Codeforces Round #358 (Div. 2) B. Alyona and Mex 水题
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
Hint
题意
给你一个序列，然后这个序列可以交换任意两个数的位置，以及减小若干个数的大小。
然后问你这个序列的mex最大能够是多少。
题解：
排序扫一遍，太水了……
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n,a[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+1+n);
int level=0;
for(int i=1;i<=n;i++)
{
if(a[i]>level)
level++;
}
cout<<level+1<<endl;
}最新文章
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