給出1~n的一個排列，統計該排列有多少個長度為奇數的連續子序列的中位數是b。中位數是指把所有元素從小到大排列後，位於中間的數。 Input 第一行為兩個正整數n和b ，第二行為1~n 的排列。 Output 輸出一個整數，即中位數為b的連續子序列個數。 Sample Input 7 4 5 7 2 4 3 1 6 Sample Output 4 Hint

第三個樣例解釋：{4}, {7,2,4}, {5,7,2,4,3}和{5,7,2,4,3,1,6}
N<=100000

map 對映就好

#include<iostream>
#include<cstdio>
#include
 
<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
 

#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned
 
 int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 1e9 + 7;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

int n, b;
int a[maxn];
map<int, int>Map;

int main()
{
	//ios::sync_with_stdio(false);
	rdint(n); rdint(b);
	int pos;
	for (int i = 1; i <= n; i++) {
		rdint(a[i]);
		if (a[i] == b)pos = i;
	}
	int cnt = 0;
	for (int i = pos; i <= n; i++) {
		if (a[i] > b)cnt++;
		if (a[i] < b)cnt--;
		Map[cnt]++;
	}
	cnt = 0;
	ll sum = 0;
	for (int i = pos; i >= 1; i--) {
		if (a[i] > b)cnt++;
		if (a[i] < b)cnt--;
		sum += (ll)Map[0 - cnt];
	}
	cout << sum << endl;
	return 0;
}