</abbr>

#include<stdio.h>
int main(){
    int weight;
    scanf("%d",&weight);
    if(weight%4==0){
        printf("YES, you can divide the watermelon into two even parts.");
    }else{
        printf("NO, you can't divide the watermelon into two even parts.");
    }
}

最后发现,原题是不要两边对半分的,没有考虑到。所以其实除了2,只需要 %2 就足够啦。

#include<stdio.h>
int main(){
    int weight;
    scanf("%d",&weight);
    if (weight%2==0&&weight!=2){
        printf("YES, you can divide the watermelon into two even parts.");
    }else{
        printf("NO, you can't divide the watermelon into two even parts.");
    }
}