CodeForces 698A —Vacations（贪心）

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本文介绍了一道关于计算最少休息日数量的算法题目。题目要求根据假期中每天的活动选项（健身房是否开放及是否有在线比赛），计算在不连续两天进行相同活动的前提下，求出最少的休息天数。

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I - 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:

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).

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

4
1 3 2 0

Output

Input

7
1 3 3 2 1 2 3

Output

Input

2
2 2

Output

// 0 健身房关门   没比赛
// 1 健身房关门   有比赛 
// 2 健身房开门   没比赛 
// 3 健身房开门   有比赛
#include <cstdio>
#include<cstring>
int a[150],b[150];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		int k=0;
		for(int i = 1 ; i  <= n ; i++)
		{
			if(a[i] == 0) 
			{
				b[i]=b[i-1]+1;
				k=0;
			}
			else
			{
				if(a[i] == 3) 
				{
					b[i]=b[i-1];
					k=3-k;
				}
				else
				{
					if(a[i] == k)
					{
						b[i]=b[i-1]+1;
						k=0;
					}
					else
					{
						b[i]=b[i-1];
						k=a[i];
					}
				}
			}
		}
		printf("%d\n",b[n]);
	}
}

代码二：

// 0 健身房关  没比赛
// 1 健身房关  有比赛 
// 2 健身房开  没比赛 
// 3 健身房开  有比赛/// 3 可以变为 2，1 

#include<cstdio>
#include<cstring>
int a[200];
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		memset(a,0,sizeof(a));
		int ans=0;
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(int i=1;i<=n;i++)
		{
			if(a[i]==0)
				ans++;
			else if(a[i]==1&&a[i-1]==1)
			{
				a[i]=0;
				ans++;
			}
			else if(a[i]==2&&a[i-1]==2)
			{
				a[i]=0;
				ans++;
			}
			else if(a[i]==3)
				a[i]=3-a[i-1];
		}
		printf("%d\n",ans);
	}
	return 0;
}