hdu4282 A very hard mathematic problem

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 991    Accepted Submission(s): 287

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

 

题意：给定 K ，求满足 X^Z + Y^Z + XYZ = K 这个等式的解。

分析：f(x)=x^b + a*b*x + a^b - k;枚举a 和b，二分找  x （x的范围[a+1,pow(k,1/b)]）;

#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

int k,maxa;

inline int f(int a,int b,int x)
{
    return pow(x*1.0,b*1.0)+pow(a*1.0,b*1.0)+a*b*x-k;
}

bool judge(int a,int b)
{
    int L,R,M,t;
    L=a+1;R=maxa;
    if(f(a,b,L)>0||f(a,b,R)<0) return false;
    while(L<=R)
    {
        M=(L+R)>>1;
        t=f(a,b,M);
        if(t==0) return true;
        else if(t>0) R=M-1;
        else L=M+1;
    }
    return false;
}

int main()
{
    int ans,a,b;
    while(scanf("%d",&k)==1&&k)
    {
        ans=0;
        for(b=2;b<=30;b++)
        {
            maxa=pow(k*1.0,1.0/b);
            for(a=1;a<=maxa;a++)
            {
                if(judge(a,b))
                    ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}