Duizi and Shunzi(贪心)

Nike likes playing cards and makes a problem of it.

Now give you n integers, ai(1≤i≤n)

We define two identical numbers (eg: 2,2) a Duizi,
and three consecutive positive integers (eg: 2,3,4) a Shunzi.

Now you want to use these integers to form Shunzi and Duizi as many as possible.

Let s be the total number of the Shunzi and the Duizi you formed.

Try to calculate max(s)

.

Each number can be used only once.

Input The input contains several test cases.

For each test case, the first line contains one integer n( 1≤n≤106).
Then the next line contains n space-separated integers ai ( 1≤ai≤n)
Output For each test case, output the answer in a line.
Sample Input

7
1 2 3 4 5 6 7
9
1 1 1 2 2 2 3 3 3
6
2 2 3 3 3 3 
6
1 2 3 3 4 5

Sample Output

Hint

Case 1（1，2，3）（4，5，6）

Case 2（1，2，3）（1，1）（2，2）（3，3）

Case 3（2，2）（3，3）（3，3）

Case 4（1，2，3）（3，4，5）

这道题看起来好像只能用dp，其实正解贪心就可以了。（还是太年轻）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

typedef long long LL;
using namespace std;
int a[1000010];
int book[1000010];
int main()
{
    int n,ans;
    while(scanf("%d",&n)!=EOF){
            ans=0;
            memset(book,0,sizeof(book));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            book[a[i]]++;
        }
        for(int i=1;i<=n;i++){
            if(book[i-1]==1&&book[i-2]==1&&book[i]>0){
                ans++;
                book[i]--;
                book[i-2]--;
                book[i-1]--;
            }//先处理前面只剩一个而且又能组成顺子的数
            ans=ans+book[i]/2;//处理对子
            book[i]=book[i]%2;
        }
        printf("%d\n",ans);
    }
    return 0;
}

Duizi and Shunzi(贪心)

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