[Educational Codeforces Round 16]B. Optimal Point on a Line

转载于 2016-08-25 14:40:00 发布 · 105 阅读

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原文链接：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/5806721.html

本文解析了Educational Codeforces Round 16中的B题——如何找到一条直线上给定点集的最优位置，使得该点到所有给定点的距离之和最小。通过排序并输出中位数的方式得出解答。

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[Educational Codeforces Round 16]B. Optimal Point on a Line

试题描述

You are given n points on a line with their coordinates x_i. Find the point x so the sum of distances to the given points is minimal.

输入

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the number of points on the line.

The second line contains n integers xi ( - 10⁹ ≤ xi ≤ 10⁹) — the coordinates of the given n points.

输出

Print the only integer x — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.

输入示例

4
1 2 3 4

输出示例

数据规模及约定

见“输入”

题解

排序输出中位数。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 300010
int n, A[maxn];

int main() {
	n = read();
	for(int i = 1; i <= n; i++) A[i] = read();
	
	sort(A + 1, A + n + 1);
	
	if(n & 1) printf("%d\n", A[(n>>1)+1]);
	else printf("%d\n", A[n>>1]);
	
	return 0;
}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5806721.html