多校The shortest problem

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原文链接：http://www.cnblogs.com/dshn/p/4750547.html

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#java

本文介绍了一种基于数字的游戏算法，玩家需要通过累加并追加数字的方式形成新的数字，并判断最终数字是否能被11整除。文章提供了完整的C++实现代码。

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",n);
        if(n%11==0)
            printf("Case #%d: Yes\n",mark);
        else
            printf("Case #%d: No\n",mark);
    }
}