E. The shortest problem

http://www.bnuoj.com/v3/contest_show.php?cid=7612#info

http://www.bnuoj.com/v3/contest_show.php?cid=7612#problem/

题解：http://bestcoder.hdu.edu.cn/blog/2015-multi-university-training-contest-7-solutions-by-uestc/

水题：

#include <iostream>
#include<stdio.h>
using namespace std;
int sum,x;
void get_sum(int n)
{
    while(n)
    {
        sum += n%10;
        n /= 10;
    }
}
void get_x()
{
    int times = 1;
    int s = sum;
    while(s)
    {
        s /= 10;
        times *= 10;
    }
    x = (x * times)+sum;
    x = x%11;
}
int main()
{
    int n,t;
    int cas = 1;
    while(cin>>n>>t && (n!=-1||t!=-1))
    {
        sum = 0;
        x = 0;
        x = n%11;
        get_sum(n);
        for(int i = 1;i <= t;i ++){
                get_x();
                get_sum(sum);
        }
        if(x == 0)
            printf("Case #%d: Yes
",cas++);
        else if(x)
            printf("Case #%d: No
",cas++);
    }
    return 0;
}