A very hard mathematic problem

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

确定要y，z对x进行二分

 

#include<iostream>
#include<stdio.h>
using namespace std;
#define ll __int64
ll lmx(ll z,ll y,ll k)
{
    ll i,sum=0,l,r,xx=1,mid;
    l=1;r=y-1;
    while(l<=r)
    {
        mid=(l+r)/2;
        xx=1;
        for(i=0;i<z;i++)
        {
            xx*=mid;
        }
        sum=xx+mid*z*y;
        if(sum==k) return 1;
        if(sum<k) l=mid+1;
        else r=mid-1;
    }
    return 0;
}
int main()
{
    ll n,i,y,z,head,rear,mid,cnt;
    while(scanf("%I64d",&n)&&n)
    {
        cnt=0;
        for(y=2;y*y<=n;y++)
        {
            for(z=2;;z++)
            {
                ll yy=1;
                for(i=0;i<z;i++) yy*=y;
                if(yy>n) break;
                if(lmx(z,y,n-yy))  cnt++;
            }
        }
        printf("%I64d\n",cnt);
    }
    return 0;
}