找区间，计算比指定数大或小的数量之差，两个数量相等的区间的中间区间 为所求答案

E1. 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 $p_1,p_2,\dots ,p_n$. A permutation of length $n$ is a sequence such that each integer between $1$ and $n$ occurs exactly once in the sequence.

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

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 $a=[4,2,7,5]$ 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 the median of $p_l,p_{l+1},\dots ,p_r$ is exactly the given number $m$.

Input

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

The second line contains a permutation $p_1,p_2,\dots ,p_n$ ($1\le p_i\le n$). Each integer between $1$ and $n$ occurs in $p$ 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: $(1,3)$, $(2,2)$, $(2,3)$ and $(2,4)$.

找一段序列中有多少个子序列使这个子序列排完序后中间元素为指定元素

暴力法  超时

#include<bits/stdc++.h>
using namespace std;
int a[200005],b[200005];
int main()
{
    int n,m,sum=0;
    cin>>n>>m;
    int i;
    for(i=0;i<n;i++)
        {
            cin>>a[i];
            b[i]=a[i];
        }
    int j;
    for(i=0;i<n;i++)
    {
        for(j=i;j<n;j++)
        {

            sort(b+i,b+j+1);
            if(b[(i+j)/2]==m)
                sum++;
            for(int k=i;k<=j;k++)
                b[k]=a[k];

        }
    }
    cout<<sum;
}

标准解答玄学。。对于m后面的数 若大于m和小于m的数量差和前面相等则这一段区间就可以，也可以加上前面的小数比大数多1这样这段区间里大数会比小数多1

#include<bits/stdc++.h>
using namespace std;
long long ans=0;
map<int,int>mp;
int n,m,cnt=0;
int main()
{

    scanf("%d%d",&n,&m);
    int i;
    mp[0]=1;
    int flag=0;
    for(i=0;i<n;i++)
    {
        int k;
        scanf("%d",&k);
        if(k==m)
            flag=1;
        else if(k>m)
            cnt++;
        else if(k<m)
            cnt--;
        if(!flag)
            mp[cnt]++;
        if(flag)
            ans=ans+mp[cnt]+mp[cnt-1];
    }
    printf("%lld",ans);

}