Codeforces 698A Vacations

Topic Link: Http://codeforces.com/problemset/problem/698/A topic to the effect:

Ah Q has a N-day holiday, there are three kinds of holidays to arrange rest, fitness, competition. There are three different options per day:

0 Gym doesn't open, no games.

1 Gym doesn't open, there's a game.

2 Gym open, no competition.

The 3 gym is open for games.

Please give AH Q reasonable arrangements for his holiday "Ah Q can not be attached to two days of fitness or a two-day match", so that a Q of the minimum number of rest days.

Problem Solving Ideas:

Ans=n, up to rest ans days

The first day was 3 a[0]=0,ans--;

As long as it is a[0]>0, then ans--;

Starting from the next day in the following day:

If the day is 1 then judge whether it was 1 days before

1: Make the current a[i]=0, rest today.

！ 1: Must exercise yesterday, ans--。

If the day is 2 then judge whether it was 2 days before

2: Make the current a[i]=0, rest today.

！ 2: Must match yesterday, ans--。

If the day is 3, make the key judgment.

If yesterday for 1 then today a[i]=2 ans--;

If yesterday for 2 then today A[i]=1 ans--;

If yesterday for 0 then today arbitrarily do a[i]=0 ans--; Focus

0 Do not judge anyway is rest ans remains unchanged.

Example:

7

1331123

1 ans--; 3 ans--, a[i]=2; 3 the day before 2 ans--, a[i]=1; 1 the day before 1 rest a[i]=0; 1 the day before was 0 ans--; 2 the day before was 1 ans--; 3 the day before was 2,a[i]=1 ans--;

Results: Ans=1.

AC Code:

1#include <bits/stdc++.h>2 using namespacestd;3 intMain ()4 {5     intN;6      while(~SCANF ("%d",&N))7     {8         intans=n,flag=0, a[n+2];9          for(intI=0; i<n; i++)Tenscanf"%d",&a[i]); One         if(a[0]>0) ans--; A         if(a[0]==3) a[0]=0; -          for(intI=1; i<n; i++) -         { the             Switch(A[i]) -             { -              Case 1: -                 if(a[i-1]!=1) ans--; +                 Elsea[i]=0; -                  Break; +              Case 2: A                 if(a[i-1]!=2) ans--; at                 Elsea[i]=0; -                  Break; -              Case 3: -                 if(a[i-1]==1) a[i]=2, ans--; -                 Else if(a[i-1]==2) a[i]=1, ans--; -                 Else if(a[i-1]==0) a[i]=0, ans--; in                  Break; -             } to         } +cout<<ans<<Endl; -     } the     return 0; *}

Codeforces 698A Vacations