POJ1840(哈希)

大意：

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

求方程解的个数。

分析；将原式变为

a1x1³+ a2x2³+ a3x3³=- a4x4³- a5x5³

对左式枚举哈希处理，再计算右式查询。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 1030307
int hashn[maxn], nextn[maxn];
int num[maxn];
using namespace std;
int main()
{
	int a, b, c, d, e;
	while (~scanf("%d%d%d%d%d", &a, &b, &c, &d, &e))
	{
		memset(hashn, -1, sizeof(hashn));
		int mm = 0;
		for (int i = -50; i <= 50; i++)
			if (i != 0)
				for (int j = -50; j <= 50; j++)
					if (j != 0)
						for (int k = -50; k <= 50; k++)
							if (k != 0)
							{
								int t = a*i*i*i + b*j*j*j + c*k*k*k;
								num[mm] = t;
								int key = t%maxn;
								key = (key + maxn) % maxn;
								nextn[mm] = hashn[key];
								hashn[key] = mm;
								mm++;
							}
		int sum = 0;
		for (int i = -50; i <= 50; i++)
			if (i != 0)
				for (int j = -50; j <= 50; j++)
					if (j != 0)
					{
						int t = -e*i*i*i - d*j*j*j;
						int key = t%maxn;
						key = (key + maxn) % maxn;
						int m = hashn[key];
						while (m != -1)
						{
							if (t == num[m])
								sum++;
							m = nextn[m];
						}
					}
		printf("%d
", sum);
	}
	return 0;
}