leetcode 快速幂

快速幂算法

• 快速幂就是快速算底数的n次幂，其时间复杂度为 O(log₂N)
• 例如： $求 7^{10}$

  $7^{10} = 7^{5+5} = 7^{5}*7^{5},而7^{5}=7^{2}*7^{2}*7$

code

• leetcode 50 Pow(x,n)
  • 实现x的n 次幂

class Solution {
public:
    double myPow(double x, int n) {
        if(n == 0 ) return 1;
        double half = myPow(x,n/2);
        if(n%2 == 0) return half*half; // even
        if(n>0) return half*half*x; // odd
        return half*half/x;  // <0, 1/x
    }
};

• leetcode 372 超级次方

class Solution {
    long long int Pow(long long int x,int n){
        if(n == 0) return 1;
        long long int v = Pow(x,n/2);
        v %= 1337;
        return (n%2 == 0 ? 1 : (x%1337) ) * (v*v)%1337;
    }
    long long int fastpower(long long int x,int n){
        long long int ans = 1;
        while(n > 0){
            if(n & 1){
                ans *= x;
                ans %= 1337;
            }
            x *=x%1337;
            n >>= 1;
        }
        return ans;
    }
public:
    int superPow(int a, vector<int>& b) {
        long long int ans = 1,tmp = a % 1337;
        int i = b.size()-1;
        while(i>=0){
            ans *= fastpower(tmp,b[i]);
            tmp = fastpower(tmp,10) % 1337;
            ans %= 1337;
            i--;
        }
        return ans;
    }
};