A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 964    Accepted Submission(s): 279

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  

Recommend liuyiding         暴力枚举z就可以了。z肯定比较小的。 当z等于2时就是平方和公式。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;

long long pow(long long a,int n)
{
    long long ret=1;
    long long temp=a;
    while(n)
    {
        if(n&1)ret*=temp;
        temp*=temp;
        n>>=1;
    }
    return ret;
}
int main()
{
    long long K;
    long long ans;
    while(scanf("%I64d",&K),K)
    {
        ans=0;
        long long temp=(long long)sqrt(K);
        if(temp*temp==K)ans+=(temp-1)/2;
        for(int z=3;z<31;z++)
        {
            for(long long x=1;;x++)
            {
                long long u=pow(x,z);
                if(u*2>=K)break;
                for(long long y=x+1;;y++)
                {
                    long long v=pow(y,z);
                    if(u+v+x*y*z>K)break;
                    if(u+v+x*y*z==K)ans++;
                }
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}