下面count_triple函数的时间复杂度为( )。
int gcd(int m, int n) { if (m == 0) return n; return gcd(n % m, m);}int count_triple(int n) { int cnt = 0; for (int v = 1; v * v * 4 <= n; v++) for (int u = v + 1; u * (u + v) * 2 <= n; u += 2) if (gcd(u, v) == 1) { int a = u * u - v * v; int b = u * v * 2; int c = u * u + v * v; cnt += n / (a + b + c); } return cnt;}