codeforcesE2. Median on Segments (General Case Edition)

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You are given an integer sequence $a_1,a_2,\dots ,a_n$.

Find the number of pairs of indices $(l,r)$ ($1\le l\le r\le n$) such that the value of median of $a_l,a_{l+1},\dots ,a_r$ is exactly the given number $m$.

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

Write a program to find the number of pairs of indices $(l,r)$ ($1\le l\le r\le n$) such that the value of median of $a_l,a_{l+1},\dots ,a_r$ is exactly the given number $m$.

Input

The first line contains integers $n$ and $m$ ($1\le n,m\le 2\cdot {10}^5$) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 2\cdot {10}^5$).

Output

Print the required number.

Examples

input

Copy

5 4
1 4 5 60 4

output

Copy

8

input

Copy

3 1
1 1 1

output

Copy

6

input

Copy

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

output

Copy

97

Note

In the first example, the suitable pairs of indices are: $(1,3)$, $(1,4)$, $(1,5)$, $(2,2)$, $(2,3)$, $(2,5)$, $(4,5)$ and $(5,5)$.

题解：

我考试的时候真是脑抽，一点都没想到用前缀和做中位数。。。

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思路：

首先我们计算出solve(m):中位数大于等于m的方案数，那么最后答案就是solve(m) - solve(m+1)

那么怎么计算sovle(m)呢？

对于一个区间[l,r]，如果它的中位数大于等于m，那么这个区间中 （大于等于m的数的个数） > （小于m的数的个数）

如果记a[i]大于等于m为+1，小于m 为 -1，即 sum(l， r)  > 0

我们枚举右端点 i ，并且同时计算sum(1, i) ，那么对于这个右端点，我们只要找到之前的 sum 中 < sum(1, i)的个数（左端点的个数），这个可以用树状数组维护

但是我们有一个O(n)的方法求，用了类似莫队的方法，记s[i]为之前的sum为i的个数，add为上一个小于sum(1, i-1)的个数，对于当前的sum，

如果它要加1，add += s[sum],  sum++

如果它要减1，sum --, add -= s[sum]

这样得出的add就是当前的小于sum(1, i)的个数

代码：

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<int,pii>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5;
int a[N], cnt[N*2], n, m;
LL solve(int m) {
    int s = n;
    mem(cnt, 0);
    cnt[s] = 1;
    LL add = 0, ans = 0;
    for (int i = 1; i <= n; i++) {
        if(a[i] >= m) add += cnt[s], s++;
        else s--, add -= cnt[s];
        cnt[s]++;
        ans += add;
    }
    return ans;
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    printf("%lld\n", solve(m) - solve(m+1));
    return 0;
}