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。
CF2180A.Carnival Wheel
Carnival Wheel
You have a prize wheel divided into sections, numbered from to . The sections are arranged in a circle, so after section , the numbering continues again from section .
Initially, the prize pointer is at section . Each time you spin the wheel, the pointer moves exactly sections forward. That is, after one spin, the pointer moves from section to section , after two spins to , and so on.
You may spin the wheel any number of times (including zero). After you stop, the section where the pointer finally lands determines your prize: you receive an amount equal to the number of that section.
What is the maximum prize you can obtain?
Here, denotes the remainder from dividing by .
Input
Each test contains multiple test cases. The first line contains the number of test cases (). The description of the test cases follows.
The first line of each test case contains three integers , and (, ).
Output
For each test case, output the maximum prize that can be obtained.
Note
In the first test case, by spinning the wheel three times and then claiming the reward, you can obtain a maximum value of . The sequence of pointer positions is:
In the second test case, the pointer will remain on section indefinitely.
In the fourth test case, with and starting from section , the pointer will iterate over all sections, including the last one.
Samples
4
5 3 2
2 0 6
8 2 4
100 0 1
4
0
6
99
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 |