[codeforce ] Vacations [贪心]

Vacations

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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.

Sample Input

Input

41 3 2 0

Output

2

Input

71 3 3 2 1 2 3

Output

0

Input

22 2

Output

1

贪心,局部最优解。

这题是要求休息天数最少的

局部最优，当a[0]==3时，令a[0]=1，b[0]=2,其他a[0]==b[0]

当遇到3 时，前一项是1或2,3变为2或1,；当遇到1或2时，和前一项一样就记为0；

当a[0]==3时，比较a,b数组0出现的次数少的

代码：

#include<cstdio>int main(){int n,a[110],b[110];while(~scanf("%d",&n)){int flag=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);if(a[0]==3){a[0]=1;b[0]=2;flag=1;}elseb[i]=a[i];}for(int i=1;i<n;i++){if(a[i]==3&&a[i-1]==2)a[i]=1;if(a[i]==3&&a[i-1]==1)a[i]=2;if(a[i]==a[i-1]&&(a[i-1]==1||a[i-1]==2))a[i]=0;}int ans=0;for(int i=0;i<n;i++){if(a[i]==0)ans++;}if(flag){for(int i=1;i<n;i++){if(b[i]==3&&b[i-1]==1)b[i]=2;if(b[i]==3&&b[i-1]==2)b[i]=1;if(b[i]==b[i-1]&&(b[i]==1||b[i]==2))b[i]=0;}int cnt=0;for(int i=0;i<n;i++){if(b[i]==0)cnt++;}if(cnt<ans)ans=cnt;}printf("%d\n",ans);}return 0;}