hdoj4908BestCoder Sequence

BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1546    Accepted Submission(s): 529


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.

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<vector>
using namespace std;
const int maxn=40010;
int num[maxn];
int vis[maxn];
int visa[maxn];
int visb[maxn];
int main()
{
int n,m,i,j,k,pos;
while(scanf("%d%d",&n,&m)!=EOF){
for(i=1;i<=n;++i){
scanf("%d",&num[i]);
if(num[i]==m)pos=i;
}
memset(vis,0,sizeof(vis));
memset(visa,0,sizeof(visa));
memset(visb,0,sizeof(visb));
int ans=1;
for(i=pos-1;i>=1;--i){
vis[i]=vis[i+1];
if(num[i]>m)vis[i]++;
if(((pos-i+1)&1)&&(vis[i]==(pos-i)/2))ans++;
int g=vis[i],s=(pos-i)-vis[i];
if(g-s>=0)visa[g-s]++;
else visb[s-g]++;
}
for(i=pos+1;i<=n;++i){
vis[i]=vis[i-1];
if(num[i]>m)vis[i]++;
if(((i-pos+1)&1)&&(vis[i]==(i-pos)/2))ans++;
int g=vis[i],s=(i-pos)-vis[i];
if(s-g>=0)ans+=visa[s-g];
else ans+=visb[g-s];
}
printf("%d\n",ans);
}
return 0;
}