BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1614    Accepted Submission(s): 566


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.
 

 

Source
 
这道题有加强版--hdu 5400 有兴趣可以做下。
今天看到给秒了，上次百度之星碰到这种题是懵逼，这种题果然要多刷才会有经验。由于是中位数，所以我们分三种情况讨论。
1.往左边区间找,当大于M的数等于小于M的数时,M肯定是中位数，计数器+1。
2.往右边区间找同理。
3。对于左右两边，我们在往左边计数时弄一个数组记录大于(小于)M的数出现num个的次数为cnt[num],当往右边计数时,碰到大于(小于)M的数有num个时,对应左边有cnt[-num]个序列,计数器+=cnt[-num]，由于数组下标不能为负，所以加个大数N。
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
const int N = 40005;
int a[N];
int cnt[2*N];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        int id = -1;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            if(a[i]==m){
                id = i;
            }
        }
        memset(cnt,0,sizeof(cnt));
        int ans = 0;
        int num = 0,j=0;
        for(int i=id-1;i>=1;i--){ ///往左计数
            j++;
            if(a[i]<m) num++;
            else num--;
            if(num==0) ans++;
            cnt[N+num]++;
        }
        num = 0,j=0;
        for(int i=id+1;i<=n;i++){
            j++;
            if(a[i]<m) num++;
            else num--;
            if(num==0) ans++;
            ans+=cnt[N-num];
        }
        printf("%d\n",ans+1);
    }
}