多校The shortest problem

 The shortest problem
Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
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 

题意：123->1236->123612->12361215，6是1 2 3的和，12是1 2 3 6的和，15是1 2 3 6 1 2的和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
#define ll long long
ll n,t,sum,ans;
//int num;

int main()
{
    int mark=0;
    int i;

    while(~scanf("%lld%lld",&n,&t))
    {
        sum=0;
        int sums;
        mark++;
        if(n==-1&&t==-1)
            break;

int nn=n;
            while(nn>0)
            {
                sum=sum+nn%10;
                nn=nn/10;
            }

        for(i=1; i<=t; i++)
        {

            int m=sum,t=0,sums=0;
            while(m>0)
            {
                sums=sums+m%10;
                m= m/10;
                t++;
            }
            int ans=n%11;
            n=ans*pow(10,t)+sum;
            sum+=sums;

        }
       // cout<<n<<endl;
        //printf("%lld
",n);
        if(n%11==0)
            printf("Case #%d: Yes
",mark);
        else
            printf("Case #%d: No
",mark);
    }
}