# BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1917    Accepted Submission(s): 635

Problem Description
Mr Potato is a coder.
Mr Potato is the BestCoder.One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence.As the best coder, Mr potato has strong curiosity, he wonder the number of consecutive sub-sequences which are bestcoder sequences in a given permutation of 1 ~ N.

Input
Input contains multiple test cases.
For each test case, there is a pair of integers N and M in the first line, and an permutation of 1 ~ N in the second line.[Technical Specification]
1. 1 <= N <= 40000
2. 1 <= M <= N

Output
For each case, you should output the number of consecutive sub-sequences which are the Bestcoder Sequences.

Sample Input
1 1
1
5 3
4 5 3 2 1

Sample Output
1
3

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<stack>

#define MAXX 41000
int a[MAXX],hashh[MAXX<<1];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int k=0,ans=0;
for(int i=0; i<n; i++)
{
scanf("%d",a+i);
if(a[i]==m)k=i;
}
memset(hashh,0,sizeof(hashh));
hashh[MAXX]=1;
for(int i=k+1,l=0; i<n; i++)
{
if(a[i]<m)l--;
else l++;
hashh[MAXX+l]++;
}
ans+=hashh[MAXX];
for(int i=k-1,l=0; i>=0; i--)
{
if(a[i]<m)l--;
else l++;
if(hashh[MAXX-l])
ans+=hashh[MAXX-l];
}
cout<<ans<<endl;
}
return 0;
}