HDOJ 5373 The shortest problem 水

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 374    Accepted Submission(s): 183

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

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long int LL;

int n,t;

int cnt(LL x)
{
	int ret=0;
	while(x) { ret+=x%10; x/=10; }
	return ret;
}

int getE(LL x)
{
	int e=0;
	while(x) { e++; x/=10; }
	return e;
}

int main()
{
	int cas=1;
	while(cin>>n>>t)
	{
		if(n==-1&&t==-1) break;

		LL sig=n%11;
		LL wei=cnt(n);

		for(int i=1;i<=t;i++)
		{
			int k=getE(wei);
			LL e=1;
			for(int j=0;j<k;j++) e=e*10;
			sig=(sig*e%11+wei)%11;

			wei+=cnt(wei);
		}

		sig%=11;
		printf("Case #%d: ",cas++);
		if(sig==0) puts("Yes");
		else puts("No");
	}
	return 0;
}