CF 862A Mahmoud and Ehab and the MEX【数组操作】

A. Mahmoud and Ehab and the MEX

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.

Dr. Evil is interested in sets, He has a set of n integers. Dr. Evil calls a set of integers evil if theMEX of it is exactly x. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0, 2, 4} is 1 and the MEX of the set{1, 2, 3} is 0 .

Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?

Input

The first line contains two integers n and x (1 ≤ n ≤ 100, 0 ≤ x ≤ 100) — the size of the set Dr. Evil owns, and the desired MEX.

The second line contains n distinct non-negative integers not exceeding 100 that represent the set.

Output

The only line should contain one integer — the minimal number of operations Dr. Evil should perform.

Examples

input

5 3
0 4 5 6 7

output

2

input

1 0
0

output

1

input

5 0
1 2 3 4 5

output

0

Note

For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.

For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.

In the third test case the set is already evil.

 

【题意】：给你n个数（从0开始）和特定的MEX数，可以增加或者删除数。求在该数组中若让MEX为最小值，需要几次增加和删除操作。

【分析】：只要让0~x之间标记没有出现的数字补位，用计数器++即可。注意若给你的MEX数在数组中已经有，那么要删除原数组的已有数，ans要+1.

【代码】：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 105;
int n,m;

int main()
{
    cin>>n>>m;
    int a[maxn];
    int mp[maxn];
    memset(mp,0,sizeof(mp));
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        mp[a[i]]=1;
    }

    int cnt=0;
    if(mp[m]==1) cnt++;
    for(int i=0;i<m;i++)
    {
        if(mp[i]==0)
            cnt++;
    }
    cout<<cnt<<endl;
    return 0;
}

#include <iostream>
using namespace std;
bool b[105];
int main()
{
    int n,x;
    scanf("%d%d",&n,&x);
    while (n--)
    {
        int a;
        scanf("%d",&a);
        b[a]=1;
    }
    int ans=b[x];
    for (int i=0;i<x;i++)
    ans+=!b[i];
    printf("%d",ans);
}