1319: water line connection

 

Description [title]

There are n

Individual queues in front of a tap water connection, if each of the water receiving time T I , please find this programming n in a sequential individual queued, so that n

The average individual waiting time to a minimum.

[Enter]

A total of two rows, the first row n- (1 ≦ n- ≤1000)

; Second line respectively denote an individual to the second n individual contact of water per time T 1, T 2, ..., T n , between each of the data 1

Spaces.

[Output]

There are two lines, the first act of the queue sequence, i.e. 1

To n

One arrangement; average waiting time in the second embodiment are arranged such behavior (outputs the result to two decimal places).

[Sample input]

10							
56 12 1 99 1000 234 33 55 99 812

[Sample Output]

3 2 7 8 1 4 9 6 10 5
291.90

// Created on 2020/2/11

/*#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <climits>*/
#include <bits/stdc++.h>

using namespace std;

//int i,j,k;

const int maxn=INT_MAX;

const int idata=1000+5;
int goal[idata],temp[idata];
int n,m;
bool flag;
long long sum,assume;

int main()
{
    int i,j;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>goal[i];
        temp[i]=i;
    }

    for(i=1;i<=n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            if(goal[i]>goal[j])
            {
                swap(goal[i],goal[j]);
                swap(temp[i],temp[j]);
            }
        }
    }

    assume=goal[1];//注意：是等待时间
    for(i=2;i<=n;i++)
    {
        sum+=assume;
        assume+=goal[i];
    }

    for(i=1;i<=n;i++)
    {
        cout<<temp[i]<<" ";
    }
    cout<<endl;
    printf("%.2lf\n",sum*1.0/n);

}