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

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

查找最大数的个数和最小数的个数相乘即得方法数（当所有数都相等时方法数为（n-1）*n/2注意n要longlong）

#include<bits/stdc++.h>
#define M 200000+100
#define LL long long
using namespace std;
int a[M];
int main()
{
int n;
while(cin>>n)
{
for(int i=0;i<n;++i)
{
scanf("%d",&a[i]);
}
sort(a,a+n);
int minx=a[0];
int maxn=a[n-1];
LL min_num=1,max_num=1;
for(int i=1;i<n;i++)
{
if(a[i]==minx)
min_num++;
else break;
}
for(int i=n-2;i>=0;i--)
{
if(a[i]==maxn)
max_num++;
else break;
}
if(min_num==n)
{
LL ans=(LL)n*((LL)n-1)/2;
printf("%d %I64d\n",maxn-minx,ans);
}
else{
LL ans=max_num*min_num;
printf("%d %I64d\n",maxn-minx,ans);
}
}

return 0;
}