Super Pow

一、问题描述

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

二、思路

1140是如何计算出来的？

phi(1337) = phi(7) * phi(191) = 6 * 190 = 1440

根据ab % 1337 = ab % phi(1337) % 1337 = ab % 1140 % 1337.

所以我们需要两个函数分别处理指数和底数分别对1140和1337求余。

三、代码

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