50. Pow(x, n) 372. Super Pow https://www.cnblogs.com/grandyang/p/5651982.html https://www.jianshu.com/p/b256bd531df0 做这个题之间先了解两个公式: 公式一:a^b mod c = (a mod c)^b mod c公式二:(ab) mod c = (a mod c)(b mod c) mod c 这道题题让我们求一个数的很大的次方对1337取余的值,即a^b mod 1337.输入…

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 Credits:Special thanks to @Stomac…

Your task is to calculate ab mod 1337 where a is a positive integer and bis an extremely large positive integer given in the form of an array. Example 1: Input: a = 2, b = [3] Output: 8 Example 2: Input: a = 2, b = [1,0] Output: 1024 你的任务是计算 ab 对 133…

你的任务是计算 ab 对 1337 取模,a 是一个正整数,b 是一个非常大的正整数且会以数组形式给出.示例 1:a = 2b = [3]结果: 8示例 2:a = 2b = [1,0]结果: 1024详见:https://leetcode.com/problems/super-pow/description/ C++: class Solution { public: int superPow(int a, vector<int>& b) { long long res = 1; f…

题目链接:https://leetcode.com/problems/super-pow/description/ Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = b = [] Result: Example2: a = b = […

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/super-pow/description/ 题目描述: Your task is to calculate a^b mod 1337 where a is a positive integer and b is an extremely large positive integer given in the fo…

问题 Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 分析 根据公式 ① ② ab mod 1337 可以优化为…

题目描述: superPow(int a, int[] b),b是一个int数组,每个元素都是正的个位数,组合起来表示一个正整数,例如b=[1,2,3]表示123,求解a^b mod 1337. 思路描述: 本题的难点是当a和b足够大时会造成溢出,因此应考虑其他算法来实现. 理论支持(转幂算法): (a^b) mod c = ((a mod c)^b) mod c ----公式1 (x*y) mod c = ((x mod c) * (y mod c)) mod c  :积的取余等于取余的积的取…

使用公式 c = ab  =>  c mod d = [a mod d * b mod d] mod d 所以a^423 mod d = (a^100)^4 * (a ^10)^2 * a^3 class Solution(object): def superPow(self, a, b): """ :type a: int :type b: List[int] :rtype: int """ ans = 1 mod = 1337 for…

▶ 指数取模运算 ab % m ▶ 参考维基 https://en.wikipedia.org/wiki/Modular_exponentiation,给了几种计算方法:暴力计算法,保存中间结果法(分为左到右的二进制法和右到左的二进制法),矩阵法,优先群法,量子计算法. ● 代码,18 ms,令 a' = a % m,b = 10 * c + d(0 ≤ d < 10),则 ab % m == ((ac % m)10 % m) * (ad % m) % m,每次将指数的个位拆出来单独算,再和高位…