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; }