POJ - 2231 - Moo Volume

最新推荐文章于 2020-01-31 14:37:53 发布

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最新推荐文章于 2020-01-31 14:37:53 发布

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针对一群牛在牧场上的不同位置发出的声音，本篇博客介绍了一种计算任意两只牛之间声音传播体积的方法，并给出了一段高效的C++代码实现。

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Moo Volume

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22767		Accepted: 6901

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.

Input

* Line 1: N

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

Output

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

Sample Input

Sample Output

Hint

INPUT DETAILS:

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

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.

Source

USACO 2005 January Silver

题意：有n只牛，每只牛在各自的位置pi，求∑(i=1,n) ∑(j=1,n) ( abs ( pi - pj ) )

思路：先对位置从小到大进行排序，然后从第二个点开始往后，每次加上该点到该点之前的所有点的距离之和dist ( i )（第i个点）

dist ( i ) 可以由 dist ( i - 1 ) + ( pos( i ) - pos( i - 1) ) * ( i - 1 )得到

不过O(n^2)的效率也可以过，注意最后的和会超 int

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#define LL long long
using namespace std;
const int N = 10000 + 10;
LL n,pos[N],ans = 0,dis = 0;;
int main()
{
    scanf("%lld",&n);
    for(LL i=0;i<n;i++) scanf("%lld",&pos[i]);
    sort(pos, pos+n);
    for(int i=1;i<n;i++){
        dis = dis + (pos[i] - pos[i-1])*(i);//当前点与其前面点之和
        ans += dis;
    }
    printf("%lld\n",ans * 2);
    return 0;
}