POJ 1840 Eqs (Hash)

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7827		Accepted: 3811

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

Source

Romania OI 2002

 解题报告：这道题的题意就是已知5元3次a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0的解为(x1, x2, x3, x4, x5)，各个x满足 -50 <= x <= 50，且x为非0整数。现给定各参数a的值，问这个5元方程有多少组解满足上述条件。思路：利用Hash，因为要枚举5个x，复杂度就是100^5，即10^10，，一般的方法肯定超时！先枚举前两个x,再枚举后三个x，最后在比对即可；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int MAX = 100007;
int hash[MAX][25], num[MAX];
int a[5];
int main()
{
    int i, j, k, ans, temp, mark, p;
    for (i = 0; i < 5; ++i)
    {
        scanf("%d", &a[i]);
    }
    ans = 0;
    for (i = -50; i <= 50; ++i)//枚举前两个x;
    {
        for (j = -50; j <= 50; ++j)
        {
            if (i && j)
            {
                temp = a[0] * i * i * i + a[1] * j * j * j;
                mark = abs(temp) % MAX;//建立hash表
                hash[mark][num[mark]] = temp;
                num[mark] ++;//处理冲突
            }
        }
    }
    for (i = -50; i <= 50; ++i)//枚举后三个x;
    {
        for (j = -50; j <= 50; ++j)
        {
            for (k = -50; k <= 50; ++k)
            {
                if (i && j && k)
                {
                    temp = a[2] * i * i *i + a[3] * j * j * j + a[4] * k * k * k;
                    mark = abs(temp) % MAX;
                    for (p = 0; p < num[mark]; ++p)
                    {
                        if (hash[mark][p] == temp)//若能成功
                        {
                            ans ++;
                        }
                    }
                }
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}