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
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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

题解:
排个序,扫一遍即可。
代码:
 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #define inf 1000000000
12 #define maxn 10000+100
13 #define maxm 100000+100
14 #define ll long long
15 using namespace std;
16 inline ll read()
17 {
18     ll x=0,f=1;char ch=getchar();
19     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
20     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
21     return x*f;
22 }
23 ll a[maxn];
24 int main()
25 {
26     freopen("input.txt","r",stdin);
27     freopen("output.txt","w",stdout);
28     ll n=read(),ans=0,sum;
29     for(int i=1;i<=n;i++)a[i]=read();
30     sort(a+1,a+n+1);sum=a[1];
31     for(int i=2;i<=n;i++)
32     {
33       ans+=(a[i]*(i-1))-sum;sum+=a[i];    
34     }
35     printf("%lld\n",2*ans);
36     return 0;
37 }
View Code

 

posted @ 2014-08-20 18:26  ZYF-ZYF  Views(165)  Comments(0Edit  收藏  举报