Description：
Consider equations having the following form: a*x1*x1 + b*x2*x2 + c*x3*x3 + d*x4*x4 = 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.
Output：
or each test case, output a single line containing the number of the solutions.
Sample Input：
1 2 3 -4
1 1 1 1
39088
0
#include<stdio.h>
#include<string.h>
int hash1[1000000] = { 0 }, hash2[1000000] = { 0 };
int main()
{
int a, b, c, d, sum;
while (scanf("%d%d%d%d", &a, &b, &c, &d) != EOF)
{
int i, j, s;
memset(hash1, 0, sizeof(hash1));
memset(hash2, 0, sizeof(hash2));
if ((a>0 && b>0 && c>0 && d>0) || (a<0 && b<0 && c<0 && d<0))
{
printf("0\n");
continue;
}
else
{
for (i = 1; i <= 100; i++)
{
for (j = 1; j <= 100; j++)
{
s = a*i*i + b*j*j;
if (s >= 0)hash1[s]++;
else hash2[-s]++;
}
}
sum = 0;
for (i = 1; i <= 100; i++)
{
for (j = 1; j <= 100; j++)
{
s = c*i*i + d*j*j;
if (s>0)sum += hash2[s];
else sum += hash1[-s];
}
}
printf("%d\n", sum * 16);
}
}
return 0;
}