A simple water problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 173    Accepted Submission(s): 112

Problem Description
Dragon is watching competitions on TV. Every competition is held between two competitors, and surely Dragon's favorite. After each competition he will give a score of either 0 or 1 for each competitor and add it to the total score of that competitor. The total
score begins with zero. Here's an example: four competitors with name James, Victoria, Penghu, and Digo. First goes a competition between Penghu and Digo, and Dragon enjoys the competition and draw both 1 score for them. Then there’s a competition between
James and Victoria, but this time Dragon draw 1 for Victoria and 0 for James. Lastly a competition between James and Digo is held, but this time Dragon really dislike the competition and give zeroes for each of them. Finally we know the score for each one:
James--0, Victoria--1, Penghu--1, Digo--1. All except James are the Winner!



However, Dragon's mom comes back home again and close the TV, driving Dragon to his homework, and find out the paper with scores of all competitors. Dragon's mom wants to know how many competition Dragon watched, but it's hard through the paper. Here comes
the problem for you, given the scores of all competitors, at least how many competitions had Dragon watched?
 
Input
The first line of input contains only one integer T(<=10), the number of test cases. Following T blocks, each block describe one test case.



For each test case, the first line contains only one integers N(<=100000), which means the number of competitors. Then a line contains N integers (a1,a2,a3,...,an).ai(<=1000000) means the score of i-th
competitor.
 
Output
Each output should occupy one line. Each line should start with "Case #i: ", with i implying the case number. Then for each case just puts a line with one integer, implying the competition at least should be watched by dragon.
 
Sample Input
1
3
2 3 4
 
Sample Output
Case #1: 5
 
Source
 

HDOJ 4974 A simple water problem的更多相关文章

  1. HDU 4974 A simple water problem(贪心)

    HDU 4974 A simple water problem pid=4974" target="_blank" style="">题目链接 ...

  2. hdu - 4974 - A simple water problem(贪心 + 反证)

    题意:N个队(N <= 100000),每一个队有个总分ai(ai <= 1000000),每场比赛比赛两方最多各可获得1分,问最少经过了多少场比赛. 题目链接:http://acm.hd ...

  3. 2014多校第十场1004 || HDU 4974 A simple water problem

    题目链接 题意 : n支队伍,每场两个队伍表演,有可能两个队伍都得一分,也可能其中一个队伍一分,也可能都是0分,每个队伍将参加的场次得到的分数加起来,给你每个队伍最终得分,让你计算至少表演了几场. 思 ...

  4. HDU-4974 A simple water problem

    http://acm.hdu.edu.cn/showproblem.php?pid=4974 话说是签到题,我也不懂什么是签到题. A simple water problem Time Limit: ...

  5. BZOJ 3489: A simple rmq problem

    3489: A simple rmq problem Time Limit: 40 Sec  Memory Limit: 600 MBSubmit: 1594  Solved: 520[Submit] ...

  6. ZOJ 3686 A Simple Tree Problem

    A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each no ...

  7. hdu4976 A simple greedy problem. (贪心+DP)

    http://acm.hdu.edu.cn/showproblem.php?pid=4976 2014 Multi-University Training Contest 10 1006 A simp ...

  8. HDU 5832 A water problem(某水题)

    p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-s ...

  9. hdu 1757 A Simple Math Problem (乘法矩阵)

    A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

随机推荐

  1. VC6.0入门使用

    软件下载地址 http://pan.baidu.com/s/1qWuqFAO 新建win console 32 project,然后新建header文件.最后新建source cpp文件.如图所看到的

  2. 算法8-4:Kruskal算法

    Kruskal算法用于计算一个图的最小生成树.这个算法的过程例如以下: 依照边的权重从小到达进行排序 依次将每条边添加到最小生成树中,除非这条边会造成回路 实现思路 第一个步骤须要对边进行排序,排序方 ...

  3. mysql主键设置成auto_increment时,进行并发性能測试出现主键反复Duplicate entry &#39;xxx&#39; for key &#39;PRIMARY&#39;

    mysql主键设置成auto_increment时,进行并发性能測试出现主键反复Duplicate entry 'xxx' for key 'PRIMARY' 解决方法: 在my.cnf的[mysql ...

  4. iOS7 文本转语音 AVSpeechSynthesizer

    OS7 的这个功能确实不错.我刚试了下,用官方提供的API ,简单的几句代码就能实现文本转语音! Xcode 5.0 工程建好后首先把AVFoundation.framework 加入到工程 AVSp ...

  5. VMware虚拟机上网络连接(network type)的三种模式--bridged、host-only、NAT

    VMware虚拟机上网络连接(network type)的三种模式--bridged.host-only.NAT VMWare提供了三种工作模式,它们是bridged(桥接模式).NAT(网络地址转换 ...

  6. Mac maven环境变量配置

    近期一直在学习使用Macbook,在这里记录一下全部遇到的问题 问题起源: 1.Macbook 安装了Eclipse,Eclipse装入插件maven & git , 可是在git中clone ...

  7. 一个Jquery特效(转)

    见证花开!!码上有花 先看效果 尝试点击,IE9+ 画心 清空 ... 请自由点击画布   创意来自于网络 可搜搜[程序员表白] 看上去像是[当耐特]的作品的加工版本 代码 当然要自己写了 能力有限, ...

  8. hdu 4885 (n^2*log(n)推断三点共线建图)+最短路

    题意:车从起点出发,每次仅仅能行驶L长度,必需加油到满,每次仅仅能去加油站或目的地方向,路过加油站就必需进去加油,问最小要路过几次加油站. 開始时候直接建图,在范围内就有边1.跑最短了,再读题后发现, ...

  9. 将ACCESS数据库迁移到SQLSERVER数据库

    原文:将ACCESS数据库迁移到SQLSERVER数据库 将ACCESS数据库迁移到SQLSERVER数据库 ACCESS2000文件 用ACCESS2007打开,并迁移到SQLSERVER2005里 ...

  10. 小米2S 中文和英文支持TWRP,真实双系统支持

    经过我几天的努力小米2S的TWRP 的功能已经完美了. 支持功能 : 中文和英文显示能相互切换 真实双系统功能已经完成95%. 刷入手机方法.由于时间原因我只制作了img文件.没有制作成卡刷包格式. ...