Median

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9201 Accepted: 3209

Description

Given N numbers, X1, X2, … , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i ＜ j ≤ N). We can get C(N,2) differences through this work, and now your task is to find the median of the differences as quickly as you can!

Note in this problem, the median is defined as the (m/2)-th smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

Input

The input consists of several test cases.
In each test case, N will be given in the first line. Then N numbers are given, representing X1, X2, … , XN, ( Xi ≤ 1,000,000,000 3 ≤ N ≤ 1,00,000 )

Output

For each test case, output the median in a separate line.

Sample Input

4
1 3 2 4
3
1 10 2

Sample Output

1
8

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中文題意：

給N數(shù)字, X1, X2, … , XN，我們計(jì)算每對數(shù)字之間的差值：∣Xi - Xj∣ (1 ≤ i ＜ j ≤ N). 我們能得到 C(N,2) 個差值，現(xiàn)在我們想得到這些差值之間的中位數(shù)。

如果一共有m個差值且m是偶數(shù)，那么我們規(guī)定中位數(shù)是第(m/2)小的差值。

輸入包含多測
每個測試點(diǎn)中，第一行有一個NThen N 表示數(shù)字的數(shù)量。
接下來一行由N個數(shù)字：X1, X2, … , XN
( Xi ≤ 1,000,000,000 3 ≤ N ≤ 1,00,000 )

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解題心得：

首先要知道的是ｎ個數(shù)字可以得到C(N,2)個數(shù)字，那么要得到中位數(shù)，如果是偶數(shù)就要得到第n/2大的數(shù)字，如果是奇數(shù)就要得到第n/2+1個數(shù)字。

具體做法可以先將給出的數(shù)字排序，然后每次二分枚舉一個中位數(shù)（O(logn)），然后驗(yàn)證比這個中位數(shù)小的數(shù)字有多少個，在驗(yàn)證比這個中位數(shù)小的數(shù)字有多少個的時候可以選擇尺取法（具體實(shí)現(xiàn)看代碼，O(n)），這樣就在O(nlogn)的復(fù)雜度內(nèi)完成。

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#include <algorithm> #include <stdio.h> #include <cstring> using namespace std; const int maxn = 1e5+100;int n,num[maxn]; long long tot;void init() {tot = 0;tot = (long long)n*(n-1)/2;//差值的個數(shù)if(tot%2 == 0) {//中位數(shù)在第幾個tot /=2;}elsetot = tot/2 + 1;for(int i=0;i<n;i++) {scanf("%d",&num[i]);}sort(num,num+n); }bool checke(int va) {int t = 0;long long sum = 0;for(int i=0;i<n;i++) {while(t < n && num[t] - num[i] <= va)//尺取法看比va小的值有多少個t++;sum += t-i-1;}if(sum < tot)return true;return false; }int binary_search() {//每次枚舉一個中位數(shù)的值int l,r;l = 0, r = 1e9+100;while(r - l > 1) {int mid = (l + r) >> 1;if(checke(mid))l = mid;elser = mid;}return r; }int main() {while(scanf("%d",&n ) != EOF) {init();int ans = binary_search();printf("%d\n",ans);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/GoldenFingers/p/9107120.html

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