题干：
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.
题目大意：
Vasya在n天中，有三件事情可以做，健身、写作或者休息，但是健身和写作不能连续两天都去做，但是连续休息两天是允许的，问题是在这n天中，Vasya最少可以休息几天？
题意2：：Vasyayo有n天的假期，他每天可以选择运动，比赛或者休息。但每天都四个状态，体育场开门没有比赛，体育场关门有比赛，体育场开门有比赛，体育场关门没比赛。问他最少休息多少天？
解题报告：
因为题目中显然有状态的转移，所以考虑dp
AC代码：
#include<bits/stdc++.h>using namespace std;
const int INF = 0x3f3f3f3f;
int dp[1000][4];//1是休息，2是比赛，3是健身 第二位代表当天做了什么（所以首先得能选择做这个决策，所以m=1时不需要更新 健身那一层），而不是
int a[1000];
int min(int x,int y,int z) {return min(x,min(y,z));
}
int main()
{int n;int minn ;cin>>n;for(int i = 1; i<=n; i++) {cin>>a[i];//1是比赛，2是健身 }//看我当天能够做什么事来选择更新什么值 for(int i = 1; i<=n; i++) {if(a[i] == 0) {minn = INF;for(int k = 1; k<=3; k++) minn = min(minn,dp[i-1][k]);
// printf("minn = %d\n",minn);dp[i][1] = minn+1;dp[i][2] = dp[i][3] = INF;}else if(a[i] == 1) {dp[i][2] = min(dp[i-1][1],dp[i-1][3]);dp[i][1] = min(dp[i-1][1] + 1,dp[i-1][2]+1,dp[i-1][3] + 1);dp[i][3] = INF;}else if(a[i] == 2) {dp[i][1] = min(dp[i-1][1]+1,dp[i-1][2]+1,dp[i-1][3]+1);dp[i][2] = INF;dp[i][3] = min(dp[i-1][2],dp[i-1][1]);}else {dp[i][1] = min(dp[i-1][1]+1,dp[i-1][2]+1,dp[i-1][3]+1);dp[i][2] = min(dp[i-1][1],dp[i-1][3]);dp[i][3] = min(dp[i-1][1],dp[i-1][2]);}}
// for(int i = 1; i<=n; i++)printf("%d\n",min(dp[n][1],dp[n][2],dp[n][3]));return 0 ;} 
