Codeforces#363 (Div. 2) C.Vacations

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本文介绍了一种使用动态规划解决最小休息日问题的方法。给定假期中每天健身房和在线比赛的状态，目标是最小化连续不活动天数，同时避免连续两天进行相同活动。

Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:

on this day the gym is closed and the contest is not carried out;
on this day the gym is closed and the contest is carried out;
on this day the gym is open and the contest is not carried out;
on this day the gym is open and the contest is carried out.

On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).

Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

Input

The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.

The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:

ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.

Output

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

to do sport on any two consecutive days,
to write the contest on any two consecutive days.

Examples

input
4
1 3 2 0

output
2

input
7
1 3 3 2 1 2 3

output
0

input
2
2 2

output
1

思路：三种情况DP

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#define INF 1e+9;
using namespace std;

int n;
int dp[100 + 5][3];

int Min(int x, int y, int z)
{
    if(x > y)
        x = y;
    if(x > z)
        x = z;
    return x;
}
int main()
{
    cin>>n;
    memset(dp,0,sizeof dp);
    for(int i = 1; i <= n; i++)
    {
        int temp;
        cin >> temp;
        if(temp == 2)
        {
            dp[i][0] = min(dp[i-1][1],dp[i-1][2]);
            dp[i][1] = INF;
            dp[i][2] = Min(dp[i-1][0],dp[i-1][1],dp[i-1][2]) + 1;
        }
        else if(temp == 0)
        {
            dp[i][0] = INF;
            dp[i][1] = INF;
            dp[i][2] = Min(dp[i-1][0],dp[i-1][1],dp[i-1][2]) + 1;
        }
        else if(temp == 1)
        {
            dp[i][1] = min(dp[i-1][0],dp[i-1][2]);
            dp[i][0] = INF;
            dp[i][2] = Min(dp[i-1][0],dp[i-1][1],dp[i-1][2]) + 1;

        }
        else
        {
            dp[i][0] = min(dp[i-1][1],dp[i-1][2]);
            dp[i][1] = min(dp[i-1][0],dp[i-1][2]);
            dp[i][2] = Min(dp[i-1][0],dp[i-1][1],dp[i-1][2]) + 1;
        }
    }
    cout<<Min(dp[n][0],dp[n][1],dp[n][2])<<endl;
}