杭电acm 4282 A very hard mathematic problem

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

解题思路：我相信大家刚看到本题时，都会想到暴力，但又看了看会觉得暴力超时呀！的确，本题直接用暴力会超时，但是，我们不妨换一种思路，当z=2时，x^z + y^z +x*y*z=(x+y)^2=k,即x+y=sqrt(k),那么只需要在1~2^16范围内来判断，并且这里拥有一个小规律，组合数ant = (sqrt(k) - 1)/2(这个仅在sqrt（k）是整数时成立，非整数时无解);

那么接下来只需要判断z>=3&&z<=31时的情况了，根据化简此时x<y<=1124(这里只要大于1024即可)，于是直接暴力，但是别忘了剪枝，否则依旧会超时。

代码如下：

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

using namespace std;

long long Pow(int x,int n)//本题需要自己编写pow函数，调用系统函数会超时。
{
long long ans = 1;
for(int i=1;i<=n;i++)
ans *= x;
return ans;
}

int main()
{
long long k;
int x,y,z;
while(scanf("%lld",&k)!=EOF && k)
{
long long ant = 0;
int a = (int)sqrt(k);
if(a*a*1.0 == k)//进行z=2的判断
ant+=(a-1)/2;
for(z=3;z<=31;z++)//对z>=3的情况进行暴力求解。
{
for(x=1;x<=1100;x++)
{
if(Pow(x,z) >= k)       break;//剪枝
for(y=x+1;y<=1100;y++)
{
long long ans = Pow(x,z)+Pow(y,z) + x*y*z;
if(ans == k)
{
ant++;
break;
}
else if(ans > k || ans < 0)   break;//剪枝
}
}
}
printf("%lld\n",ant);
}
return 0;
}