哈希表应用：POJ1840 公式

题目描述：

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.

输入：

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

输出：

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

输入样例：

37 29 41 43 47

输出样例：

654

解题思路：

        可以直接使用穷举，通过5轮for循环找出能使公式成立的解的个数，不过太消耗时间。可以通过移项，使左边3个项、右边两个项；降低了时间复杂度。把右边的所有结果放到哈希表中，再计算左边，如果有相同解就累加。

代码实现：

#include <iostream>
using namespace std;
#define maxn 100000000

int a1, a2, a3, a4, a5;
int hush_t[maxn];

int main()
{
    cin >> a1 >> a2 >> a3 >> a4 >> a5;
    int temp = 0;
    for (int i = -50; i <= 50; i++)
    {
        for (int j = -50; j <= 50; j++)
        {
            if (i == 0 || j == 0)
            {
                continue;
            }
            temp = (a1 * i * i * i + a2 * j * j * j) * (-1);
            if (temp < 0)
            {
                temp = temp + 25000000;
            }
            hush_t[temp]++;
        }
    }
    int ans = 0;
    for (int i = -50; i <= 50; i++)
    {
        for (int j = -50; j <= 50; j++)
        {
            for (int k = -50; k <= 50; k++)
            {
                if (i == 0 || j == 0 || k == 0)
                {
                    continue;
                }
                temp = a3 * pow(i, 3) + a4 * pow(j, 3) + a5 * pow(k, 3);
                if (temp < 0)
                {
                    temp += 25000000;
                }
                if (hush_t[temp])
                {
                    ans += hush_t[temp];
                }
            }
        }
    }
    cout << ans << endl;

    return 0;
}