题解 | 牛牛的计划

#include <stdio.h>
//有点取巧了，不过比较简洁
int main() {
    int y,m,d,y1,m1,d1;
    scanf("%d %d %d\n%d %d %d",&y,&m,&d,&y1,&m1,&d1);   
    int a = y*10000+m*100+d;
    int a1 = y1*10000+m1*100+d1;
    if(a>a1){
        printf("no");
    }
    else{
        printf("yes");
    }
    
    return 0;
}