开始的时候一直执着于<abbr title="%4==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.");
}
} 
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