AtCoder Regular Contest 095 D - Binomial Coefficients

Problem Statement

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

Constraints

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

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Input

Input is given from Standard Input in the following format:

n
a1 a2 … an

Output

Print ai and aj that you selected, with a space in between.

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Sample Input 1

Copy

5
6 9 4 2 11

Sample Output 1

Copy

11 6

comb(ai,aj) for each possible selection is as follows:

• comb(4,2)=6
• comb(6,2)=15
• comb(6,4)=15
• comb(9,2)=36
• comb(9,4)=126
• comb(9,6)=84
• comb(11,2)=55
• comb(11,4)=330
• comb(11,6)=462
• comb(11,9)=55

Thus, we should print 11 and 6.

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Sample Input 2

Copy

2
100 0

Sample Output 2

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100 0

题意:找最大的组合数

思路：由组合数的性质可得如果ak>ai,则comb(ak,aj)>comb(ai,aj).则可知ai应选最大值，aj则选距ai/2最近的值，即为解；

#include 
 
using namespace std;
const int N = 100005;
typedef long long ll;
int a[N];
double b[N];
int main()
{
     int n,maxa=-1;
     cin>>n;
     for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    sort(a+1,a+1+n);
    double mid=1.0*a[n]/2,minb=1000000009.0;
    int ans;
    for(int i=1;i