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;
}
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