上台阶，上三下二

一个寺庙有N个台阶，一个人站在下面，他要去到第M个台阶，假设他只有2个方案可以选择 1. 向上走3台阶 2. 向下走2个台阶。请问他到达第M个台阶最少要走多少步。给出核心算法并分析复杂度

// taijie_up3_down2.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <vector>

using namespace std;

#define N 200
#define MAX 0xFFFF



int min (int a, int b)
{
	if (a <= b)
		return a;

	return b;
}

int getSuccessedDP(int i, vector<int> &dp)
{
	int idx = i;
	if (dp[idx] == 0xFFFF)
	{
		dp[i] = min(dp[i-3], getSuccessedDP(i+2, dp))+1;
	}

	return dp[i];

}


int minStep(int M)
{
	if (M%3 == 0)
		return M/3;

	vector<int> dp(N+1);

	for (int i=0; i<=N; i++)
	{
		dp[i] = MAX;
		if (i%3 == 0)
			dp[i] = i/3;
	}

	dp[1] = 2;
	dp[2] = 4;
	dp[4] = 3;

	for (int i=5; i<=M; i++)
	{
		dp[i] = min(dp[i-3], getSuccessedDP(i+2, dp))+1;
	}

	return dp[M];
}


int _tmain(int argc, _TCHAR* argv[])
{
	int ret = minStep(47);
	return 0;
}