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）

题意：给出一个长度为n的序列，要求求出对子数与顺子数总数最大值，2个相同的数字称为对子，3个相邻的数字称为顺子（如：2,3,4）

思路：贪心，因为组成对子的扑克数量比组成顺子的扑克数量大，要想总数最大，就尽可能的组成对子

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1e6+5;
int vis[N];
int main(){
	int n;
	while(scanf("%d",&n)!=EOF){
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=n;i++){
			int x;
			scanf("%d",&x);
			vis[x]++;
		}
		int ans=0;
		for(int i=1;i<=n;i++){
			if(vis[i]>=2){
				ans+=vis[i]/2;
				vis[i]%=2;
			}
			if(vis[i]&&vis[i+1]%2==1&&vis[i+2]){
				ans++;
				vis[i]--;
				vis[i+1]--;
				vis[i+2]--;
			}
			  
		}
		printf("%d\n",ans);
	}
	return 0;
}

Duizi and Shunzi （贪心）

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