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

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.

题目意思：有 n 朵 flowers，每朵flower有相应的 beauty，求出最大的beauty 差 和 要达到这个最大的差 的取法有多少种。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
long long maxnum=INT_MIN,minnum=INT_MAX,maxcnt,mincnt;
int n,tmp;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&tmp);
if(tmp>maxnum)
{
maxnum=tmp;
maxcnt=1;
}
else if(tmp==maxnum)
maxcnt++;
if(tmp<minnum)
{
minnum=tmp;
mincnt=1;
}
else if(tmp==minnum)
mincnt++;
}
long long delta=maxnum-minnum;
long long idea;
if(delta==0)
idea=maxcnt*(maxcnt-1)/2;
else
idea=maxcnt*mincnt;
printf("%I64d %I64d\n",delta,idea);
return 0;
}