Eqs hash表存储，注意hash要用char类型

Problem 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

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hash表的应用

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#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
const int maxn=20000000;//防止出现负；
char hash[40000000];//此处用char节省内存
int a1,a2,a3,a4,a5;
int i,j,k;
int main()
{
    int sum;
    while(scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)!=EOF)
    {
        sum=0;
        memset(hash,0,sizeof(hash));
        //前两个
        for(i=-50;i<=50;i++)
         for(j=-50;j<=50;j++)
          if(i!=0&&j!=0)
           hash[i*i*i*a1+j*j*j*a2+maxn]++;//hash表存储
        //后三个
        for(i=-50;i<=50;i++)
         for(j=-50;j<=50;j++)
          for(k=-50;k<=50;k++)
           if(i!=0&&j!=0&&k!=0)
             sum+=hash[i*i*i*a3+j*j*j*a4+k*k*k*a5+maxn];//由结果的对称性，可得
        printf("%d
",sum);
    }
}