hdu 5373 The shortest problem 2015多校联合训练赛#7 模拟

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 532    Accepted Submission(s): 271

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 
   
   
   
   

 

Source

2015 Multi-University Training Contest 7

 

对于s，算出s所有位上的数字之和f ，把f加入s后面，求t次操作后的数字能否被11整除。

模拟：s1记录得到数字对11的模数，s记录所有数字之和。

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

int main(){
    int n,t,tt=1;

    while(scanf("%d%d",&n,&t),n!=-1){
        int s = 0,s1=n%11;
        while(n){
            s += n%10;
            n /= 10;
        }
        for(int i = 0;i < t; i++){
            int f = s,f1=s;
            while(f){
                s += f%10;
                s1 = s1*10%11;
                f/=10;
            }
            s1 = (s1 + f1) % 11;
        }
        printf("Case #%d: ",tt++);
        if(s1 == 0) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}