D - Binomial Coefficients

Time limit : 2sec / Memory limit : 256MB

Score : <var>400</var> points

Problem Statement

Let <var>comb(n,r)</var> be the number of ways to choose <var>r</var> objects from among <var>n</var> objects, disregarding order. From <var>n</var> non-negative integers <var>a1,a2,…,an</var>, select two numbers <var>ai>aj</var> so that <var>comb(ai,aj)</var> is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

Constraints

  • <var>2≤n≤105</var>
  • <var>0≤ai≤109</var>
  • <var>a1,a2,…,an</var> are pairwise distinct.
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

<var>n</var>
<var>a1</var> <var>a2</var> <var>…</var> <var>an</var>

Output

Print <var>ai</var> and <var>aj</var> that you selected, with a space in between.

Sample Input 1

Copy 
5
6 9 4 2 11

Sample Output 1

Copy 
11 6

题意：给定一些个数，求出comb(a[i],a[j])的最大值，其中a[j]>a[i]

思路:

组合数存在性质：comb(n + 1, m) > comb(n, m)

证明如下：comb(n + 1, m) = comb(n, m) + comb(n, m − 1) 杨辉三角形

那么必定有，n=maxelement(Array),m-> maxelement(Array)/2。即Comb(n,m)为最大值

时间复杂度:O(n) or O(nlogn)

#include<bits/stdc++.h>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
const int N=2e5+5;
const int MOD=1e9+7;
ll a[N];
int main(void){
//    ios::sync_with_stdio(false);
//    cin.tie(0);cout.tie(0);

    int n;
    cin >> n;
    for(int i=1;i<=n;++i)    cin >> a[i];
    sort(a+1,a+1+n);//写了个stl,sort偷懒，也可以O(n)扫一遍找最大值
    ll ans,dis=1e18;
    for(int i=1;i<=n-1;i++){
        if(a[n]%2==0){//偶数的中位数是a[n]/2
            if(abs(a[i]-a[n]/2)<dis)    ans=a[i],dis=abs(a[i]-a[n]/2);
        }
        else{奇数的中位数是a[n]/2+1或者是a[n]/2
            if(abs(a[i]-a[n]/2)<dis)    ans=a[i],dis=abs(a[i]-a[n]/2);
            if(abs(a[i]-a[n]/2-1)<dis)    ans=a[i],dis=abs(a[i]-a[n]/2-1);
        }
    }
    cout << a[n] <<" "<<ans<<endl;

    return 0;
}