从前往后扫，找到一出现次数为0的数，从后面找一个出现不为0的数转化而来。设置两指针l, r来处理。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 1000008, INF = 0x3F3F3F3F;
int a[N], cnt[N], rem[N];
int main(){
int n, s;
cin >> n >> s;
for(int i = 1; i <= n; i++){
scanf("%d", &a[i]);
}
int ans = 0;
if(a[s]){
ans++;
a[s] = 0;
}
for(int i = 1; i <= n; i++){
cnt[a[i]]++;
}
cnt[n] += cnt[0] - 1;
int l = 1, r = n;
while(l < r){
while(cnt[l] && l < r){
l++;
}
while(cnt[r] == 0 && l < r){
r--;
}
if(l < r){
cnt[l]++;
cnt[r]--;
ans++;
}
}
cout<<ans<<endl; return 0;
}