Joey Takes Money

本文介绍: Jo e y 很穷，因此他的朋友 Chandler 想要借给他一些钱。但是 Jo e y 的自尊心很强，为了不让他的自尊心受挫又能给他钱，Chandler 打算和他玩一个游戏。采用贪心的思想，计算每一个数字的连乘，并放到第一个位置，其他位置为1，此时可以达到最大。共一行，一个整数，表示 Jo e y 最多可以得到的钱乘以 2022 后的值。综上所述， Jo e y 可以得到的最多的钱即为。即 Joe y 所得的钱。，即 Joey 最多可以得到的钱，并输出。因为我们再也见不到它了！最后， Joey 将得到的钱就是。

x=12,;y=1[12,1,1]

综上所述， Joey 可以得到的最多的钱即为

12+1+1=14

$12 + 1 + 1 = 14$ 元，所以输出应为

2022

28308

14times 2022 = 28308

$14 \times 2022 = 28308$ .

题目 描述

Joey i s low on m oney. His fri end Chandler wants to len d Joey some money, bu t can’t give hi m directly, as Joey is too proud of hi m self to accept it. So, in order to tr ic k hi m, Chandler asks Joey to play a g a me.

In this g a me, Chandler giv es Joey an array $ a_1, a_2, dots, a_n $ ( $ n geq 2 $ ) of positive integers ( $ a_i ge 1 $ ).

Joey can per form the follo win g operation on the array any number of times:

Take two in dic es $ i $ and $ j $ ( $ 1 le i < j le n) $ .
Choose two integers $ x $ and $ y $ ( $ x, y ge 1 $ ) su ch that $ x cdot y = a_i cdot a_j $ .
Replace $ a_i $ by $ x $ and $ a_j $ by $ y $ .

In the end, Joey will get the money equal to the sum of elements of the final array.

Find the maximum amount of money $ mathrm{ans} $ Joey can get but print $ 2022 cdot mathrm{ans} $ . Why multi plied by $ 2022 $ ? Because we are never go nna see it again!

It is guar ant ee d that the product of all the elements of the array $ a $ doesn’t exc ee d $ 10^{12} $ .

输入 格式

Each test contains multi ple test cases. The first line contains the number of test cases $ t $ ( $ 1 leq t leq 4000 $ ). Descript ion of the test cases follows.

The first line of each test case contains a single integer $ n $ ( $ 2 leq n leq 50 $ ) — the length of the array $ a $ .

The second line contains $ n $ integers $ a_1, a_2, dots, a_n $ ( $ 1 leq a_i leq 10^6 $ ) — the array itself.

It’s guar ant ee d that the product of all $ a_i $ doesn’t exc eed $ 10^{12} $ (i. e. $ a_1 cdot a_2 cdot ldots cdot a_n le 10^{12} $ ).

输出 格式

For each test case, print the maximum money Joey can get multi plied by $ 2022 $ .

样例 #1

样例 输入 #1

3
3
2 3 2
2
1 3
3
1000000 1000000 1

样例 输出 #1

28308
8088
2022000000004044

提示

In the first test case, Joey can do the follo wing:

He choos es $ i = 1 $ and $ j = 2 $ (so he has $ a[i] cdot a[j] = 6 $ ), choos es $ x = 6 $ and $ y = 1 $ and makes $ a[i] = 6 $ and $ a[j] = 1 $ . $ $ $KaTeX p ar se error : Can’t use function ‘$’ in math mode at position 66: \dots= 2} [6, 1, 2] $̲ $ </ li >< li > He\dots$ $.

solution

采用贪心的思想，计算每一个数字的连乘，并放到第一个位置，其他位置为1，此时可以达到最大

//
// Created by Gowi on 2023/12/2.
//

#include <iostream>

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        long long n, s = 1;
        cin >> n;
        for (int i = 0; i < n; ++i) {
            long long a;
            cin >> a;
            s *= a;
        }
        s = s + n - 1;
        cout << 2022 * s << endl;
    }
}