Equations

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

Total Submission(s): 5226    Accepted Submission(s): 2079

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

哈希表基础题：

#include<stdio.h>
#include<algorithm>
#define M 2000004
int hash[M];
int main ()
{
	int a,b,c,d;
	int i,j;
	int sum,s[101];
	for(i=1;i<=100;i++)
		   s[i]=i*i;
	while(~scanf("%d%d%d%d",&a,&b,&c,&d))
	{
		if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
		{
			printf("0\n");
			continue;
		}
		sum=0;
		memset(hash,0,sizeof(hash));
		for(i=1;i<=100;i++)
			for(j=1;j<=100;j++)
				hash[a*s[i]+b*s[j]+M/2]++;  // M/2 是防止下标为负数。

			for(i=1;i<=100;i++)
				for(j=1;j<=100;j++)
				   sum+=hash[-(c*s[i]+d*s[j])+M/2];  // 亏大神们想得出啊！

				printf("%d\n",sum*16);
	}
	return 0;
}

hdu 1469 Equations