CodeForces 682B

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121183#problem/B

B - Alyona and Mex

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Someone gave Alyona an array containing n positive integers a₁, a₂, ..., a_n. 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, ..., b_n such that 1 ≤ b_i ≤ a_i 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, ..., a_n (1 ≤ a_i ≤ 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

Input

5
1 3 3 3 6

Output

5

Input

2
2 1

Output

3

题意:给出一串数可以对该串进行处理(对每一个数都可以任意减(只要保证最后得到的数是正数) ),也可以不处理,找到串中不存在的最小正数(mex)( Mex  of an array in this problem is the  minimum positive  integer that doesn't appear in this array.)任务是找到这些mex的最大值

思路:把a数组处理成连续的非降的数组,找到这个b数组中的最大值 加1即为所求

#include
using namespace std;
typedef long long ll;
const int maxn=100000+5;
int a[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i

@1:注意maxx从0开始和从1开始的区别