• 題目
• 思路
• 程式碼

題目

\(n\)個數\(a_{1\ldots n}\),兩兩做差(大減小),得到陣列\(b\),求\(b\)的中位數.

思路

對原陣列排序,二分列舉中位數\(mid\),求有多少個做差陣列中比\(mid\)大/小的數分別由多少個,時間複雜度是\(O(n\log^2n)\),可以卡過去.

正解是雙指標?

程式碼

//#pragma GCC optimize(2)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>

#define int long long
using namespace std;
int read() {
    int re = 0;
    char c = getchar();
    bool negt = false;
    while(c < '0' || c > '9')negt |= (c == '-') , c = getchar();
    while(c >= '0' && c <= '9')re = (re << 1) + (re << 3) + c - '0' , c = getchar();
    return negt ? -re : re;
}

const int N = 2e5 + 10;
int abs_(int a) {
    return a < 0 ? -a : a;
}

int n , m;
int a[N];

vector <int> ans;
int check(int val) { //return 0(too small) 1(too big) 2(find answer)
    int sm = 0 , bg = 0;
    int v , l , r;
    for(int i = 1 ; i <= n ; ++i) {
        v = a[i] + val;
        l = lower_bound(a + 1 , a + n + 1 , v) - a , r = upper_bound(a + 1 , a + n + 1 , v) - a;
        if(a[l] >= v || l == n + 1)--l;
        sm += max(0ll , l - i) ;
        bg += n - r + 1;
    }
    int equ = m - sm - bg;
    if(sm >= (m + 1) / 2)return 1;
    if(sm + equ < (m + 1) / 2)return 0;
    if(equ == 0)return 0;
    ans.push_back(val);
    return 2;
}
int check2(int val) { //return 0(too small) 1(too big) 2(find answer)
    int sm = 0 , bg = 0;
    int v , l , r;
    for(int i = 1 ; i <= n ; ++i) {
        v = a[i] + val;
        l = lower_bound(a + 1 , a + n + 1 , v) - a , r = upper_bound(a + 1 , a + n + 1 , v) - a;
        if(a[l] >= v || l == n + 1)--l;
        sm += max(0ll , l - i) ;
        bg += n - r + 1;
    }
    int equ = m - sm - bg;
    if(sm >= m / 2 + 1)return 1;
    if(sm + equ < m / 2 + 1)return 0;
    if(equ == 0)return 0;
    ans.push_back(val);
    return 2;
}
signed main() {
	// freopen("mid.in" , "r" , stdin);
	// freopen("mid.out" , "w" , stdout);
	
    n = read() , m = n * (n - 1) / 2;
    for(int i = 1 ; i <= n ; i++)
        a[i] = read();
    sort(a + 1 , a + n + 1);
    
    
    int l = 0 , r = 1ll << 31;
    while(l <= r) {
        int mid = (l + r) / 2;
        int tmp = check(mid);
        if(tmp == 2)break;
        if(tmp == 0)l = mid + 1;
        else r = mid - 1;
    }
    if((m & 1) == 0) {//有兩個中位數
    	l = 0 , r = 1ll << 31;
	    while(l <= r) {
	        int mid = (l + r) / 2;
	        int tmp = check2(mid);
	        if(tmp == 2)break;
	        if(tmp == 0)l = mid + 1;
	        else r = mid - 1;
	    }
	}
	int sum = 0;
	for(int i : ans)
		sum += i;
	cout << sum / ans.size();
    return 0;
}