一维点距的中位数 （分治）

Posted 2021-02-08 812-xiao-wen

题面

(solution:)

(code:)

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>

#define ll long long
#define db double
#define inf 0x7fffffff
#define rg register int

using namespace std;

int n,m,k,ans;
int a[50005];

inline int qr(){
    char ch;
    while((ch=getchar())<'0'||ch>'9');
    int res=ch^48;
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+(ch^48);
    return res;
}

inline bool check(ll x,int k){
    int tot=0;
    for(rg i=1,j=2;i<=n;++i){
        while(a[j]-a[i]<=x&&j<=n)++j;
        tot+=j-i-1;
    }
    return tot<k?0:1;
}

inline int fen(int k){
    ll l=1,r=inf,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(check(mid,k))r=mid-1;
        else l=mid+1;
    }return l;
}

int main(){
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
    n=qr(); m=(ll)n*(n-1)/2;
    for(rg i=1;i<=n;++i)a[i]=qr();
    sort(a+1,a+n+1);
    ans=fen(m/2+1);
    if(!(m&1))ans=((ll)ans+(ll)fen(m/2))/2;
    printf("%d
",ans);
    return 0;
}