CodeForces 682BAlyona and Mex

http://codeforces.com/problemset/problem/682/B

B. Alyona and Mex time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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 every1 ≤ 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.

Examples input 5 1 3 3 3 6 output 5 input 2 2 1 output 3 Note

In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.

To reach the answer to the second sample case one must not decrease any of the array elements.

题意：

没看懂题目。

思路：

参考了官方的题解，引入变量 cur=1 对排序后的数组进行遍历，如果a[i] > cur -> cur++；

Code：

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MYDD=1103+1e5; int a[MYDD]; int main() { int n; while(scanf("%d",&n)!=EOF) { for(int j=0; j<n; j++) scanf("%d",&a[j]); int cur=1;; sort(a,a+n); for(int j=0; j<n; j++) { if(a[j]>=cur) cur++; } printf("%d\n",cur); } return 0; } /* By:Shyazhut */