Eqs（hash初步）

Eqs

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
题目大意：
给出一个5元3次方程，输入其5个系数，求它的解的个数
其中系数 ai∈[-50,50] 自变量xi∈[-50,0）∪（0,50]

#include<iostream>
using namespace std;
const int maxn=25000000;
int a1,a2,a3,a4,a5;
short hash[maxn+1];
int main()
{
    cin>>a1>>a2>>a3>>a4>>a5;
    for(int x1=-50;x1<=50;x1++)
    {
        if(!x1) continue;
        for(int x2=-50;x2<=50;x2++)
        {
            if(!x2) continue;
            int num=a1*x1*x1*x1+a2*x2*x2*x2;
            if(num<0)
            num+=maxn;
            hash[num]++;
        }
    }
    int ans=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 num=a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;
                if(num<0)
                num+=maxn;
                if(hash[num])
                ans+=hash[num];
            }
        }
    }
    cout<<ans;
    return 0;
}