372. Super Pow

▶ 指数模运算。求 c 使得 c ≡ a^b (mod m)，其中 b 用 vector<int> 的形式保存（长整数），m = 1337

● 代码，18 ms，令 b=10c+d，则 a^b % m == ((a^c % m)¹⁰ % m) * (a^d % m) % m，递归，每次将指数的个位拆出来

class Solution
{
    const int mod = 1337;
    int powMod(int a, int k) //a^k mod 1337 where 0 <= k <= 10
    {
        a %= mod;
        int i, result;
        for (i = 0, result = 1; i < k; result = (result * a) % mod, i++);
        return result;
    }
public:
    int superPow(int a, vector<int>& b)
    {
        if (b.empty())
            return 1;
        int last_digit = b.back();
        b.pop_back();
        return powMod(superPow(a, b), 10) * powMod(a, last_digit) % mod;
    }
};

● 代码，10 ms，与上面蕾丝是的算法，二进制优化了指数运算

class Solution
{
public:
    const int mod = 1337;
    int powMod(int a, int b)// a ^ b % mod
    {
        int result;
        for (result = 1; b; b >>= 1)
        {
            if (b & 1)
                result = (result * a) % mod;
            a = (a * a) % mod;
        }
        return result;
    }    
    int superPow(int a, vector<int>& b)
    {
        int i, result;              
        for (a %= mod, i = b.size() - 1, result = 1; i >= 0; i--)
        {
            if (i < b.size() - 1)
                a = powMod(a, 10);
            result = (result * powMod(a, b[i])) % mod;
        }
        return result;
    }
};

● 代码，8 ms，使用欧拉函数，注意到 a^b % m == a^{b % φ(m)} % m，φ(1337)=1140

class Solution
{
public:
    const int mod = 1337;
    int superPow(int a, vector<int>& b)
    {
        int p = 0, ret;
        for (int i : b)
            p = (p * 10 + i) % 1140;
        if (p == 0)
            p += 1140;
        for (ret = 1, a %= mod; p > 0; a = a * a % mod, p >>= 1)
        {
            if (p & 1)
                ret = ret * a % mod;
        }
        return ret;
    }
};