hdu4282 A very hard mathematic problem-----天津网络赛

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 913    Accepted Submission(s): 268

Problem Description

　　Haoren is very good at solving mathematic problems. Today he is working a problem like this:
　　Find three positive integers X, Y and Z (X < Y, Z > 1) that holds
　　 X^Z + Y^Z + XYZ = K
　　where K is another given integer.
　　Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.
　　Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?
　　Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.
　　Now, it’s your turn.

Input

　　There are multiple test cases.
　　For each case, there is only one integer K (0 < K < 2^31) in a line.
　　K = 0 implies the end of input.
　　

Output

　　Output the total number of solutions in a line for each test case.

Sample Input

   
   
   
   

    
    
    
    9
53
6
0
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    1
1
0
　　

    
    
    
    
 
     
 
      Hint 
     
 9 = 1^2 + 2^2 + 1 * 2 * 2 53 = 2^3 + 3^3 + 2 * 3 * 3 
    
    
    
    
     
   
   
   
   

Source

2012 ACM/ICPC Asia Regional Tianjin Online

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liuyiding

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
using namespace std;
#define ll __int64
ll power(ll x,int y)
{
    ll temp=x;
    for(int i=2;i<=y;i++)
    temp*=x;
    return temp;
}
int main()
{
    ll n;
    while(scanf("%I64d",&n)&&n)
    {
        int ans=0;
        int temp=sqrt(n);
        if(temp*temp==n) ans+=(temp-1)/2;
        for(int i=3;i<31;i++)//z
        {
         
          for(ll j=1;;j++)
          {
              ll y=power(j,i);
              if(y>=n/2) break;
              for(ll k=j+1;;k++)//先枚举小的
              {
                  ll x=power(k,i);
                  ll uu=x+y+i*j*k;
                  if(uu>n) break;
                  else if(uu==n){ans++;break;}
              }
          }
        }
        cout<<ans<<endl;

    }
}