Let median of some array be the number which would stand in the middle of this array if it was sorted beforehand. If the array has even length let median be smallest of of two middle elements. For example, median of the array [10,3,2,3,2][10,3,2,3,2] is 33 (i.e. [2,2,3–,3,10][2,2,3_,3,10]). Median of the array [1,5,8,1][1,5,8,1] is 11 (i.e. [1,1–,5,8][1,1_,5,8]).
Let array be mm-good if its median is greater or equal than mm.
Let the partition of array [a1,a2,…,an][a1,a2,…,an] be a set of subarrays {b1,b2,…,bk}{b1,b2,…,bk} such that b1=[a1,a2,…,ai1]b1=[a1,a2,…,ai1], b2=[ai1+1,ai1+2,…,ai2]b2=[ai1+1,ai1+2,…,ai2], ..., bk=[aik−1+1,aik−1+2,…,an]bk=[aik−1+1,aik−1+2,…,an]. For example, array [10,3,2,3,2][10,3,2,3,2] can be partitioned as follows: {[10,3,2,3,2]}{[10,3,2,3,2]} or {[10],[3],[2],[3],[2]}{[10],[3],[2],[3],[2]}, or {[10],[3,2,3,2]}{[10],[3,2,3,2]}, or {[10,3],[2],[3,2]}{[10,3],[2],[3,2]} and so on.
You are given array aa of length nn and integer mm. Find the partition of aa into maximum number of subarrays such that each subarray is mm-good.
Input
The first line contains two integers nn and mm (1≤n≤50001≤n≤5000, 1≤m≤50001≤m≤5000) — length of array aa and constant mm.
The second line contains nn integers a1a1, a2a2, ..., anan (1≤ai≤50001≤ai≤5000)— array aa.
Output
If there is no valid partition of array aa into mm-good subarrays, print 00. Otherwise print maximum number of subarrays in partition of array aa such that each subarray is mm-good.
Examples
Input
5 2 10 3 2 3 2
Output
5
Input
5 3 10 3 2 3 2
Output
1
Input
5 4 10 3 2 3 2
Output
0
题意：最多能分成多少个连续子序列都满足中位数大于等于m
题解：一个序列中位数大于等于m，当然大于等于m的数量 - 小于m的数量 >= 1 ，所以我们就让他等于1即可，就让每个序列大于等于m的数量比小于m的数量多一个，为什么一定符合呢，因为若当前序列不符合肯定要再增加两边的数，增加后若符合即可，若增加后还不符合就继续增加，但此时大于等于m的数量 是小于等于 小于m的数量。因此我们就统计一下大于等于m的数量和小于m的数量即可
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
int n,m;
int main()
{
while(~scanf("%d%d",&n,&m))
{
int x;
int s1=0,s2=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
if(x<m) s1++;
else s2++;
}
printf("%d\n",max(0,s2-s1));
}
return 0;
}