POJ 3579

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3528		Accepted: 1001

Description

Given N numbers, X₁, X₂, ... , X_N, let us calculate the difference of every pair of numbers: ∣X_i - X_j∣ (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, ... , X_N, ( X_i≤ 1,000,000,000 3 ≤ N ≤ 1,00,000 )

Output

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

Sample Input

Sample Output

1

8

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

二分中位数

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 #define maxn 100005

 #define INF 200005

 typedef long long ll;

 int n;

 int a[maxn],dis[maxn];

 ll num;

 bool judge(int x) {

     int pos,i = ,now = ;

     ll sum = ;

     while(i < n) {

             pos = upper_bound(dis + ,dis + n + ,x + now) - dis;

             sum += n - (pos);

             now = dis[i];

             if((n - ) - pos +  == ) break;

             ++i;

     }

     //printf("x = %d sum = %lld\n",x,sum);

     return num - sum - (num % ) >= sum;

 }

 void solve() {

     num = (ll)n * (n - ) / ;

     int l = INF,r = a[n] - a[];

     for(int i = ; i < n; ++i) {

             dis[i] = a[i + ] - a[];

             l = min(l,a[i + ] - a[i]);

            // printf("%d ",dis[i]);

     }

     dis[n] = INF;

     //printf("l = %d r = %d\n",l,r);

     while(l < r) {

             int mid = (l + r) >> ;

             if(judge(mid)) {

                     r = mid;

             } else {

                     l = mid + ;

             }

     }

     printf("%d\n",l);

 }

 int main()

 {

    // freopen("sw.in","r",stdin);

     while(~scanf("%d",&n)) {

             for(int i =  ; i <= n; ++i) {

                     scanf("%d",&a[i]);

             }

             sort(a + ,a + n + );

             solve();

     }

     return ;

 }

巴特西

POJ 3579

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