Codeforces Round #261 (Div. 2) B. Pashmak and Flowers【水】

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:

The maximum beauty difference of flowers that Pashmak can give to Parmida.

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:

choosing the first and the second flowers;

choosing the first and the fifth flowers;

choosing the fourth and the second flowers.

choosing the fourth and the fifth flowers.

略坑的水题，直接排序，第1和最后个差就是最大差，分别统计和第1个、最后个 相同的花的数目，乘积就是组合情况，即答案（乘积可能爆int)
坑在于全部数一样的时候，答案是C(n,2)，注意爆int。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
#define NMAX 500001
#define FOR(i,n) for(int i=0;i<n;i++)

int a[200002];

int main()
{
int n;
ll mid,ans,first,last;
int dis;
//freopen("in.txt","r",stdin);
while(scanf("%d",&n)==1)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
dis = a[n-1] - a[0];
if(dis == 0)
ans = (ll)n*(n-1)/2;
else
{
first = last = 1;
for(int i=1;i<n;i++)
{
if(a[i]!=a[0])
break;
first++;
}
for(int i=n-2;i>=0;i--)
{
if(a[i]!=a[n-1])
break;
last++;
}
ans = first*last;
}
printf("%d %I64d\n",dis,ans);
}
return 0;
}