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\n",cas++);
else if(x)
printf("Case #%d: No\n",cas++);
}
return 0;
}