BZOJ1679: [Usaco2005 Jan]Moo Volume 牛的呼声

1679: [Usaco2005 Jan]Moo Volume 牛的呼声

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 723  Solved: 346
[Submit][Status]

Description

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.

    约翰的邻居鲍勃控告约翰家的牛们太会叫．

    约翰的N(1≤N≤10000)只牛在一维的草场上的不同地点吃着草．她们都是些爱说闲话的奶牛，每一只同时与其他N-1只牛聊着天．一个对话的进行，需要两只牛都按照和她们间距离等大的音量吼叫，因此草场上存在着N（N-1）/2个声音．  请计算这些音量的和．

Input

* Line 1: N * Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

    第1行输入N，接下来输入N个整数，表示一只牛所在的位置．

Output

* Line 1: A single integer, the total volume of all the MOOs.

    一个整数，表示总音量．

Sample Input

5
1
5
3
2
4

INPUT DETAILS:

There are five cows at locations 1, 5, 3, 2, and 4.

Sample Output

40

OUTPUT DETAILS:

Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3
contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4
contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.

HINT

 

Source

Silver

题解：

排个序，扫一遍即可。

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#define inf 1000000000
#define maxn 10000+100
#define maxm 100000+100
#define ll long long
using namespace std;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
ll a[maxn];
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    ll n=read(),ans=0,sum;
    for(int i=1;i<=n;i++)a[i]=read();
    sort(a+1,a+n+1);sum=a[1];
    for(int i=2;i<=n;i++)
    {
      ans+=(a[i]*(i-1))-sum;sum+=a[i];    
    }
    printf("%lld
",2*ans);
    return 0;
}