The shortest problem

The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 969 Accepted Submission(s): 491

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

比赛的时候别人直接暴力就可以过,我们为什么过不了QAQ

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;

const int INF =0x3f3f3f3f;

int k;

int a[52];

int aa(int m)
{
    int sum=0;
    k=0;
    while(m)
    {
        sum+=m%10;
        a[k++]=m%10;
        m/=10;

    }
    return sum;
}

int main()
{
    int n,m;
    int sum;
    int w=1;
    int ans;
    while(scanf("%d %d",&n,&m))
    {
        if(n==-1&&m==-1)
        {
            break;
        }
        if(n==0)
        {
            printf("Case #%d: Yes\n",w++);
            continue;
        }
        sum=aa(n);
        ans=0;
        int v=1;
        while(k--)
        {
            if(v)
            {
                ans+=a[k];
                v=0;
            }
            else
            {
                ans-=a[k];
                v=1;
            }
        }
        while(m--)
        {
            sum+=aa(sum);
            while(k--)
            {
                if(v)
                {
                    ans+=a[k];
                    v=0;
                }
                else
                {
                    ans-=a[k];
                    v=1;
                }

            }
        }
        if(ans<0)
        {
            ans=-ans;
        }
        if(ans%11)
        {
            printf("Case #%d: No\n",w++);
        }
        else
        {
            printf("Case #%d: Yes\n",w++);
        }
    }
    return 0;
}