杭电 HDU 4282 A very hard mathematic problem 2012 ACM/ICPC 天津网赛

            A very hard mathematic problem

         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
           Total Submission(s): 2268    Accepted Submission(s): 683

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 
    
    
    
    
    

        
    
    
    
    
    

        
    
    
    
    
    
 
     
#include<iostream>
#include<cmath>
using namespace std;

int main()
{
	int x,y,z,k,ans;
	while(cin>>k,k)
	{
		ans=0;
		for(z=2;z<31;z++)
		{
			y=(int)(pow((double)k,1.0/z));//求解k开n次方    
			for(x=1;x<y;)
			{
				__int64 temp=x*y*z+(__int64)(pow((double)x,(double)z)+pow((double)y,(double)z)+0.5);
				if(temp>k)
					y--;	
				else if(temp<k)
					x++;
				else
				{x++;y--;ans++;}
			}	
		}
		printf("%d\n",ans);	
	}
	return 520;	
}