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
Hint：
For the second case, {3},{5,3,2},{4,5,3,2,1} are Bestcoder Sequence.
题意：有一个1~n的序列，现在我们要找到这个序列以m为中位数（eg：1 2 3 4 5 || 5 4 3 2 1，这是两个以3为中位数的序列）的连续子序列有多少个。
这里先来看一下1 2 3 4 5这个序列，首先我们可以得到它的sum值（请结合下面的代码来看）：
a[i] 1 2 3 4 5
sum[i] -1 -2 -2 -1 0
由于我们下标都是从1开始的，那么我们可以加上a[0]和sum[0]，变成
a[i] 0 1 2 3 4 5
sum[i] 0 -1 -2 -2 -1 0
离散化之后（变成正数，便于模拟，这里假如加上的是40000）
a[i] 0 1 2 3 4 5
sum[i] 0 -1 -2 -2 -1 0
sum[i]+40000 40000 39999 39998 39998 39999 40000
首先从0~x-1遍历，这时b[40000] = 1， b[39999] =1， b[39998]=1；那我们在遍历x~n时由上面分析得也会遇见这三个值，那么我们最终就会发现满足条件的序列就是三个。（注意b数组的下标是sump[i]+40000）
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;
const int N=1e6+10;
const int M=50000;
const int INF=0x3f3f3f3f;
int a[N], sum[N], b[N];
int main ()
{
int n, m, i, x, ans;
while (scanf("%d%d", &n, &m) != EOF)
{
memset(sum, 0, sizeof(sum));
memset(b, 0, sizeof(b));
ans = 0;
for (i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
sum[i] = sum[i-1];
if (a[i] < m) sum[i]--; ///比中位数小--
if (a[i] > m) sum[i]++; ///比中位数大++
if (a[i] == m)
x = i;
}
for (i = 0; i < x; i++)
b[sum[i]+M]++; ///由于我们的sum值可能是负数，那么我们需要将所有的数变成正数，为了结果不受影响，加上一个比较大的数
for (i = x; i <= n; i++)
ans += b[sum[i]+M]; ///首先我们要明白当两个位置的sum值相等时，那么这中间就会产生一个符合题意的序列
printf("%d\n", ans);
}
return 0;
}