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

 

哈希只会一丢丢。就不写详解了。自己体会，O(∩_∩)O哈哈~

 

对于最后的结果要乘16，是因为，计算出来的结果可正可负，一共有 2^4 种情况

 

using namespace std;const int maxn = 2000005;const int h = 1000000;int hash1[maxn];int a, b, c, d, sum, k;//主思路： 分两段计算 最后乘16（ 2^4 ）int main(){    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 )        {            cout << "0" << endl ;            continue ;        }        memset(hash1,0,sizeof(hash1));        sum = 0 ;        for(int i=1; i<=100; i++)            for(int j=1; j<=100; j++)                hash1[ a * i * i + b * j * j + h ] ++ ;        for(int i=1; i<=100; i++)            for(int j=1; j<=100; j++)                sum += hash1[ h - c * i * i - d * j * j ] ;        cout << sum*16 << endl ;    }    return 0;}