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

題目：

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

題意分析：

感覺題目意思沒啥好說的，注意一下會爆int的問題就好了。

代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <iostream>

#define INF 0x3f3f3f3f

using namespace std;


const int MAXN = 200000+5;

int a[MAXN];
map<int, __int64> p;

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        p.clear();
        int i;
        int minf = INF, maxf = 0;
        for(i = 0;i < n;i++)
        {
            scanf("%d", &a[i]);
            minf = min(minf, a[i]);
            maxf = max(maxf, a[i]);
            p[a[i]]++;
        }
        int d = maxf-minf;
        printf("%d ", d);
        __int64 ans = 0;
        if(d == 0)
        {
            __int64 t = p[minf]-1;
            ans = (1+t)*t/2;
        }
        else
        {
            ans = p[minf]*p[maxf];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}