POJ 2231 Moo Volume

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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

5
1
5
3
2
4

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.

 

暴力做的，之前没给牛排序，超时；按位置排序之后就过了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 10005
int c[N];
int main()
{
    int n,i,j;
    long long ans;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
    for(i=0;i<n;i++)
    {
        scanf("%d",&c[i]);
    }
    sort(c,c+n);
    for(i=0;i<n-1;i++)
    {
        for(j=i+1;j<n;j++)
        {
            ans+=2*abs(c[i]-c[j]);
        }
    }
    printf("%lld
",ans);}
    return 0;
}