題目：
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).
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
n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
a1, a2, ..., an (0 ≤ ai) separated by space, where:
- aiequals 0, if on thei-th day of vacations the gym is closed and the contest is not carried out;
- aiequals 1, if on thei-th day of vacations the gym is closed, but the contest is carried out;
- aiequals 2, if on thei-th day of vacations the gym is open and the contest is not carried out;
- aiequals 3, if on thei-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
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.
本題的題意是給出n天假期，每天的數字代表每天這個人可以做的事情。
當為0時這天沒事幹就去休息，為1時可以去做測試，為2時可以去鍛煉，為3時可以鍛煉可以做測試。
但這個人不能連續兩天做同一件事，求出這段假期裡這個人最少能休息多少天。
這是一個動态規劃題，設出dp[i][j] 其中i代表第i天，j代表這天做的事情為0（休息） 1（做測試） 2（鍛煉）
dp[i][j]代表這個人在第i天做j的話能休息最少的天數
算出每天做每件事能休息的最少天數
代碼如下：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 105
#define M 1e9
int dp[N][N],a[N];
int main()
{
int i,j,n,m;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
cin>>a[i];
memset(dp,N,sizeof(dp));
dp[0][0]=0;
for(i=1;i<=n;i++)
{
dp[i][0]=min(dp[i-1][0],min(dp[i-1][1],dp[i-1][2]))+1;
if(a[i]==1)
{
dp[i][1]=min(dp[i-1][0],min(dp[i-1][1]+1,dp[i-1][2]));
dp[i][2]=min(dp[i-1][0],min(dp[i-1][1],dp[i-1][2]))+1;
}
else if(a[i]==2)
{
dp[i][1]=min(dp[i-1][0],min(dp[i-1][1],dp[i-1][2]))+1;
dp[i][2]=min(dp[i-1][0],min(dp[i-1][1],dp[i-1][2]+1));
}
else if(a[i]==3)
{
dp[i][1]=min(dp[i-1][0],min(dp[i-1][1]+1,dp[i-1][2]));
dp[i][2]=min(dp[i-1][0],min(dp[i-1][1],dp[i-1][2]+1));
}
}
int MIN=M;
for(i=0;i<=2;i++)
{
if(dp[n][i]<MIN)
MIN=dp[n][i];
}
cout<<MIN<<endl;
}
return 0;
}