vacations
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.
这个是动态规划。这个思想我还是不太熟练，想了好长时间才理解别人的代码。
/*
0 gym closed contest closed
1 gym closed contest open
2 gym open contest closed
3 gym open contest open
dp[i][j]表示第i天做j事最少休息的天数。
dp[i+1][0]=min{dp[i][0]+1,dp[i][1]+1,dp[i][2]+1}第i+1天休息
dp[i+1][1]=min{dp[i][0],dp[i][2]}第i+1天可以运动
dp[i+1][2]=min{dp[i][0],dp[i][1]}第i天可以打比赛
res=min{dp[n][0],dp[n][1],dp[n][2]}
dp[i][0]表示第i天休息的情况下当前i天最少能休息多少天
dp[i][1]表示第i天比赛的情况下当前i天最少能休息多少天
dp[i][2]表示第i天健身的情况下当前i天最少能休息多少天
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][0]=min(dp[i-1][0],dp[i-1][1],dp[i-1][2])+1
其他的到不了的状态直接复制INF
*/
#include
#include
#include
using namespace std;
#define INF 1e9+7
const int maxn = 105;
int dp[maxn][3];
int main()
{
int n, a[105];
scanf ("%d", &n);
for (int i = 1; i <= n; i++)
scanf ("%d", a+i);
for (int i = 0; i < 3; i++)
dp[0][i] = 0;
for (int i = 1; i <= n; i++){
dp[i][0] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2])) + 1;
//这一天休息的情况下最少能休息一天
if (a[i] == 1 || a[i] == 3)
dp[i][1] = min(dp[i-1][0], dp[i-1][2]);
//这一天做运动的情况下他休息的最少天数就是之前休息或者打比赛的最少天数
else
dp[i][1] = INF;
if (a[i] == 2 || a[i] == 3)
dp[i][2] = min(dp[i-1][0], dp[i-1][1]);
else
dp[i][2] = INF;
if (a[i] == 0){
dp[i][1] = INF;
dp[i][2] = INF; // 这一天必须休息
}
}
printf("%d\n", min(dp[n][0], min(dp[n][1], dp[n][2])));
return 0;
}