CodeForces - 698A Vacations 贪心 dp

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:

1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. 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, ..., a_n (0 ≤ a_i ≤ 3) separated by space, where:

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

Example

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.

    这个题我的一个想法就是贪心，，， 从开始不是0，或者3的地方开始，1212121212，或者21212121（3可以为1也可以为2）这两种情况尽可能多

#include<cstdio>
using namespace std;
int main()
{
    int t;
    int a[110],b[110];
    while(~scanf("%d",&t))
    {
        for(int i=1;i<=t;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=0,i;
        for(i=1;i<=t;i++)
        {
            if(a[i]==0)
                ans++;
            else
                if(a[i]==3)
                continue;
            else
                break;
        }
        int n=a[i];
        for(i+=1;i<=t;i++)
        {
            if(n==a[i]&&n!=3)
            {
                ans++;
                n=0;
            }
            if(n!=a[i]&&a[i]!=0)
            {
                n=3-n;
            }
            if(a[i]==0)
            {
                ans++;
                n=3;
            }
            if(n==0&&a[i]!=0)
            {
                n=a[i];
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

用dp 解法

#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int dp[110][4];
int main()
{
    int t,m;
    while(~scanf("%d",&t))
    {
        memset(dp,inf,sizeof(dp));
        dp[0][0]=0;
        for(int i=1;i<=t;i++)
        {
           scanf("%d",&m);
            dp[i][0]=min(min(dp[i-1][1],dp[i-1][2]),dp[i-1][0])+1;
           if(m==1)
           {
               dp[i][1]=min(dp[i-1][0],dp[i-1][2]);
           }
           if(m==2)
           {
               dp[i][2]=min(dp[i-1][0],dp[i-1][1]);
           }
           if(m==3)
           {
               dp[i][1]=min(dp[i-1][0],dp[i-1][2]);
               dp[i][2]=min(dp[i-1][0],dp[i-1][1]);
           }
        }
        printf("%d\n",min(min(dp[t][1],dp[t][2]),dp[t][0]));
    }
    return 0;
}