[Practice]快速幂求模递归法

概述

离散书中，出现了此算法。Recursive Modular Exponentiation，感觉还是挺快的，有点像二分查找法？

原理

ab mod c = (a mod c) * (b mod c)

  对a^b mod c而言，

当b = 0时, a^0 mod c = 1 mod c = 1(base condition)

（当然，b = 1时作为基线条件也是可以的）

当b是偶数时，a ^ b mod c = (a^(b/2) * a^(b/2)) mod m

= (a^(b/2) mod m)^2 mod m

当b是奇数时，a ^ b mod c =(a ^ (b-1)) mod c * (a mod c)

.................................................偶数情况...........................

实现

#include <iostream>
#include <cmath>
using namespace std;

int power(int num)
{
	return num * num;
}

int powerMod(int a, int b, int c)
{
	if (a <= 0 || b < 0 || c <= 0)
		return -1;
	if (!b)	return 1;

	int ans = 1;
	if (b % 2)
		ans *= (a % c);
	return (ans * power(powerMod(a, b/2, c)) % c);
}

int main(int argc, char const *argv[])
{
	int a = 0, b = 0, c = 0;
	cout << "Please enter a,b,c such that (a^b mod c):" << endl;
	cin >> a >> b >> c;
	cout << powerMod(a, b, c);
	return 0;
}

[Practice]快速幂求模递归法,布布扣,bubuko.com

[Practice]快速幂求模递归法

原文：http://blog.csdn.net/tbz888/article/details/23134925