2015多校联合第六场 hdu5373 The shortest problem 水

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我是靠谱客的博主要减肥菠萝，最近开发中收集的这篇文章主要介绍2015多校联合第六场 hdu5373 The shortest problem 水，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

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=%dn",n,t);
        if(m==0)
        {

            if(n%11==0) printf("Case #%d: Yesn",cas++);
            else printf("Case #%d: Non",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: Yesn",cas++);
        else  printf("Case #%d: Non",cas++);

    }
    return 0;
}