codeforce 698A Vacations——贪心——动态规划

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：每天有三种选择 运动 打比赛和休息 给出每天可以做的事情 相邻两天不能做相同的事情 求总共的最少休息天数


①贪心
思路：遍历每一天 与前一天作比较 相同就计数++并且把这天设为0 表示这天休息了（忘了归0 WA了两发） 不同就依据前一天推现在这一天要干什么

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#define max_ 100010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n,cnt=0;;int num[110];int main(int argc, char const *argv[]){scanf("%d",&n);int i;for(i=1;i<=n;i++)scanf("%d",&num[i]);if(num[1]==0)cnt++;for(i=2;i<=n;i++){if(num[i]==0)cnt++;if(num[i]==1&&num[i-1]==1)cnt++,num[i]=0;if(num[i]==2&&num[i-1]==2)cnt++,num[i]=0;if(num[i]==3){if(num[i-1]==1)num[i]=2;if(num[i-1]==2)num[i]=1;}}printf("%d\n",cnt);return 0;}

②动态规划

思路：用dp[i][j]表示第i天时选择j会有的休息天数 每一天有三种选择 分三个情况

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#define max_ 100010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n;int num[110],dp[110][4];int main(int argc, char const *argv[]){scanf("%d",&n);int i,j;memset(dp,0x3f,sizeof(dp));memset(dp[0],0,sizeof(dp[0]));for(i=1;i<=n;i++)scanf("%d",&num[i]);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(num[i]==1||num[i]==3)dp[i][1]=min(dp[i-1][0],dp[i-1][2]);if(num[i]==2||num[i]==3)dp[i][2]=min(dp[i-1][0],dp[i-1][1]);}printf("%d\n",min(dp[n][0],min(dp[n][1],dp[n][2])));return 0;}