URAL 1493. One Step from Happiness

浏览数：45 / 时间：2015年06月11日

1493. One Step from Happiness

Time limit: 1.0 second
Memory limit: 64 MB

Vova bought a ticket in a tram of the 13th route and counted the sums of the first three and the last three digits of the ticket‘s number (the number has six digits). It turned out that the sums differed by one exactly. "I‘m one step from happiness," Vova thought, "either the previous or the next ticket is lucky." Is he right?

Input

The input contains the number of the ticket. The number consists of six digits, some of which can be zeros. It is guaranteed that Vova counted correctly, i.e., that the sum of the first three digits differs from the sum of the last three digits by one exactly.

Output

Output "Yes" if Vova is right and "No" otherwise.

Samples

input	output
715068	Yes
445219	No
012200	Yes

Notes

All tram tickets have exactly six digits. A ticket is considered lucky if the sum of its first three digits equals the sum of its last three digits.

题意：判断六位数字的的前，后三位数字之和是否相等。

解析：直接暴力判断即可。

AC代码：

#include <cstdio>

int sum(int n){
    int ans = 0;
    while(n){
        ans += n % 10;
        n /= 10;
    }
    return ans;
}

bool check(int n){
    int a = n % 1000, b = n / 1000;
    return sum(a) == sum(b);
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif //sxk

    int n;
    while(scanf("%d", &n)==1){
        puts(check(n) || check(n-1) || check(n+1) ? "Yes" : "No");      //前一个和后一个也可以
    }
    return 0;
}

郑重声明：本站内容如果来自互联网及其他传播媒体，其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享，并不代表本站赞同其观点和对其真实性负责，也不构成任何其他建议。

URAL 1493. One Step from Happiness