(Jizhong) 2200. [Training] GDKOI card games (card) [greedy]

(File IO): input: card.in output: card.out
time limit: 1000 ms space constraints: 262144 KB specific restrictions
Goto ProblemSet

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Title Description
$Bessie$ is very much like a card cows, though she did not thumb, but she has a near-obsession with the love of playing cards. Unfortunately, other cattle in the herd are not good opponent. Their level is really bad. They are always in a completely predictable manner to play cards! despite this, $Bessie$ still can choose how to win.
$Bessie$ and her friend Elsie playing a simple card game, they have got to pay a $2n$ cards cards, digital numbers on the card is $1- 2n$ , and are divided into two, to a card $Bessie$ and a card to $Elsie$ .
Then they began to play cards, a total of conduct $n$ wheel, in each round, $Bessie$ 和 $Elsie$ are playing a card, the cards who have a big points.
magical $Bessie$ can be predicted $Elsie$ cards sequence, and the like as in winning. please confirm $Bessie$ the maximum number of points can be earned.

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Enter
the first line an integer $N (1≤N≤50,000)$ .
The next $N$ line is $Elsie$ to round out the continuous game cards. Please note that it is easy to determine from the information $Bessie$ card.

Output
line is given $Bessie$ can score the maximum number of points.

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Sample input
. 3
. 1
. 6
. 4

Sample Output
2

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Data range limit

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Tip
Bessie hands of cards are $2,3，5$ , according to her $2，3，5$ sequential 2 points cards can be obtained.

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Problem-solving ideas
greedy.
Prepare two orderly queues, with the principle of Tian Ji's horse.

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int n,m,a[100010],b[50010],c[50010],x,ans,k,k1,j;
int main(){
 freopen("card.in","r",stdin);
  freopen("card.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        a[x]=1;
    }
    k=0,k1=0;
    for(int i=1;i<=n*2;i++)
        if(!a[i])
            b[++k]=i;
        else
            c[++k1]=i;
    j=1,ans=0;
    for(int i=1;i<=n;i++)
    {
        if(c[j]<b[i])
        {
            j++;
            ans++;
            if(j>n)
            {
                printf("%d",ans);
                return 0;
            }
        }
    }
    printf("%d",ans);
}