欢迎来到起遇信息学
起遇信息学正处于上线筹建阶段,以下功能已全部开放免费体验: ✅ 完整题库浏览与代码提交评测(C / C++ / Python / Java 等) ✅ 入门到进阶的系列课程试读、作业与考试 ✅ AI 提示、AI 作业分析等智能助教功能 ✅ 赛事模拟与个人能力报告 ✅ 邮箱注册开放 ⏳ 付费课程订阅与微信/支付宝支付通道 ⏳ 手机号登录,微信扫码登录、微信公众号绑定 使用中如遇任何问题,欢迎通过页面底部 **"联系我们"** 与我们沟通。
CF1933B.Turtle Math: Fast Three Task
Turtle Math: Fast Three Task
You are given an array .
In one move, you can perform either of the following two operations:
- Choose an element from the array and remove it from the array. As a result, the length of the array decreases by ;
- Choose an element from the array and increase its value by .
You can perform any number of moves. If the current array becomes empty, then no more moves can be made.
Your task is to find the minimum number of moves required to make the sum of the elements of the array divisible by . It is possible that you may need moves.
Note that the sum of the elements of an empty array (an array of length ) is equal to .
Input
The first line of the input contains a single integer () — the number of test cases.
The first line of each test case contains a single integer ().
The second line of each test case contains integers ().
The sum of over all test cases does not exceed .
Output
For each test case, output a single integer: the minimum number of moves.
Note
In the first test case, initially the array . One of the optimal ways to make moves is:
- remove the current th element and get ;
As a result, the sum of the elements of the array will be divisible by (indeed, ).
In the second test case, initially, the sum of the array is , which is divisible by . Therefore, no moves are required. Hence, the answer is .
In the fourth test case, initially, the sum of the array is , which is not divisible by . By removing its only element, you will get an empty array, so its sum is . Hence, the answer is .
Samples
8
4
2 2 5 4
3
1 3 2
4
3 7 6 8
1
1
4
2 2 4 2
2
5 5
7
2 4 8 1 9 3 4
2
4 10
1
0
0
1
1
2
1
1
在线编程 IDE
建议全屏模式获得最佳体验
| 进入全屏编程 | Alt+E |
| 递交评测 | Ctrl+Enter |
| 注释/取消注释 | Ctrl+/ |
| 缩放字体 | Ctrl+滚轮 |