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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
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<math.h> 6 #include<stdlib.h> 7 using namespace std; 8 #define N 100006 9 int n,m; 10 int a[N]; 11 bool solve(int mid){ 12 int cnt=0; 13 for(int i=0;i<n;i++){ 14 int tmp=n-(lower_bound(a,a+n,a[i]+mid)-a);//a[i]加上mid看看有多少个 15 cnt+=tmp; 16 } 17 if(cnt>m) return true;//如果太多，则要调整mid向上，>还是>=有时候要调整 18 return false; 19 } 20 int main() 21 { 22 while(scanf("%d",&n)==1){ 23 m=n*(n-1)/4; 24 for(int i=0;i<n;i++){ 25 scanf("%d",&a[i]); 26 } 27 sort(a,a+n); 28 int low=0; 29 int high=a[n-1]-a[0]; 30 while(low<high){ 31 int mid=(low+high)>>1; 32 if(solve(mid)){ 33 low=mid+1; 34 } 35 else{ 36 high=mid; 37 } 38 } 39 printf("%d\n",low-1);//有时要调整，减1或不减 40 } 41 return 0; 42 }
poj 3579 Median （二分搜索之查找第k大的值）
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原文地址：http://www.cnblogs.com/UniqueColor/p/4782212.html
