A. Mahmoud and Ehab and the MEX

最新推荐文章于 2019-03-22 14:16:10 发布

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最新推荐文章于 2019-03-22 14:16:10 发布

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原文链接：http://www.cnblogs.com/zllwxm123/p/7562377.html

本文介绍了一个有趣的算法问题：Dr.Evil希望通过最少的操作使他的整数集合达到特定的MEX值。文章提供了完整的代码实现，并解释了通过标记数组检查并补全缺失值的方法。

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 the MEX 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

Input

1 0
0

Output

Input

5 0
1 2 3 4 5

Output

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.

弄一个标记数组,然后合适补位就行了.

 1 #include <bits/stdc++.h>
 2 #define N 105
 3 using namespace std;
 4 int n,x;
 5 int k[N],m[N];
 6 int main(){
 7     cin>>n>>x;
 8     for(int i=0;i<n;i++){
 9       cin>>k[i];
10       m[k[i]]=1;
11     }
12     int cnt=0;
13     if(m[x]==1){
14       cnt++;
15     }
16     for(int i=0;i<x;i++){
17       if(m[i]==0)
18         cnt++;
19     }
20     cout<<cnt<<endl;
21   return 0;
22 }