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。
CF1780A.Hayato and School
Hayato and School
Today Hayato came home from school with homework.
In the assignment, Hayato was given an array of length . The task was to find numbers in this array whose sum is odd. At school, he claimed that there are such numbers, but Hayato was not sure, so he asked you for help.
Answer if there are such three numbers, and if so, output indices , , and such that is odd.
The odd numbers are integers that are not divisible by : , , , and so on.
Input
The first line contains a single integer () — the number of test cases.
For each test case, the first line contains one integer () — the length of .
The second line contains integers () — the array .
It is guaranteed that the sum of over all test cases does not exceed .
Output
For each test case, in the first line print one word "YES" (without quotes) if there are numbers with an odd sum or "NO" (without quotes) if there are no such numbers.
If the answer exists, then on the second line print distinct integers () — the indices of the numbers. If there are several answers, output any.
Note
In the first test case, there is one way to choose numbers, and since , this triple is fine for us.
In the second test case, you need to choose the numbers , since .
In the third test case, there is one way to choose three numbers, but is an even number, so the required triple does not exist.
In the fifth test case, no matter what three numbers we choose, their sum is even.
Samples
6
3
1 1 1
4
1 1 2 2
3
1 2 3
5
1 4 5 1 2
4
2 6 2 4
5
5 6 3 2 1
YES
1 2 3
YES
3 4 1
NO
YES
1 3 4
NO
YES
1 3 5
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 |