▶ 指数模运算。求 c 使得 c ≡ ab (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 ab % m == ((ac % m)10 % m) * (ad % m) % m，递归，每次将指数的个位拆出来
1 class Solution 2 { 3 const int mod = 1337; 4 int powMod(int a, int k) //a^k mod 1337 where 0 <= k <= 10 5 { 6 a %= mod; 7 int i, result; 8 for (i = 0, result = 1; i < k; result = (result * a) % mod, i++); 9 return result; 10 } 11 public: 12 int superPow(int a, vector<int>& b) 13 { 14 if (b.empty()) 15 return 1; 16 int last_digit = b.back(); 17 b.pop_back(); 18 return powMod(superPow(a, b), 10) * powMod(a, last_digit) % mod; 19 } 20 };
● 代码，10 ms，与上面蕾丝是的算法，二进制优化了指数运算
1 class Solution 2 { 3 public: 4 const int mod = 1337; 5 int powMod(int a, int b)// a ^ b % mod 6 { 7 int result; 8 for (result = 1; b; b >>= 1) 9 { 10 if (b & 1) 11 result = (result * a) % mod; 12 a = (a * a) % mod; 13 } 14 return result; 15 } 16 int superPow(int a, vector<int>& b) 17 { 18 int i, result; 19 for (a %= mod, i = b.size() - 1, result = 1; i >= 0; i--) 20 { 21 if (i < b.size() - 1) 22 a = powMod(a, 10); 23 result = (result * powMod(a, b[i])) % mod; 24 } 25 return result; 26 } 27 };
● 代码，8 ms，使用欧拉函数，注意到 ab % m == ab % φ(m) % m，φ(1337)=1140
1 class Solution 2 { 3 public: 4 const int mod = 1337; 5 int superPow(int a, vector<int>& b) 6 { 7 int p = 0, ret; 8 for (int i : b) 9 p = (p * 10 + i) % 1140; 10 if (p == 0) 11 p += 1140; 12 for (ret = 1, a %= mod; p > 0; a = a * a % mod, p >>= 1) 13 { 14 if (p & 1) 15 ret = ret * a % mod; 16 } 17 return ret; 18 } 19 };
