Tourist Problem

CodeForces - 340C

Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.

Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.

The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.

Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.

Input

The first line contains integer n (2 ≤ n ≤ 105). Next line contains n distinct integers a1a2, ..., an (1 ≤ ai ≤ 107).

Output

Output two integers — the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.

Examples

Input
3
2 3 5
Output
22 3

Note

Consider 6 possible routes:

  • [2, 3, 5]: total distance traveled: |2 – 0| + |3 – 2| + |5 – 3| = 5;
  • [2, 5, 3]: |2 – 0| + |5 – 2| + |3 – 5| = 7;
  • [3, 2, 5]: |3 – 0| + |2 – 3| + |5 – 2| = 7;
  • [3, 5, 2]: |3 – 0| + |5 – 3| + |2 – 5| = 8;
  • [5, 2, 3]: |5 – 0| + |2 – 5| + |3 – 2| = 9;
  • [5, 3, 2]: |5 – 0| + |3 – 5| + |2 – 3| = 8.

The average travel distance is  =  = .

sol:小学奥数题.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
ll s=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n;
ll a[N],ans=;
ll Jiec[N],Invj[N];
inline ll gcd(ll a,ll b)
{
return (b==)?a:(gcd(b,a%b));
}
int main()
{
int i;
ll S=,GG;
R(n);
for(i=;i<=n;i++) R(a[i]);
sort(a+,a+n+);
for(i=;i<=n;i++)
{
S+=a[i-];
ans+=1ll*(i-)*a[i]-S;
}
S+=a[n];
ans=1ll*(ans*2ll+S);
GG=gcd(ans,n);
W(ans/GG); Wl(n/GG);
return ;
}
/*
Input
3
2 3 5
Output
22 3 Input
4
1 5 77 2
Output
547 4 Input
40
8995197 7520501 942559 8012058 3749344 3471059 9817796 3187774 4735591 6477783 7024598 3155420 6039802 2879311 2738670 5930138 4604402 7772492 6089337 317953 4598621 6924769 455347 4360383 1441848 9189601 1838826 5027295 9248947 7562916 8341568 4690450 6877041 507074 2390889 8405736 4562116 2755285 3032168 7770391
Output
644565018 5
*/

codeforces340C的更多相关文章

随机推荐

  1. JavaSE基础知识之继承

    一.概述 继承描述的是事物之间的所属关系,这种关系是: is-a 的关系.例如,图中的兔子属于食草动物,食草动物又属于动物.继承可以使多种事物之间形成一种关系体系,让父类更通用,子类更具体. 1.1  ...

  2. (三)创建基于maven的javaFX+springboot项目创建

    创建基于maven的javaFx+springboot项目有两种方式,第一种为通过非编码的方式来设计UI集成springboot:第二种为分离用户界面(UI)和后端逻辑集成springboot,其中用 ...

  3. pycharm问题

    Pycharm 出现Unresolved reference '' 错误的解决方法:http://www.mamicode.com/info-detail-2190842.html

  4. JS 中的跨域请求

    跨域请求并不仅仅只是 Ajax 的跨域请求,而是对于一个页面来说,只要它请求了其他域名的资源了,那么这个过程就属于跨域请求了. 比如,一个带有其他域名的 src 的 <img> 标签,以及 ...

  5. 常用的Java工具类——十六种

    常用的Java工具类——十六种 在Java中,工具类定义了一组公共方法,这篇文章将介绍Java中使用最频繁及最通用的Java工具类.以下工具类.方法按使用流行度排名,参考数据来源于Github上随机选 ...

  6. 第五篇python进阶之深浅拷贝

    目录 第五篇python进阶之深浅拷贝 一.引言 1.1可变 和不可变 二.拷贝(只针对可变数据类型) 三.浅拷贝 四.深拷贝 第五篇python进阶之深浅拷贝 一.引言 1.1可变 和不可变 id不 ...

  7. 分布式消息中间件之kafka设计思想及基本介绍(一)

    Kafka初探 场景->需求->解决方案->应用->原理 我该如何去设计消息中间件--借鉴/完善 场景 跨进程通信(进程间生产消费模型) 需求 基本需求 实现消息的发送和接收. ...

  8. 十一,k8s集群访问控制之ServicAccount

    目录 认证安全 连接Api-Server的两类账号 ServiceAccount 创建 使用admin 的SA 测试 URL访问kubernetes资源 APIserver客户端定义的配置文件 kub ...

  9. 在RecyclerView中集成QQ汽泡一

    上次已经实现了QQ汽泡的自定义View的效果[http://www.cnblogs.com/webor2006/p/7726174.html],接着再将它应用到列表当中,这样才算得上跟QQ的效果匹配, ...

  10. Summer training #6

    A:水.看0多还是1多就行 B:模拟二进制运算 ,,卡了好久 不应该 #include <bits/stdc++.h> #include <cstring> #include ...