HDU 5373 The shortest problem (递归调用，详解)

题目地址

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

The shortest problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 935    Accepted Submission(s): 479

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 
   
   
   
   

 

思路：

判断这个数能否被11整除，当数十分大的时候会爆掉，简单的方法就是奇数位数的总和减去偶数位数的总和能否被11整除。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<math.h>
#include<stdlib.h>
using namespace std;
int tot;
__int64 ji,ou,sum;
void solve(int x){
    if(x>=10)    solve(x/10);
    if(tot%2==0){
        ou+=x%10;
    }
    else ji+=x%10;
    sum+=x%10;
    tot++;
    return;
}
int main()
{
    int ncase=0,n,t;
    while(cin>>n>>t)
    {
        ncase++;
        if(n==-1&&t==-1)
            break;
        ji=0,ou=0;
        sum=0,tot=0;
        solve(n);
       // printf("%d %d %d\n",ji,ou,sum);
        for(int i=0;i<t;i++){
            solve(sum);
        }
        printf("Case #%d: ",ncase);
      //  printf("%d %d %d\n",ji,ou,sum);
        if(abs(ji-ou)%11==0)
            printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

代码用C++交就1400ms+险过，用G++交700ms+就过了