openjudge-Moo Volume

排序+简单公式推导

描述

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.

输入

* Line 1: N

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

输出

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

样例输入

5
1
5
3
2
4

样例输出

40

提示

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.

题目是比较好理解的，关键是找出规律，并且推导出公式

简单题。
题目大意是说求n个牛相互之间距离和，如hint所描述的一样。
首先将这些牛的位置排序。
然后得出有n头牛时，相邻两头牛间距离相加的次数。（如下）
例如5头牛时：设这5头牛为abcde则：
a   b   c   d   e
  4   3   2   1      >>a叫的时候  (注意，这里:  a->e ==  a->b + b->c + c->d + d->e  )
  1   3   2   1      >>b叫的时候
  1   2   2   1      >>c叫的时候
  1   2   3   1      >>d叫的时候
  1   2   3   4      >>e叫的时候
求和得：
  8   12   12   8
通过尝试，得到如下规律：
n=2                2
n=3                4    4
n=4                6    8    6
n=5                8    12  12  8
n=6                10  16  18  16  10 
。。。。。。。。。。。。。。。。。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 100000000;
long long n,i,j,sum;//注意要用long long ,poj好像不能用__int64，这好像有点坑爹，害我wrong了一遍.....
long long a[maxn];
int main()
{
    while(~scanf("%lld",&n))
    {
        for(i=0;i<n;i++)
            scanf("%lld",&a[i]);
        sum=0;
        sort(a,a+n);
        for(j=1;j<n;j++)
        {
            sum=sum+(a[j]-a[j-1])*j*(n-j)*2;
        }
        printf("%lld\n",sum);
    }
    return 0;
}