poj1840 Eqs

使用hash實現，原式：a1*x1^3+a2*x2^3+a3*x3^3+a4*x4^3+a5*x5^3=0

可以推出：a1*x1^3+a2*x2^3 = -(a3*x3^3+a4*x4^3+a5*x5^3) 把O(n^5)轉化爲O(n^3)+O(n^2)了

#include<iostream>
using namespace std;
short h[25000001];
int main()
{
	int a1,a2,a3,a4,a5;
	while(cin>>a1>>a2>>a3>>a4>>a5)
	{
		memset(h,0,sizeof(h));

		for(int x1=-50;x1<=50;x1++)
		{
			if(!x1)
				continue;
			
			for(int x2=-50;x2<=50;x2++)
			{
				if(!x2)
					continue;
				int sum=(a1*x1*x1*x1 + a2*x2*x2*x2)*(-1);
				if(sum<0)
					sum+=25000000;
				h[sum]++;
			}
		}

		int ans=0;

		for(int x3=-50;x3<=50;x3++)
		{
			if(!x3)
				continue;
			for(int x4=-50;x4<=50;x4++)
			{
				if(!x4)
					continue;
				for(int x5=-50;x5<=50;x5++)
				{
					if(!x5)
						continue;
					int sum=a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;
					if(sum<0)
						sum+=25000000;
					if(h[sum])
						ans+=h[sum];
				}
			}
		}
		cout<<ans<<endl;
	}				
	return 0;
}