Shooting

题意翻译

#### 题目大意 有n个罐子，第i（从1到n编号）个罐子耐久度为ai。设x为已经打到的罐子数量，则达到第i个罐子需射击（ai⋅x+1）次，每射击一个罐子需将其击倒后才能切换目标。问要将所有罐子击倒的最少次数及击倒序列。 #### 输入格式 第一行一个整数表示n。 第二行n个整数分别为a1、a2...an。 #### 输出格式 第一行输出最少需要的射击次数。 第二行表示罐子击倒的序号，若有多种方法，输出任意一种。

题目描述

Recently Vasya decided to improve his pistol shooting skills. Today his coach offered him the following exercise. He placed $ n $ cans in a row on a table. Cans are numbered from left to right from $ 1 $ to $ n $ . 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 $ i $ -th can is $ a_i $ . It means that if Vasya has already knocked $ x $ cans down and is now about to start shooting the $ i $ -th one, he will need $ (a_i \cdot x + 1) $ shots to knock it down. You can assume that if Vasya starts shooting the $ i $ -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 $ n $ given cans down exactly once is minimum possible.

输入输出格式

输入格式

The first line of the input contains one integer $ n $ $ (2 \le n \le 1\,000) $ — the number of cans. The second line of the input contains the sequence $ a_1, a_2, \dots, a_n $ $ (1 \le a_i \le 1\,000) $ , where $ a_i $ is the durability of the $ i $ -th can.

输出格式

In the first line print the minimum number of shots required to knock each of the $ n $ given cans down exactly once. In the second line print the sequence consisting of $ n $ distinct integers from $ 1 $ to $ n $ — the order of indices of cans that minimizes the number of shots required. If there are several answers, you can print any of them.

输入输出样例

输入样例 #1

3
20 10 20

输出样例 #1

43
1 3 2

输入样例 #2

4
10 10 10 10

输出样例 #2

64
2 1 4 3

输入样例 #3

6
5 4 5 4 4 5

输出样例 #3

69
6 1 3 5 2 4

输入样例 #4

2
1 4

输出样例 #4

3
2 1

说明

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 \cdot 1 + 1 = 21 $ times. After that only second can remains. To knock it down Vasya shoots $ 10 \cdot 2 + 1 = 21 $ times. So the total number of shots is $ 1 + 21 + 21 = 43 $ . In the second example the order of shooting does not matter because all cans have the same durability.