CF1005E1 Median on Segments (Permutations Edition) 思维

Median on Segments (Permutations Edition)

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p1,p2,…,pnp1,p2,…,pn. A permutation of length nn is a sequence such that each integer between 11 and nn occurs exactly once in the sequence.

Find the number of pairs of indices (l,r)(l,r) (1≤l≤r≤n1≤l≤r≤n) such that the value of the median of pl,pl+1,…,prpl,pl+1,…,pr is exactly the given number mm.

The median of a sequence is the value of the element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4,2,7,5]a=[4,2,7,5] then its median is 44 since after sorting the sequence, it will look like [2,4,5,7][2,4,5,7] and the left of two middle elements is equal to 44. The median of [7,1,2,9,6][7,1,2,9,6] equals 66 since after sorting, the value 66 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l,r)(l,r) (1≤l≤r≤n1≤l≤r≤n) such that the value of the median of pl,pl+1,…,prpl,pl+1,…,pr is exactly the given number mm.

Input

The first line contains integers nn and mm (1≤n≤2⋅1051≤n≤2⋅105, 1≤m≤n1≤m≤n) — the length of the given sequence and the required value of the median.

The second line contains a permutation p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n). Each integer between 11 and nn occurs in pp exactly once.

Output

Print the required number.

Examples

input

Copy

5 4
2 4 5 3 1

output

Copy

4

input

Copy

5 5
1 2 3 4 5

output

Copy

1

input

Copy

15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9

output

Copy

48

Note

In the first example, the suitable pairs of indices are: (1,3)(1,3), (2,2)(2,2), (2,3)(2,3) and (2,4)(2,4).

题意：给你n个数和m，问在这n个数中以m为中位数的区间有多少个？

因为要使m为中位数，肯定是m的值位于这些数的中间的部分，即要有比m大的数和比m小的数，且大的数和小的数要相等或者大的数比小的数多一

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + ;
const int mod = 1e9 + ;
typedef long long ll;
ll a[maxn], vis[maxn];
int main() {
    ll n, m;
    while( cin >> n >> m ) {
        map<ll,ll> mm;
        ll  pos;
        for( ll i = ; i <= n; i ++ ) {
            cin >> a[i];
            if( a[i] == m ) {
                pos = i;
            }
        }
        ll cnt = ;
        for( ll i = pos; i <= n; i ++ ) {
            if( a[i] > m ) {
                cnt ++;
            } else if( a[i] < m ) {
                cnt --;
            }
            mm[cnt] ++;
        }
        ll ans = ;
        cnt = ;
        for( ll i = pos; i >= ; i -- ) {
            if( a[i] > m ) {
                cnt ++;
            } else if( a[i] < m ) {
                cnt --;
            }
            ans += mm[-cnt];
            ans += mm[-cnt];  //个数为偶数，中位数在中间两位的左边一位
        }
        cout << ans << endl;
    }
    return ;
}