Codeforces Round #363 (Div. 2) C. Vacations （DP）

C. Vacations

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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

Note

In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.

In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.

In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

题意：一个人每天可能会有三种状态，休息、健身、考试，有n天

，当天所做事情有限制：

0：不能健身，不能考试

1: 不能健身，可以考试

2: 可以健身，不能考试

3: 可以健身，可以考试

连续两天不能做同一件事情，但是可以同时休息，问你，这n天里，这个人最少休息几天。

思路：明显的DP，根据当天限制进行DP，dp[i][1]表示当天休息，前i天最多不休息的天数，dp[i][2]表示当天健身，前i天最多不休息的天数，dp[i][3]表示当天考试，前i天最多不休息的天数，具体详见代码。

ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int dp[200][5];
int main()
{
int n;scanf("%d",&n);
memset(dp, 0, sizeof(dp));
for(int i=1;i<=n;i++)
{
int a;scanf("%d",&a);
if(a==0)
{
int k=max(dp[i-1][1],max(dp[i-1][2],dp[i-1][3]));
dp[i][1]=k;dp[i][2]=k;dp[i][3]=k;
}
else if(a==1)
{
dp[i][1]=max(dp[i-1][1],max(dp[i-1][2],dp[i-1][3]));
dp[i][2]=dp[i][1];
dp[i][3]=max(dp[i-1][1],dp[i-1][2])+1;
}
else if(a==2)
{
dp[i][1]=max(dp[i-1][1],max(dp[i-1][2],dp[i-1][3]));
dp[i][2]=max(dp[i-1][1],dp[i-1][3])+1;
dp[i][3]=dp[i][1];
}
else if(a==3)
{
dp[i][1]=max(dp[i-1][1],max(dp[i-1][2],dp[i-1][3]));
dp[i][2]=max(dp[i-1][1],dp[i-1][3])+1;
dp[i][3]=max(dp[i-1][1],dp[i-1][2])+1;
}
}
int ans=max(dp
[1],max(dp
[2],dp
[3]));
printf("%d\n",n-ans);
return 0;
}