2013
12-11

# Equations

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.

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.

For each test case, output a single line containing the number of the solutions.

1 2 3 -4
1 1 1 1

39088
0

/*187ms,8044KB*/

#include<cstdio>
#include<cstring>
const int maxn = 1000000;

int hash[2 * maxn + 5];

int main()
{
int a, b, c, d, i, j;
long long ans;
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)
{
puts("0");
continue;
}
memset(hash, 0, sizeof(hash));
for (i = 1; i <= 100; ++i)
for (j = 1; j <= 100; ++j)
++hash[a * i * i + b * j * j + maxn];
ans = 0L;
for (i = 1; i <= 100; ++i)
for (j = 1; j <= 100; ++j)
ans += hash[-c * i * i - d * j * j + maxn];
printf("%I64d\n", ans << 4);
}
}

1. 这道题目的核心一句话是：取还是不取。
如果当前取，则index+1作为参数。如果当前不取，则任用index作为参数。