Problem Description
In this problem, we should solve an interesting game. At first, we have an integer n, then we begin to make some funny change. We sum up every digit of the n, then insert it to the tail of the number n, then let the new number be the interesting number n. repeat it for t times. When n=123 and t=3 then we can get 123->1236->123612->12361215.
 

Input
Multiple input.
We have two integer n (0<=n<= 104 ) , t(0<=t<= 105) in each row.
When n==-1 and t==-1 mean the end of input.
 

Output
For each input , if the final number are divisible by 11, output “Yes”, else output ”No”. without quote.
 

Sample Input
35 2 35 1 -1 -1
 

Sample Output
Case #1: Yes Case #2: No
 


第一次做 把每一位存到数组中 逐次往前挪 华丽丽的超时了 看标程 发现可以在源数字的基础上只操作数字 好了 用log10()又超时了orz 

得 照着标称做吧 改了一上午加一晚上一直WA 把变量名都换过去了 还不对 开头特判0写成1 一直没看出来！！！长点心吧

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
  //  freopen("cin.txt","r",stdin);
    int n,m,k,l,cas=1;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1) break;
       // printf("n=%d t=%d\n",n,t);
        if(m==0)
        {

            if(n%11==0) printf("Case #%d: Yes\n",cas++);
            else printf("Case #%d: No\n",cas++);
            continue;
        }
        k=n;
        l=0;
        while(k>0)
        {
            l+=k%10;
            k/=10;
        }
        while(m--)
        {
            k=l;
            while(k>0)
            {
                k/=10;
                n*=10;

            }
            n=(n+l)%11;
            k=l;
            while(k>0)
            {
                l+=(k%10);
                k/=10;
            }
        }

      //  printf("%d   ",n);
         if(n==0) printf("Case #%d: Yes\n",cas++);
        else  printf("Case #%d: No\n",cas++);

    }
    return 0;
}