G - Equations

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

题目给出一个方程组，给出a, b, c, d,问在[-100，100]区间内有多少组解，可以先移项，分成两部分，a*x1*x1+b*x2*x2 和c*x3*x3+d*x4*x4两部分，先对一个部分进行处理，在哈希数组里进行标记，然后再对另外一部分进行枚举求解，最后的答案需要乘以16， 因为两边的值都可以有正负，且两边都是两个未知数。

#include
#include
#include
using namespace std;
char vis[2500000];
int main()
{
    int a, b, c, d;
    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;
        }
        long long ans = 0;
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= 100; i ++)
        {

            for(int j = 1; j <= 100; j ++)
            {
                int cnt = a*i*i+b*j*j;
                if(cnt < 0)
                {
                    cnt += 2500000;
                }
                vis[cnt] ++;
            }
        }
        for(int i = 1; i <= 100; i ++)
        {

            for(int j = 1; j <= 100; j ++)
            {
                    int cnt = -1*c*i*i+(-1)*d*j*j;
                    if(cnt < 0)
                    {
                        cnt += 2500000;
                    }
                    if(vis[cnt])
                    {
                        ans += vis[cnt];
                    }
            }
        }
        ans *= 16;
        printf("%lld\n", ans);
    }
    return 0;
}