HDU - 4282 A very hard mathematic problem

问题描述：

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.

输入说明：

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.

输出说明：

Output the total number of solutions in a line for each test case.

SAMPLE INPUT：

9
53
6
0

SAMPLEOUTPUT：

1
1
0

Hints：

9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3

思路：

题目的要求是让我们解一个方程，并求出一共有多少组解，输出解的数量。
乍一看很难想到怎么用二分去做，仔细想一下，我们可以发现一共有3个未知数，显然，如果对他们都进行二分或者三分是不现实的，所以我们选择枚举x和z，二分搜索合适的y值。此外还有一些需要注意的点，一是z 的取值范围，z是作为上标出现的，其最大值是有限制的（小于32）二是在乘方运算时需要使用快速幂，否则无论时循环求解还是pow求解，时间复杂度都比较高，会tle，因此要使用快速幂的方法。

AC代码：

#include <bits/stdc++.h>
using namespace std;
long long pow(long long a,long long n)
{long long  sum=1;long long  temp=a;while(n){if(n&1)sum*=temp;temp*=temp;n>>=1;}return sum;
}
int main()
{long long k;long long ans;while(scanf("%d",&k)&&k){ans=0;long long temp=(long long)sqrt(k);if(temp*temp==k)ans+=(temp-1)/2;for(long long z=3;z<31;z++){for(long long x=1;;x++){long long u=pow(x,z);if(u*2>=k)break;for(long long y=x+1;;y++){long long v=pow(y,z);if(u+v+x*y*z>k)break;if(u+v+x*y*z==k)ans++;}}}cout<<ans<<endl;}return 0;
}