codeforce B. Alyona and Mex

最新推荐文章于 2021-03-15 08:27:18 发布

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本文介绍了一种算法，用于解决给定整数数组通过减少操作后能达到的最大MEX值问题。MEX值定义为未出现在数组中的最小正整数。算法通过排序并遍历数组来确定这一值。

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.
Examples
Input

5
1 3 3 3 6

Output

Input

2
2 1

Output

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.
贪心一下，让最后的数成为缺的数，就是最大的！

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
int a[100005];
int main()
{
    int n;
    while(cin>>n)
    {

        for(int i=0;i<n;i++)
            cin>>a[i];#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
int a[100005];
int main()
{
    int n;
    while(cin>>n)
    {

        for(int i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        int k=1;
        for(int i=1;i<n;i++)
        {
            if(a[i]-k>1)
                k++;
            else k=a[i];
        }
        k++;
        cout<<k<<endl;
    }
    return 0;
}