校赛 选修课网址 1097: Meeting

1097: Meeting

Time Limit: 2 Sec Memory Limit: 64 MB

Submit: 10 Solved: 7

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Description

There n peoples, they want to meet for discussing something about NBUT ACM Laboratory, so they need to decide a location where they meeting.

But they are lazy, they want a location where the summary of each one’s moving

distance is least, now you know the n peoples’ position, can you help them to calculate the minimum summary of each one’s moving distance.

Input

Input starts with an integer T(1 <= T <= 20), denoting the number of test case.

For each test case, first line contains an integer n(1 <= n <= 100000), denoting the number of people.

Next line contains n integers Pi(0 <= Pi <= 109), denoting each one’s position. Note that maybe someone stay at the same place.

Output

For each test case, print the minimum summary of each one’s moving distance.

Sample Input

1

5

1 2 3 4 5

Sample Output

6

HINT

You can set the location at 3, so the minmum summary of each one’s moving distance is 6(2+1+0+1+2).

找最短路径 sort一下 相对最大的减去相对最小的 只要重复的线段最小就是答案 最短的一般都是在中间

#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<sstream>
using namespace std;
int main()
{
long long n,m,i,k,sum,max,min;
int a[100010];
ios::sync_with_stdio(false);
cin>>k;
while(k--)
{
sum=0;
cin>>n;
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
cin>>a[i];
sort(a,a+n);
max=n-1;
min=0;
while(max>=min)
{
sum+=a[max]-a[min];
max--;
min++;
}
cout<<sum<<endl;
}
return 0;
}