cf----698A vacations

vacations

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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:

   0，on this day the gym is closed and the contest is not carried out;
   1，on this day the gym is closed and the contest is carried out;
   2，on this day the gym is open and the contest is not carried out;
   3，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：就是分类，然后对判断过的数据进行处理一下就行了。

#include<stdio.h>
int main()
{
    int n,i,num=0,a[110];
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    if(a[0]==0)     //对首位进行特判 
        num++;
    for(i=1;i<n;i++)    //对接下来的进行判定 
    {
        if(a[i]==0)  
            num++;
        if(a[i]==1 && a[i-1]==1)    //当其和前一位相同时加一，记住要归零 
            a[i]=0,num++;
        if(a[i]==2 && a[i-1]==2)
            a[i]=0,num++;
        if(a[i]==3){        //对三的情况分类，当其前一位为零时不用管 
            if(a[i-1]==1)
                a[i]=2;
            else if(a[i-1]==2)
                a[i]=1;
        }
    }
    printf("%d\n",num);
    return 0;
}

//下面的是样例6，答案是16，在这挂了很多次。
/*100
3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1
2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 3 1 1
3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3
3 3 2 3 1 3 3 3 3 2 2 2 1 3 1 3 3 3 3 1
3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3
*/