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.

 1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector>10 #include <queue>11 #include <map>12 #include <set>13 using namespace std;14 typedef long long LL;15 #define RG register16 const int MAXN = 50011;17 int n;18 LL a[MAXN];19 LL l,r,ans,ans2;20 LL N,cnt,zhong;21 22 inline int getint()23 {24        RG int w=0,q=0; char c=getchar();25        while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); if(c==‘-‘) q=1,c=getchar(); 26        while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w;27 }28 29 inline bool check(LL x){30     cnt=0; LL now;31     for(RG int i=1;i<=n;i++) {32     now=lower_bound(a+i,a+n+1,a[i]+x)-a;33     cnt+=(LL)n-now+1;34     }35     if(cnt>zhong) return true; return false;36 }37 38 inline void work(){39     while(scanf("%d",&n)!=EOF) {40     for(RG int i=1;i<=n;i++) a[i]=getint();41     sort(a+1,a+n+1); l=0; r=a[n]-a[1]+1; a[n+1]=(1<<30);42     N=(LL)n*(n-1)/2; LL mid; 43     zhong=N/2; ans=0;44     while(l<=r) {45         mid=(l+r)/2;46         if(check(mid)) ans=mid,l=mid+1;47         else r=mid-1;48     }49     printf("%lld\n",ans);50     }51 }52 53 int main()54 {55   work();56   return 0;57 }