USACO2011 November Gold Above the Median

问题 C: Above the Median

时间限制: 1 Sec  内存限制: 64 MB
提交: 9  解决: 5
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题目描述

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.

输入

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

输出

* 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.

样例输入

4 6

10

5

6

2

样例输出

7

提示

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

这个题目在落谷有连接：https://www.luogu.org/problemnew/show/3031

题目的意思就是：给出一串数字，问中位数大于等于X的连续子串有几个。（这里如果有偶数个数，定义为偏大的那一个而非中间取平均）

树状数组的方法可以看下：http://blog.csdn.net/kanosword/article/details/52607044

不知道是不是数据太少  O(nlogn) 和 O(n) 的时间一样 

#include<stdio.h>#include<iostream> #include <algorithm>#include<string.h>#include<vector>#include<math.h>#include<queue>#include<deque>#include<set>#define ll long long#define INF 0x3f3f3f3fconst int  mod = 1e9+7;using namespace std;int KGCD(int a,int b){if(a==0)return b;if(b==0)return a;if(~a&1){ if(b&1) return KGCD(a>>1,b);else return KGCD(a>>1,b>>1) <<1; } if(~b & 1)  return KGCD(a, b>>1);  if(a > b) return KGCD((a-b)>>1, b);return KGCD((b-a)>>1, a);}  int LCM(int a,int b){ return a/KGCD(a,b)*b; } inline ll qpow(ll n,ll m){n%=mod;ll ans=1;while(m){if(m%2) ans=(ans*n)%mod;m/=2;n=(n*n)%mod;}return ans;}inline ll inv(ll b){return b==1?1:(mod-mod/b)*inv(mod%b)%mod;}inline ll inv2(ll b){return qpow(b,mod-2);}int dir[5][2]={0,1,0,-1,1,0,-1,0};using namespace std;int sum[1000005];//记录前边所有的大小于x的和 int num[1000005];//记录那种i有多少个 int main(){int n,m,x;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&x);if(x>=m)//有多少符合题目的  从1-N sum[i]=sum[i-1]+1;elsesum[i]=sum[i-1]-1;} long long ans=0; int pre=0;//记录i-1的区间和 num[0]=1; //已经存在0的时候sum[0]=0  所以num[0]=1 for(int i=1;i<=n;i++) {if(sum[i]>sum[i-1])//如果当前可以符合题意pre=pre+num[sum[i]]+1;//加上i-1符合题意的区间 和 和sum[i]一样高的区间数目 再加上 当前这个else//不符合题意 {pre=pre-num[sum[i-1]]+1; //如果不符合题意 pre减去上一个区间添加的(后边详细注解） }num[sum[i]]++;ans=ans+pre; } printf("%lld\n",ans);} //因为它每次的波动最大就是1 所以如果当前小了，说明sum[i]==sum[i-2]所以直接翻转回去就OK了 //玄学思路 膜拜大佬