Codeforces Round #358 (Div. 2) B. Alyona and Mex

B. Alyona and Mex

题目连接：

http://www.codeforces.com/contest/682/problem/B

Description

Someone gave Alyona an array containing n positive integers a1, a2, ..., an. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.

Formally, after applying some operations Alyona will get an array of n positive integers b1, b2, ..., bn such that 1 ≤ bi ≤ ai for every 1 ≤ i ≤ n. Your task is to determine the maximum possible value of mex of this array.

Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of elements in the Alyona's array.

The second line of the input contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.

Sample Input

5
1 3 3 3 6

Sample Output

5

题意：给你一个序列，可以变换两个数的位置，也可以将一个数变为比其本身小的数，问最后形成的序列的最大值+1

思路：排序，然后扫一遍就可以了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int a[MAXN];
int main()
{
    int n;scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    sort(a,a+n);
    int ans=0,num=1;
    for(int i=0;i<n;i++)
    {
        if(a[i]<num)
            ans++;
        else
            num++;
    }
    printf("%d\n",num);
    return 0;
}