C - A very hard mathematic problem

      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

    思路：充分利用题目中的条件，0<x<y  说明y最小的是2，所以进一步可以知道z最大的也只能是31，然后进行枚举z,枚举x,因

               为要用到二分法，所以要将y 的最大值确定下来，也就是的=( int) pow( double (n), 1.0/( double) z);      注意：因为,为

                了避免超时，在这里我采用了快速幂的算法，还有为了避免某些数的溢出我用了long long ;

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
long long fun(long long a,long long b){
    long long  sum= 1,z=a;
    while(b)
    {
        if(b% 2) sum*=z;
        z=z*z;
        b/= 2;
    }
    return sum;
}
///X^Z + Y^Z + XYZ = K
int main(){
    int n;
    while( scanf( "%d",&n)== 1)
    {
        if(n== 0)
            break;
        int Z= 0,sum= 0;
        long long y;
        for( y= 2; y<=n; y=y* 2)
            Z++;
        int k= sqrt(n);
        if(k*k==n) //z等于2 的情况
            sum=(k -1)/ 2;
        for( int z= 3; z<=Z; z++)
        {
            for( int x= 1; ; x++)
            {
                long long  al=fun(x,z);
                if(al>n/ 2) break; //y总是大于x,所以x^z总是小于n/2;
                int q=x+ 1, w=( int) pow( double (n), 1.0/( double) z);

                while(q<=w)
                {
                    y=(q+w)/ 2;
                    if(al+fun(y,z)+z*x*y>n)
                        w=y -1;
                    else if(al+fun(y,z)+z*x*y<n)
                        q=y+ 1;
                    else
                    {
                        sum++;
                        break;
                    }
                }
            }
        }
        printf( "%d\n",sum);
    }
    return 0;
}