The shortest problem（hdu，多校

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

 

 #include <iostream>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 //int main(int argc, char** argv) 

 #define N 100005

 int main()
 {
     int n, m, d, a, sum, r, nu, t = ;

     while(scanf("%d%d", &n, &m), n != - || m != -)
     {
         r = n % ;
         sum = ;

         while(n)   // 不知道为什么非得这么写才对~
         {
             sum += n % ;
             n /= ;
         }

         while(m--)
         {
             a = ;
             nu = n = sum;  // nu，存当前次数之前出现的根的总和
             sum = ;

             while(n)
             {
                 sum += n % ;  // sum存目前的根总和的根
                 n /= ;
                 a *= ;
             }

             r = (r * a + nu) % ;
             sum += nu;
         }

         if(r == )
             printf("Case #%d: Yes\n", t++);
         else
             printf("Case #%d: No\n", t++);
     }
     return ;
 }