# BSOJ3686 -- 【USACO 2011 November Gold】Above the Median【树状数组统计】

dzy posted @ 2013年8月07日 20:24 in BSOJ with tags usaco 树状数组 , 2472 阅读

### Description

Farmer John has lined up his N (1 <= N <= 100,000) cows in a row to measure their heights; cow i has height H_i (1 <= H_i <= 1,000,000,000) nanometers--FJ believes in precise measurements! He wants to take a picture of some contiguous subsequence of the cows to submit to a bovine photography contest at the county fair.

The fair has a very strange rule about all submitted photos: a photograph is only valid to submit if it depicts a group of cows whose median height is at least a certain threshold X (1 <= X <= 1,000,000,000).

For purposes of this problem, we define the median of an array A[0...K] to be A[ceiling(K/2)] after A is sorted, where ceiling(K/2) gives K/2 rounded up to the nearest integer (or K/2 itself, it K/2 is an integer to begin with). For example the median of {7, 3, 2, 6} is 6, and the median of {5, 4, 8} is 5.

Please help FJ count the number of different contiguous subsequences of his cows that he could potentially submit to the photography contest.

### Input

* Line 1: Two space-separated integers: N and X.
* Lines 2..N+1: Line i+1 contains the single integer H_i.

### Output

* Line 1: The number of subsequences of FJ's cows that have median at least X. Note this may not fit into a 32-bit integer.

### Sample Input

4 6
10
5
6
2

INPUT DETAILS:

FJ's four cows have heights 10, 5, 6, 2. We want to know how many contiguous subsequences have median at least 6.

### Sample Output

7

OUTPUT DETAILS:

There are 10 possible contiguous subsequences to consider. Of these, only 7 have median at least 6. They are {10}, {6}, {10, 5}, {5, 6}, {6, 2}, {10, 5, 6}, {10, 5, 6, 2}.

（新本本好好用。呵呵呵。）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=210000,d=100010;
int n,m,a,bit[maxn+10];
void modify(int x){
for(;x<=maxn;x+=x&-x)	bit[x]++;
}
int r=0;
for(;x;x-=x&-x)	r+=bit[x];
return r;
}
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
if(a[i]>=m)	a[i]=1;	else a[i]=-1;
}
int sum=0;
long long An=0;
modify(d);
for(int i=1;i<=n;i++){
sum+=a[i];
modify(sum+d);
}
cout << An << endl;
return 0;
} Zachary Chelmsford 说:
2019年2月18日 16:22

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