[BZOJ] 1679: [Usaco2005 Jan]Moo Volume 牛的呼声

Posted 2020-10-06 Lev今天学习了吗

1679: [Usaco2005 Jan]Moo Volume 牛的呼声

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 1101  Solved: 573
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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

Silver

 

Analysis

 水水水题？？

不是很清楚怎么优化，只把空间优化到线性

反正随便就过了qwq

递推什么的不会啊qwq

 

Code

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 long long pos[100000],cnt[100000],tot,n;
 7 
 8 int main(){
 9     scanf("%lld",&n);
10     for(int i = 1;i <= n;i++) scanf("%lld",&pos[i]);
11     sort(pos+1,pos+1+n);
12     
13     for(int i = 1;i <= n-1;i++){
14         for(int j = i;j <= n-1;j++){
15             cnt[j] += n-j;
16         }tot += cnt[i]*(pos[i+1]-pos[i]);
17     }
18     
19     printf("%lld",tot*2);
20     
21     return 0;
22 }

n^2