CF_459B_PashmakAndFlowers

B. Pashmak and Flowers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Pashmak decided to give Parmida a pair of flowers from the garden. There are 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

The first line of the input contains 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

In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:

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.

簡單的模拟題

題目說那個姑娘？隻喜歡美麗值差别最大的兩朵花

問送給她這兩朵花的方法有多少種

唯獨需要注意的就是如果所有花都一樣

那麼這兩種花的取法與其他情況不同

#include <iostream>
#include <stdio.h>

using namespace std;
int main()
{
    int n;
    int be;
    int mi=1e9+1,ma=0;
    long long nmi=0,nma=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&be);
        if(be==mi)
            nmi++;
        if(be<mi)
        {
            mi=be;
            nmi=1;
        }

        if(be==ma)
            nma++;
        if(be>ma)
        {

            ma=be;
            nma=1;
        }

        //cout<<mi<<" "<<ma<<" "<<nmi<<" "<<nma<<endl;
    }
    if(ma==mi)
        printf("0 %I64d\n",nma*(nma-1)/2);
    else
        printf("%d %I64d\n",ma-mi,nma*nmi);
    return 0;
}