标签：
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
Source
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
static int n;
static int a[] = new int[100000];
static int m;
static int solve(int x,int y,int val,int flag){
while(x<=y){
int mid=(x+y)/2;
if(val-a[mid]<=flag)
y=mid-1;
else
x=mid+1;
}
return y;
}
static boolean ok(int val){
int j,k=0;
for(int i=2;i<=n;i++){
j=solve(1,i-1,a[i],val);
k+=i-j-1;
}
if(k<m)
return false;
else
return true;
}
public static void main(String[] args) throws IOException {
//Scanner scan = new Scanner(System.in);
StreamTokenizer st = new StreamTokenizer(new BufferedReader(
new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
while (st.nextToken()!=StreamTokenizer.TT_EOF) {
n =(int)st.nval;
if(n*(n-1)/2%2!=0){
m=(n*(n-1)/2+1)/2;
}
else
m=n*(n-1)/2/2;
for (int i = 1; i <=n; i++) {
st.nextToken();
a[i] = (int)st.nval;
}
Arrays.sort(a,1,n+1);
int l = 0, r = a[n]-a[1];
while(l<=r){
int mid=(l+r)/2;
if(ok(mid))
r=mid-1;
else
l=mid+1;
}
out.println(l);
out.flush();
}
}
}版权声明：本文博主原创文章，博客，未经同意不得转载。
标签：
原文地址：http://www.cnblogs.com/lcchuguo/p/4820394.html
