hdu 4282A very hard mathematic problem(枚举+二分)

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3578    Accepted Submission(s): 1052

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 
    
    
    
    
     
   
   
   
   

 

题目大意很简单，解题思路也比较容易想的，先枚举z再枚举x然后二分y即可，时间用了200ms-。自己写的时候竟然因为用了库函数自带的pow函数然后有精度的损失，记得以前碰过这样的。一直以为是自己二分或者是范围枚举出错了。。。

题目地址：A very hard mathematic problem

AC代码：

#include<iostream>
#include<cstring>
#include<cmath>
#include<string>
using namespace std;

long long pow1(long long a,long long p)
{
    long long s=1;
    while(p)
    {
        if(p&1)
            s*=a;
        a*=a;
        p>>=1;
    }
    return s;
}

int main()
{
    long long k;
    long long x,y,z;
    long long res,flag;
    while(cin>>k&&k)
    {
        res=0;
        for(z=2;z<=30;z++)  //枚举z
        {
            long long tmp=pow(k/2.0,1.0/z);
            long long tmp1=pow(k*1.0,1.0/z);
            for(x=1;x<=tmp;x++)  //枚举x
            {
                flag=0;
                long long l,r;
                l=x+1,r=tmp1;
                while(l<=r)   //二分y
                {
                    y=(l+r)>>1;
                    long long s=pow1(x,z)+pow1(y,z)+x*y*z;
                    if(s>k) r=y-1;
                    else if(s<k) l=y+1;
                    else
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag) {res++;}
            }
        }

        cout<<res<<endl;
    }
    return 0;
}

/*
625
*/

//187MS