codefrces 1203 E. Boxers(贪心）

E. Boxers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

There are ?
boxers, the weight of the ?-th boxer is ??. Each of them can change the weight by no more than 1

before the competition (the weight cannot become equal to zero, that is, it must remain positive). Weight is always an integer number.

It is necessary to choose the largest boxing team in terms of the number of people, that all the boxers’ weights in the team are different (i.e. unique).

Write a program that for given current values ??

will find the maximum possible number of boxers in a team.

It is possible that after some change the weight of some boxer is 150001

(but no more).
Input

The first line contains an integer ?
(1≤?≤150000) — the number of boxers. The next line contains ? integers ?1,?2,…,??, where ?? (1≤??≤150000) is the weight of the ?

-th boxer.
Output

Print a single integer — the maximum possible number of people in a team.
Examples
Input
Copy

4
3 2 4 1

Output
Copy

Input
Copy

6
1 1 1 4 4 4

Output
Copy

Note

In the first example, boxers should not change their weights — you can just make a team out of all of them.

In the second example, one boxer with a weight of 1
can be increased by one (get the weight of 2), one boxer with a weight of 4 can be reduced by one, and the other can be increased by one (resulting the boxers with a weight of 3 and 5, respectively). Thus, you can get a team consisting of boxers with weights of 5,4,3,2,1.

题意：给你一组数，你可以对一个数＋１或者－１也可以不操作，问你可以生成几个不同的数字；

思路：直接贪心，想好优先级，尽量－１，然后是不变，最后是＋１；

#include <iostream>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define ULL unsigned long long
using namespace std;
const int maxn=150000+10;
const int inf=0x3f3f3f3f;
int a[maxn];
int vis[maxn];
int main()
{int i,n,m,j,k,l;scanf("%d",&n);for(i=0;i<=n-1;i++){scanf("%d",&a[i]);}sort(a,a+n);int ans=0;for(i=0;i<=n-1;i++){if(a[i]-1>0&&vis[a[i]-1]==0){ans++;vis[a[i]-1]=1;}else if(a[i]>0&&vis[a[i]]==0){ans++;vis[a[i]]=1;}else if(vis[a[i]+1]==0){ans++;vis[a[i]+1]=1;}}printf("%d\n",ans);return 0;
}