HDU 4282 A very hard mathematic problem（暴力优化）

原题：http://acm.hdu.edu.cn/showproblem.php?pid=4282

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3871    Accepted Submission(s): 1133

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

题意：求出符合上面式子的个数

AC代码：

#include 
#include 
#define LL __int64
/*
author:YangSir
time:2014/5/9
*/
LL qpow(LL x,LL y) 
{ 
    LL temp=x,i; 
    for(i=2;i<=y;i++) 
		temp*=x; 
    return temp; 
} 

int main()
{
	LL n,m,x,y,z,k,s,num;
	while(~scanf("%I64d",&k)&&k)
	{
		num=0;
		s=sqrt(k);
		if(s*s==k)
		{
			num+=(s-1)/2;//当z为2时，式子符合完全平方，只要求组成s的x、y的个数（x=k/2)//因为想xk)//大于显然不符合
						break;
					if(qpow(x,z)+qpow(y,z)+x*y*z==k)
					{
						num++;
						break;//只有一次
					}
				}
			}
		}
		printf("%I64d\n",num);
	}
	return 0;
}