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

A very hard mathematic problemTime 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

95360

Sample Output

110　　Hint9 = 1^2 + 2^2 + 1 * 2 * 253 = 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