Equations（哈希）

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6254 Accepted Submission(s): 2531



Problem Description
Consider equations having the following form:

a*x1^2+b*x2^2+c*x3^2+d*x4^2=0

a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.

It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.

Determine how many solutions satisfy the given equation.



Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.

End of file.


Output
For each test case, output a single line containing the number of the solutions.



Sample Input

1 2 3 -4
1 1 1 1

Sample Output

39088
0

Author
LL


这一题如果不使用哈希表存的话也可以三重暴力！



费时间

用哈希的话可以两个二重循环的时间复杂度！



特别好，记录一下！同时也说明了哈希的用处！

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int hash1[3000050];//50*100*100*2=1000000;
int main()
{
    int a,b,c,d,ans,i,j;
    while(cin>>a>>b>>c>>d)
    {
        if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)
        {
            cout<<"0"<<endl;
            continue;
        }
        ans=0;
        memset(hash1,0,sizeof(hash1));
        for(i=1;i<=100;i++)
        {
            for(j=1;j<=100;j++)
            {
                hash1[1000000+i*i*a+j*j*b]++;
            }
        }//哈希先打表
        for(i=1;i<=100;i++)
        {
            for(j=1;j<=100;j++)
            {
                ans+=hash1[-(i*i*c+j*j*d)+1000000];
            }
        }//比较他们的值是否相等！
        cout<<ans*16<<endl;
    }
}