信院政〔2015〕17号
各班级、社团：
为激发我院大学生的科研兴趣，进一步提高我院大学生计算机专业知识的应用能力，鼓励广大学子积极参与科学创新实践活动，为学生提供一个相互学习、充分展示自己创新能力的平台，特决定开展2015大学生程序设计竞赛，现将有关事项通知如下：
一、竞赛组织
主办单位：294俄罗斯专享会
承办单位：计算机与网络科技协会
二、竞赛模式和环境
竞赛模式：采用ICPC竞赛模式
操作系统：Windows
语言：C/C++、Java
开发工具：Dev C++、Eclipse、codeblock、JCreator、VC++6.0
三、参赛对象：
信息学院在校学生
四、关键日期
报名截止日期：2015年5月8日（以班级为统计单位，注明年级、专业、团队负责人、联系方式，通过邮件发jsjds@chzu.edu.cn）
竞赛举办日期：2015年5月10日下午
五、竞赛地点
信息楼机房
六、竞赛规则
（一）参赛队伍组成
比赛以队为参赛单位，队的组成不超过三名参赛队员，每个年级专业可派≤3支代表队。
（二）比赛开展
竞赛中每队三人拥有一台计算机，竞赛命题5题，比赛时间为3个小时。
参赛队员可以携带诸如书、手册、程序清单等参考资料。参赛队员不能携带任何可用计算机处理的软件或数据(不允许任何私人携带的U盘、磁盘或计算器)。
参赛队员不能携带任何类型的通讯工具，包括无线电接收器、移动电话。
在竞赛中，参赛队员不得和竞赛技术委员会指定工作人员以外的人交谈；系统支持人员可以回答和系统相关的问题，例如解释系统错误信息。
竞赛的预定时间为3小时，但当竞赛进行一定时间后，竞赛技术委员会可以因为出现不可预见的事件而调整比赛时间长度，一旦比赛时间长度发生改变，将会以及时并且统一的方式通告所有参赛队员。
比赛过程中，选手若要上厕所，必须征得裁判批准，且有裁判陪同。
（三）竞赛评分
试题的解答提交裁判运行，每一次运行会被判为正确或者错误，判决结果会及时通知参赛队伍。
参赛队员有权利通过提交解释请求，针对试题描述中的不明确或错误的部分提问。如果裁判确认试题中确实存在不明确或错误的部分，将会通告所有参赛队伍进行声明或更正。
竞赛裁判主要负责判定解答提交是否正确，根据正确解答的解题数目进行排名，如果多支队伍解题数量相同，则根据总用时加上惩罚时间进行排名。总用时和惩罚时间由每道解答正确的试题的用时加上惩罚时间而成。每道试题用时将从竞赛开始到试题解答被判定为正确为止，其间每一次错误的运行将被加罚20分钟时间，未正确解答的试题不记时。
七、结果应用
学院将参考比赛结果综合考核，择优推荐参加安徽省程序设计大赛。
附件：赛题参考格式
294俄罗斯专享会
计算机与网络科技协会
2015-5-5
赛题参考格式如下：
Problem C: Biorhythms
Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier.
Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
You will be given a number of cases. The input for each case consists of one line of four integers p, e, i, and d. The values p, e, and i are the number of days from the beginning of the current year at which the physical, emotional, and intellectual cycles peak, respectively. The value d is the given date and may be smaller than any of p, e, or i. All values are non-negative and at most 365, and you may assume that a triple peak will occur within 21252 days of the given date. The end of input is indicated by a line in which p = e = i = d = -1.
Output
For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form:
Case 1: the next triple peak occurs in 1234 days.
Use the plural form ``days'' even if the answer is 1.
Sample Input
1
0 0 0 0
0 0 0 100
5 20 34 325
4 5 6 7
283 102 23 320
203 301 203 40
-1 -1 -1 -1
Sample Output
Case 1: the next triple peak occurs in 21252 days.
Case 2: the next triple peak occurs in 21152 days.
Case 3: the next triple peak occurs in 19575 days.
Case 4: the next triple peak occurs in 16994 days.
Case 5: the next triple peak occurs in 8910 days.
Case 6: the next triple peak occurs in 10789 days.
终审人：王松劼