CF 459B(Pashmak and Flowers-数列最大2数差及能取到最大数差的取法数)

B. Pashmak and Flowers

time limit per test

memory limit per test

input

output

n flowers in the garden and the i-th of them has a beauty number bi. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!

Your task is to write a program which calculates two things:

1. The maximum beauty difference of flowers that Pashmak can give to Parmida.
2. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.

Input

n (2 ≤ n ≤ 2·105). In the next line there are n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109).

Output

The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.

Sample test(s)

input

2 1 2

output

1 1

input

3 1 4 5

output

4 1

input

5 3 1 2 3 1

output

2 4

Note

2 and there are 4

1. choosing the first and the second flowers;
2. choosing the first and the fifth flowers;
3. choosing the fourth and the second flowers;
4. choosing the fourth and the fifth flowers.

排序直接算，注意特判“数列中所有数均相同”的情况

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,b[MAXN];
int main()
{
//  freopen("Flowers.in","r",stdin);
//  freopen(".out","w",stdout);
  scanf("%d",&n);
  For(i,n) scanf("%d",&b[i]);
  sort(b+1,b+1+n);
  cout<<b[n]-b[1]<<' ';
  if (b[1]==b[n]) 
  {
    cout<<((ll)n*ll(n-1))/2<<endl;
    return 0;
  }
  int t=1;
  while (t<n&&b[t+1]==b[t]) ++t;
  int r=n;
  while (r>1&&b[r-1]==b[r]) --r;
  
  cout<<(ll)t*(ll)(n-r+1)<<endl;
  
  
  return 0;
}