A23846. 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 * …

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;
}