Codeforces Round #435 (Div. 2) A. Mahmoud and Ehab and the MEX(思路)

题目：

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
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

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.

思路：

给出一个集合，定义一个集合的MEX为不存在于这个集合中的最小非负整数，问需要最少进行几次操作可以使一个集合的MEX为样例给出的数。把集合中出现过的先标记一下，然后遍历输出

代码：

#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<iostream>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100000+20#define ll longlong
using namespace std;
int a[100+20];
int vis[100+20];
int main()
{
int n,x;
mem(vis,0);
scanf("%d%d",&n,&x);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
vis[a[i]]=1;
}
sort(a,a+n);
if(x==0)
{
if(a[0]==0)
puts("1");
else
puts("0");
}
else
{
int sum=0;
for(int i=0; i<x; i++)
{
if(!vis[i])
sum++;
}
if(vis[x])
sum++;
printf("%d\n",sum);
}
return 0;
}