time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Recently Vasya decided to improve his pistol shooting skills. Today his coach offered him the following exercise. He placed nn cans in a row on a table. Cans are numbered from left to right from 11 to nn. Vasya has to knock down each can exactly once to finish the exercise. He is allowed to choose the order in which he will knock the cans down.
Vasya knows that the durability of the ii-th can is aiai. It means that if Vasya has already knocked xx cans down and is now about to start shooting the ii-th one, he will need (ai⋅x+1)(ai⋅x+1) shots to knock it down. You can assume that if Vasya starts shooting the ii-th can, he will be shooting it until he knocks it down.
Your task is to choose such an order of shooting so that the number of shots required to knock each of the nn given cans down exactly once is minimum possible.
Input
The first line of the input contains one integer nn (2≤n≤1000)(2≤n≤1000) — the number of cans.
The second line of the input contains the sequence a1,a2,…,ana1,a2,…,an (1≤ai≤1000)(1≤ai≤1000), where aiai is the durability of the ii-th can.
Output
In the first line print the minimum number of shots required to knock each of the nn given cans down exactly once.
In the second line print the sequence consisting of nn distinct integers from 11 to nn — the order of indices of cans that minimizes the number of shots required. If there are several answers, you can print any of them.
Examples
input
Copy
3 20 10 20
output
Copy
43 1 3 2
input
Copy
4 10 10 10 10
output
Copy
64 2 1 4 3
input
Copy
6 5 4 5 4 4 5
output
Copy
69 6 1 3 5 2 4
input
Copy
2 1 4
output
Copy
3 2 1
Note
In the first example Vasya can start shooting from the first can. He knocks it down with the first shot because he haven't knocked any other cans down before. After that he has to shoot the third can. To knock it down he shoots 20⋅1+1=2120⋅1+1=21 times. After that only second can remains. To knock it down Vasya shoots 10⋅2+1=2110⋅2+1=21 times. So the total number of shots is 1+21+21=431+21+21=43.
In the second example the order of shooting does not matter because all cans have the same durability.
解题说明：此题是一道贪心题，按照数字大小排序，这样确保所需要的射击次数最少，可以使用pair来记录数字位置和大小，最后输出即可。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include<iostream>
#include<algorithm>
#include <bits/stdc++.h>
using namespace std;
int n, x, ans;
pair <int, int>a[1005];
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> x;
a[i].first = x;
a[i].second = i;
}
sort(a + 1, a + n + 1);
for (int i = n; i >= 1; i--)
{
ans += a[i].first * (n - i) + 1;
}
cout << ans << '\n';
for (int i = n; i >= 1; i--)
{
cout << a[i].second << " ";
}
cout << '\n';
return 0;
}