HDU 3466 Moo Volume

时间：2016-08-31 00:45:35 阅读：192 评论：0 收藏：0 [点我收藏+]

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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

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.

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\n",ans);}
    return 0;
}

HDU 3466 Moo Volume

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原文地址：http://www.cnblogs.com/Annetree/p/5824102.html

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