p1627 [CQOI2009]中位数

传送门

分析

https://www.luogu.org/blog/user43145/solution-p1627

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define li long long
const int N = 1e5;
li a[2*N+10],b[2*N+10],Ans=1;
int d[N+10];
int main(){
    int n,m,i,j,k;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;i++){
      scanf("%d",&d[i]);
      if(d[i]<k)d[i]=-1;
        else if(d[i]>k)d[i]=1;
        else d[i]=0,j=i;
    }
    m=0;
    for(i=j-1;i>0;i--){
      m+=d[i];
      a[m+N]++;
    }
    m=0;
    for(i=j+1;i<=n;i++){
      m+=d[i];
      b[m+N]++;
    }
    Ans+=a[N]+b[N];
    for(i=0;i<=2*N;i++)
      Ans+=a[i]*b[2*N-i];
    cout<<Ans;
    return 0;
}

标签：10,int,中位数,long,li,CQOI2009,p1627,include
来源： https://www.cnblogs.com/yzxverygood/p/10354465.html