# Poj2231 Moo Volume 2017-03-11 22:58 30人阅读 评论(0) 收藏 2022年11月23日
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Moo Volume

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22104 Accepted: 6692

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

`515324`

Sample Output

`40`

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.

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`#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <stack>#include <string>#include <set>#include<vector>#include <map>using namespace std;#define inf 0x3f3f3f3f#define LL long longLL a;int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        sort(a,a+n);        LL ans=0;        for(int i=1;i<n;i++)        {            ans+=(a[i]-a[i-1])*i*(n-i)*2;        }printf("%lld\n",ans);    }    return 0;}`

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