HDOJ4908 - BestCoder Sequence

题目出处：HDOJ4908

BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1514    Accepted Submission(s): 516

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

1 115 34 5 3 2 1

 

Sample Output

13
Hint
For the second case, {3},{5,3,2},{4,5,3,2,1} are Bestcoder Sequence.
 

 

中位数，先向左遍历，再向右遍历，比中位数大的count＋＋，小的count－－，把结果存到数组或者map，如果出现和为0，即相反数，代表一定包含中位数的字串，统计字串个数。the accepted code like this:

////  main.cpp//  B////  Created by IntelliegeWither on 16/1/27.//  Copyright © 2016年 IntelliegeWither. All rights reserved.//#include <iostream>#include <stdio.h>using namespace std;const int MAX=40010;int main(){    int i;    int a[MAX];    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        int index[MAX]={0};        int sign = 0;        for(i=1;i<=n;i++)        {            scanf("%d",a+i);            if(a[i]==m)            {                sign=i;            }        }        int sum=0;        int count=0;        for(i=sign;i>0;i--)        {            if(a[i]>m)            {                count++;            }            if(a[i]<m)            {                count--;            }            index[MAX/2+count]++;        }        sum=0;        count=0;        for(i=sign;i<=n;i++)        {            if(a[i]>m)            {                count++;            }            if(a[i]<m)            {                count--;            }            sum+=index[MAX/2-count];        }        cout<<sum<<endl;    }}