poj-1840 Eqs（hash）

Eqs

Time Limit:5000MS Memory Limit:65536KB
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

题意：输入5个数(a1,a2,a3,a4,a5)∈[-50,50]，求满足a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 的多项式的个数(x1,x2,x3,x4,x5)∈[-50,50]。

由于范围太大不能用暴力方法，可以转换成a1x13+ a2x23=-a3x33+ a4x43+ a5x53，先求出所有a1x13+ a2x23不同值的个数存入表中，（数组无法表示负值，可开范围最小50*50*50*50*3*2=37500000）。然后求s=-a3x33+
a4x43+ a5x53的所有值，如果之前该值出现过（即f[s]!=0） sum+=f[s];

#include<stdio.h>
#include<string.h>
#define size 18750000
short f[37500001];
//char f[37500001];
//short类型 56260K	297MS
//char类型 29868K	329MS
int main()
{
	int i,j,k,a1,a2,a3,a4,a5;
	int sum=0;
	scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
	memset(f,0,sizeof(0));
	for(i=-50;i<=50;i++)
	{
		if(i==0) continue;
		for(j=-50;j<=50;j++)
		{
			if(j==0) continue;
			f[a1*i*i*i+a2*j*j*j+size]++;
		}
	}
	for(i=-50;i<=50;i++)
	{
		if(i==0) continue;
		for(j=-50;j<=50;j++)
		{
			if(j==0) continue;
			for(k=-50;k<=50;k++)
			{
				if(k==0) continue;
				int s=-(a3*i*i*i+a4*j*j*j+a5*k*k*k)+size;
				if(f[s]) sum+=f[s];
			}
		}
	}
	printf("%d\n",sum);
	return 0;
}