HDU4908 BestCoder Sequence

BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1955    Accepted Submission(s): 653

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.
 

题意：给一个数字序列1~n（n个数字各不相同），和一个数字m，求m作为连续的自序列的中位数的序列个数。

题解：组合数学。先找到m，开一个数组f[2][N]，分别遍历m的左右两边，如果数字大于m，则cnt++，小于则cnt--，然后在f数组中记录下cnt的个数，即记录m左右两边区间比m大的和小的差值的个数，再相乘。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long ll;#define N 40005 int a[N];long long f[2][N*2];void fun(int n, int m, int pos){memset(f, 0, sizeof(f));int cnt = 0;for(int i = pos-1; i >= 0; i--){if(a[i] > m)cnt++;elsecnt--;f[0][cnt+n]++;}cnt = 0;for(int i = pos+1; i < n; i++){if(a[i] > m)cnt++;elsecnt--;f[1][cnt+n]++;}}int main(){int n, m, x;while(scanf("%d%d", &n, &m) != EOF){ll ans = 0;int pos;for(int i = 0; i < n; i++){scanf("%d", &a[i]);if(a[i] == m)pos = i;}fun(n, m, pos);for(int i = 0; i <= n; i++){ans += f[0][i+n] * f[1][n-i] + f[1][i+n] * f[0][n-i];}ans += f[0][n] + f[1][n] - f[0][n]*f[1][n];printf("%lld\n", ans + 1);}return 0;}