CF 817B Makes And The Product（思维+组合数）

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn’t been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i,  j,  k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!
Input
The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains n positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.
Output
Print one number — the quantity of triples (i,  j,  k) such that i,  j and k are pairwise distinct and ai·aj·ak is minimum possible.
Examples
Input
4
1 1 1 1
Output
4
Input
5
1 3 2 3 4
Output
2
Input
6
1 3 3 1 3 2
Output
1
Note
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there’s only one way to choose indices.

a[1]=a[2]=a[3],三元素相同,可以统计与这三元素相等的个数，假设cnt个，结果是从cnt中找3个
a[1]=a[2]<a[3]，统计后续等于a[3]的元素个数，假设cnt个，结果就是cnt
a[1]<a[2]=a[3]，统计后续等于a[3]的元素个数，假设cnt个，结果是从cnt中找两个
a[1]<a[2]<a[3]，统计后续等于a[3]的元素个数，假设cnt个，结果也是cnt

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm> 
using namespace std;
typedef long long ll;

ll a[100005];
ll n, ans;
ll func(int n, int m)
{ 
  ll ans=1;
  ll i;
  if (m<n-m)
    m=n-m;
  for (i=m+1; i<=n; i++)
    ans*=i;
  for (i=1; i<=n-m; i++)
    ans/=i;
  return ans;
}
int main()
{
  scanf("%lld", &n);
  for (int i=1; i<=n; i++)
    scanf("%lld", &a[i]);
  sort(a+1, a+n+1);
  if (n==3){
    printf("1");
    return 0;
  }
  if (a[1]==a[2] && a[2]==a[3]){
    for (ll i=4; i<=n; i++){
      if (a[i]==a[3]) ans++;
      else  break;
    }
    ans=func(ans+3, 3);
  }else if (a[1]==a[2] && a[2]<a[3]){
    for (ll i=4; i<=n; i++){
      if (a[i]==a[3]) ans++;
      else  break;
    }
    ans=ans+1;
  }else if (a[1]<a[2] && a[2]==a[3]){
    for (ll i=4; i<=n; i++){
      if (a[i]==a[3]) ans++;
      else  break;
    }
    ans=func(ans+2, 2);
  }else{
    for (ll i=4; i<=n; i++){
      if (a[i]==a[3]) ans++;
      else  break;
    }
    ans=ans+1;
  }
  printf("%lld", ans);
  return 0;
}