[BestCoder Round #3] hdu 4908 BestCoder Sequence (计数）

BestCoder Sequence

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

field=problem&key=BestCoder%20Round%20#3&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">BestCoder Round #3

解题思路：

题意为 给定一个1 -N的排列，再给定一个数M（1<=M<=N)。问有多少连续的长度为奇数子序列，使得M在当中为中位数（M在子序列中）。

比方例子

5 3

4 5 3 2 1 N=5， M=3

{3},{5,3,2},{4,5,3,2,1} 为符合题意的连续子序列....

当时做的时候把包括M的全部长度为0,1,2.......的连续子序列都枚举了出来。然后推断推断M是否是中位数。结果
果断超时.......

贴一下题解思路：

写了一堆字，CSDN的排版太....了，贴图片把.

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=40010;
int sum[maxn];
int cnt[maxn*2];
int n,m;
int val,p;
 
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1];
            scanf("%d",&val);
            if(val<m)
                sum[i]--;
            else if(val>m)
                sum[i]++;
            else
                p=i;
        }
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<p;i++)
            cnt[sum[i]+maxn]++;
        int ans=0;
        for(int i=p;i<=n;i++)
            ans+=cnt[sum[i]+maxn];
        printf("%d\n",ans);
    }
    return 0;
}