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

2022-04-24 16:16:06

一开始直接 O( n² ) 暴力..结果就 A 了... USACO 数据是有多弱 = =

先sort , 然后自己再YY一下就能想出来...具体看code

-----------------------------------------------------------------------------------------

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#define rep( i , n ) for( int i = 0 ; i < n ; i++ )

#define clr( x , c ) memset( x , c , sizeof( x ) )

using namespace std;

const int maxn = 10000 + 5;

int h[ maxn ];

int main() {

// freopen( "test.in" , "r" , stdin );

int n;

cin >> n;

rep( i , n )

scanf( "%d" , h + i );

sort( h , h + n );

long long ans = 0;

rep( i , n - 1 )

ans += 1LL * ( i + 1 ) * ( n - i - 1 ) * ( h[ i + 1 ] - h[ i ] );

cout << ans * 2 << "\n";

return 0;

}

-----------------------------------------------------------------------------------------

1679: [Usaco2005 Jan]Moo Volume 牛的呼声

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 798 Solved: 385
[Submit][Status][Discuss]

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

码农公寓