众所周知，快速幂可以这么写：

int power(int a, int b, int p)
{
    if (b == 1)return a;
    int ans = power(a, b / 2, p);
    ans = (long long)ans * ans % p;
    if (b % 2)ans = (long long)ans * a % p;
    return ans;
}

那么，如果我们每次把问题规模缩小到$\frac{1}{3}$，会不会更快呢？快速幂能不能这样写：

int power(int a, int b, int p)
{
    if (b == 0)return 1;
    if (b == 1)return a;
    int ans = power(a, b / 3, p);
    ans = (unsigned long long)(ans * ans % p) * ans % p;
    if (b % 3 == 1)ans = (unsigned long long)ans * a % p;
    else if (b % 3 == 2)ans = (unsigned long long)(ans * a % p) * a % p;
    return ans;
}

求助大佬