CodeForces-698A Vacations（水题）

题目链接：http://codeforces.com/problemset/problem/698/A点击打开链接

A. Vacations

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., 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

41 3 2 0

output

2

input

71 3 3 2 1 2 3

output

0

input

22 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.

题意：0休息 1健身或休息 2运动或休息 3健身或运动或休息 不能两天同时健身或两天同时休息

如果当前健身 并且下一天为3 则下一天改为运动 反之亦然 如果下一天与当前相同 下一天改为0

#include <stdio.h>#include <stdlib.h>#include <iostream>#include<algorithm>#include <math.h>#include <string.h>#include <limits.h>#include <string>#include <queue>#include <stack>using namespace std;int a[100];int main(){    int n=0;int num=0;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%d",&a[i]);    for(int i=0;i<n-1;i++)    {        if(a[i]==0)            num++;        else if(a[i]==1)        {            if(a[i+1]==1)                a[i+1]=0;            else if(a[i+1]==3)                a[i+1]=2;        }        else if(a[i]==2)        {            if(a[i+1]==2)                a[i+1]=0;            else if(a[i+1]==3)                a[i+1]=1;        }    }    if(a[n-1]==0)        num++;    printf("%d",num);    }

又写了一遍 这次是对当前节点往前判断 忘了修改当前的值 wa了

如果只记录不修改的话 会导致连续的 1 1 1 这样的多算 因此要改当前的数据

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>#define maxn 100010using namespace std;int a[maxn];int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    a[0]=0;    int cnt=0;    for(int i=1;i<=n;i++)    {        if(a[i]==1)        {            if(a[i-1]==1)            {                a[i]=0;                cnt++;            }        }        else if(a[i]==2)        {            if(a[i-1]==2)            {                a[i]=0;                cnt++;            }        }        else if(a[i]==3)        {            if(a[i-1]==2)                a[i]=1;            else if(a[i-1]==1)                a[i]=2;        }        else if(a[i]==0)        {            cnt++;        }    }    printf("%d",cnt);}