The shortest problem（hdu5373+11的倍数）

The shortest problem

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

Total Submission(s): 503    Accepted Submission(s): 257

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

##1005. The shortest problem

本场最水的题，被11整除的性质是奇偶位的和之差能被11整除，然后模拟一下就好了。 然后如果用longlong的话可能会T（写的好就不会），用int就好了。（经过测试用int时间为longlong的1/4）

转载请注明出处：寻找&星空の孩子 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5373

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
#define ll long long
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

int n,t;
int x,y,k;

int main ()
{
    int a,b,c,d,e,ii=1;
    while (scanf ("%d%d",&n,&t)==2)
    {
        if (n==-1&&t==-1)
            break;
        a = n/10000;
        b = (n/1000)%10;
        c = (n/100)%10;
        d = (n/10)%10;
        e = n%10;
        //if (d!=0){k++; if(c!=0)k++; if(b!=0)k++; if(a!=0)k++;}
        y = d+b;
        x = c+a+e;

        while (t--)
        {
            k = 0;
            int p=0,q=0,m=x+y;
            while (m)
            {
                k++;
                if (k%2)
                    p += m%10;
                else
                    q += m%10;
                m /= 10;
            }
            //cout<<p<<" "<<q<<endl;cout<<x<<" "<<y<<endl;
            if (k%2)
            {
                x += q;
                y += p;
                swap(x, y);
            }
            else
            {
                x += p;
                y += q;
            }
        }
        if ((x-y)%11)
            printf ("Case #%d: No\n",ii++);
        else
            printf ("Case #%d: Yes\n",ii++);
    }
    return 0;
}