BZOJ1679: [Usaco2005 Jan]Moo Volume 牛的呼声

2022-10-29 11:18:52

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

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=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }

 ll a[maxn];

 int main()

 {

     freopen("input.txt","r",stdin);

     freopen("output.txt","w",stdout);

     ll n=read(),ans=,sum;

     for(int i=;i<=n;i++)a[i]=read();

     sort(a+,a+n+);sum=a[];

     for(int i=;i<=n;i++)

     {

       ans+=(a[i]*(i-))-sum;sum+=a[i];

     }

     printf("%lld\n",*ans);

     return ;

 }

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