哈希表

Eqs

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Consider equations having the following form: 
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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

首先将等式做一个简单的变换：  -(a1*x1^3 + a2*x2^3) = a3*x3^3+a4*x4^3+a5*x5^3      ，然后运用哈希表！

#include <iostream>
#include <cstdlib>
using namespace std;

#define sum 25000001
short a[sum];

int main()
{
	int a1, a2, a3, a4, a5;
	cin >> a1>> a2>> a3>> a4>> a5;
	memset(a, 0, sum);
	for (int x1 = -50; x1<=50; x1++)
	{
		if ( !x1 )
			continue;
		for (int x2 = -50; x2<=50; x2++)
		{
			if ( !x2 )
				continue;
			int sum1 = (a1*x1*x1*x1 + a2*x2*x2*x2) *(-1);
			if (sum1 < 0)
				sum1 = sum1 + 25000000;
			a[sum1] ++ ;
		}
	}
	int num = 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 sum2 = a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;
				if (sum2 < 0)
					sum2 = sum2 + 25000000;
				if (a[sum2])
					num = num + a[sum2];
			}
		}
	}
	cout << num << endl;
	return 0;
}