程序设计C 实验三 题目九 方程式(0300)

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[] = {  }, hash2[] = {  };
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, , sizeof(hash1));
        memset(hash2, , sizeof(hash2));
        if ((a> && b> && c> && d>) || (a< && b< && c< && d<))
        {
            printf("0\n");
            continue;
        }
        else
        {
            for (i = ; i <= ; i++)
            {
                for (j = ; j <= ; j++)
                {
                    s = a*i*i + b*j*j;
                    if (s >= )hash1[s]++;
                    else hash2[-s]++;
                }
            }
            sum = ;
            for (i = ; i <= ; i++)
            {
                for (j = ; j <= ; j++)
                {
                    s = c*i*i + d*j*j;
                    if (s>)sum += hash2[s];
                    else sum += hash1[-s];
                }
            }
            printf("%d\n", sum * );
        }
    }
    return ;
}