Equations（Hash）

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分类专栏： Hash 文章标签： hash ACM

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本文介绍了一种用于解决特定形式的四元二次方程的高效算法。该算法通过预计算和哈希表来快速查找满足条件的整数解，特别适用于系数范围限制在[-50,50]且变量取值范围在[-100,100]的情况。

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Equations

Time limit:1000 ms Memory limit:32768 kB

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

Code：

#include <iostream>
#include <cstdio>
#include <cstring>
#define mem(a,b) memset(a,b,sizeof(a))

int pin[101];
int hash[2000003];

int main()
{
    int a,b,c,d;
    for(int i=1;i<=100;i++)
        pin[i]=i*i;
    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
    {
        if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
        {
            printf("0\n");
            continue;
        }
        mem(hash,0);
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
                hash[a*pin[i]+b*pin[j]+1000000]++;
        int ans=0;
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
                ans+=hash[-(c*pin[i]+d*pin[j])+1000000];
        printf("%d\n",ans<<4);
    }
    return 0;
}