HDU 1496 Equations 【整数Hash】

Equations

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

Total Submission(s): 5751 Accepted Submission(s): 2308



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

//整数Hash问题.常规解法需要4层for循环，显然超时。
//这里用到了 两层循环+Hash存储查找

#include<cstdio>
#include<cstring>
int hash[2000002];
int main()
{
    int a,b,c,d,i,j,sum;
    int p[102];
    for(i=1; i<=100; i++)
    {
        p[i]=i*i;
    }
    while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
    {
        memset(hash,0,sizeof(hash));
        if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
        {
            printf("0\n");
            continue;
        } 
        for(i=1; i<=100; i++)
        {
            for(j=1; j<=100; j++)
            {
                hash[p[i]*a+p[j]*b+1000000]++;
            }
        }
        for(i=1,sum=0; i<=100; i++)
        {
            for(j=1; j<=100; j++)
            {
                sum+=hash[-(p[i]*c+p[j]*d)+1000000];
            }   
        }
        printf("%d\n",sum*16); 
    }
    return 0;
}