POJ 1840：Eqs 哈希求解五元方程

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 14169		Accepted: 6972

Description

Consider equations having the following form: 

a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 

The coefficients are given integers from the interval [-50,50]. 

It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 



Determine how many solutions satisfy the given equation. 

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

654

给定方程是a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 ，给了a1，a2，a3，a4，a5。求解 解的数量。x都在-50到50之间。

暴搜肯定GG，就得把方程变形将a1x1³+ a2x2³
 移到方程左边，作为结果，然后暴搜x3，x4，x5。时间复杂度就降下来了。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

short hash1[25000000];

int main()
{
	int a1, a2, a3, a4, a5;
	int x1, x2, x3, x4, x5;
	int temp;
	long long sum;

	scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);

	memset(hash1, 0, sizeof(hash1));
	for (x1 = -50; x1 <= 50; x1++)
	{
		if (x1 == 0)continue;
		for (x2 = -50; x2 <= 50; x2++)
		{
			if (x2 == 0)continue;

			temp = -1*(a1*x1*x1*x1 + a2*x2*x2*x2);

			if (temp < 0)
			{
				temp += 25000000;
			}
			hash1[temp]++;
		}
	}
	sum = 0;
	for (x3 = -50; x3 <= 50; x3++)
	{
		if (x3 == 0)continue;
		for (x4 = -50; x4 <= 50; x4++)
		{
			if (x4 == 0)continue;
			for (x5 = -50; x5 <= 50; x5++)
			{
				if (x5 == 0)continue;

				temp = a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;

				if (temp < 0)
				{
					temp += 25000000;
				}
				if (temp >= 0 && temp < 25000000)
				{
					sum += hash1[temp];
				}
			}
		}
	}
	cout << sum << endl;
	return 0;
}

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