现场赛的时候错了十四次。

。。。

Comparison of Android versions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 76    Accepted Submission(s): 60

Problem Description
As an Android developer, itˇs really not easy to figure out a newer version of two kernels, because Android is updated so frequently and has many branches. Fortunately, Google identifies individual builds with a short build code, e.g. FRF85B.



The first letter is the code name of the release family, e.g. F is Froyo. The code names are ordered alphabetically. The latest code name is K (KitKat).

The second letter is a branch code that allows Google to identify the exact code branch that the build was made from, and R is by convention the primary release branch.



The next letter and two digits are a date code. The letter counts quarters, with A being Q1 2009. Therefore, F is Q2 2010. The two digits count days within the quarter, so F85 is June 24 2010.



Finally, the last letter identifies individual versions related to the same date code, sequentially starting with A; A is actually implicit and usually omitted for brevity.



Please develop a program to compare two Android build numbers.
 
Input
The first line is an integer n (1 <= n <= 2000), which indicates how many test cases need to process.



Each test case consists of a single line containing two build numbers, separated by a space character.
 
Output
For each test case, output a single line starting with ¨Case #: 〃 (# means the number of the test case). Then, output the result of release comparison as follows:



● Print "<" if the release of the first build number is lower than the second one;

● Print "=" if the release of the first build number is same as he second one;

● Print ">" if the release of the first build number is higher than the second one.



Continue to output the result of date comparison as follows:

● Print "<" if the date of the first build number is lower than the second one;

● Print "=" if the date of the first build number is same as he second one;

● Print ">" if the date of the first build number is higher than the second one.



If two builds are not in the same code branch, just compare the date code; if they are in the same code branch, compare the date code together with the individual version.
 
Sample Input
2
FRF85B EPF21B
KTU84L KTU84M
 
Sample Output
Case 1: > >
Case 2: = <
 
Source
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; char s1[10],s2[10];
char ONE,TWO; void BJ1()
{
if(s1[0]<s2[0]) ONE='<';
else if(s1[0]==s2[0]) ONE='=';
else ONE='>';
} void BJ2()
{
if(s1[2]<s2[2]) TWO='<';
else if(s1[2]==s2[2])
{
int day1=(s1[3]-'0')*10+s1[4]-'0';
int day2=(s2[3]-'0')*10+s2[4]-'0';
if(day1<day2) TWO='<';
else if(day1==day2)
{
if(s1[1]==s2[1])
{
if(s1[5]<s2[5]) TWO='<';
else if(s1[5]==s2[5]) TWO='=';
else TWO='>';
}
else TWO='=';
}
else TWO='>';
}
else TWO='>';
} int main()
{
int T_T;
int cas=1;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%s%s",s1,s2);
if(strlen(s1)<6) s1[5]='A';
if(strlen(s2)<6) s2[5]='A';
BJ1();BJ2();
printf("Case %d: %c %c\n",cas++,ONE,TWO);
}
return 0;
}

HDOJ 5099 Comparison of Android versions 坑题的更多相关文章

  1. 模拟 HDOJ 5099 Comparison of Android versions

    题目传送门 /* 题意:比较型号的大小 模拟:坑点在长度可能为5,此时设为'A' */ #include <cstdio> #include <algorithm> #incl ...

  2. HDU5099——Comparison of Android versions(简单题)(2014上海邀请赛重现)

    Comparison of Android versionsProblem DescriptionAs an Android developer, itˇs really not easy to fi ...

  3. Comparison of Android versions(strcmp的应用)

    Description As an Android developer, itˇs really not easy to figure out a newer version of two kerne ...

  4. hdoj-5099-Comparison of Android versions

    Comparison of Android versions Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (J ...

  5. Android开发如何准备技术面试(含Android面试押题)

    今年毋庸置疑是找工作的寒冬,每一个出来找工作的同学都是值得尊敬的.现在找工作,虽然略难,但是反过来看也会逼迫我们成为更加优秀的自己. 但是不管是旺季还是寒冬,有一些优秀的同学找工作还是挺顺利的.所以说 ...

  6. Android测试基础题(三)

    今天接着给大家带来的是Android测试基础题(三).    需求:定义一个排序的方法,根据用户传入的double类型数组进行排序,并返回排序后的数组 俗话说的好:温故而知新,可以为师矣 packag ...

  7. 【BZOJ-1952】城市规划 [坑题] 仙人掌DP + 最大点权独立集(改)

    1952: [Sdoi2010]城市规划 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 73  Solved: 23[Submit][Status][ ...

  8. hdu 5455 Fang Fang 坑题

    Fang Fang Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5455 ...

  9. String类面试坑题

    1.面试坑题F:\SHJT\JavaWorkspace\JavaSE\workspace\day13ezra\src\cn\itcast\sh\classcode\BTStringLastIndexO ...

随机推荐

  1. python spark 求解最大 最小 平均

    rdd = sc.parallelizeDoubles(testData); Now we’ll calculate the mean of our dataset.   1 LOGGER.info( ...

  2. CodeForces 660A

    Description You are given an array of n elements, you must make it a co-prime array in as few moves ...

  3. NEU操作系统实验课4——线程同步

    实验要求: 创建两个线程按数字顺序打印10以下自然数,其中一个线程打印1-3及8-10:另一个线程打印4-6.要求使用线程同步机制实现上述打印顺序. 看网上的资料学习了两种写法 C++11的,使用了s ...

  4. php pdo具体操作

    0x01:测试PDO是否安装成功 运行如下代码,如果提示参数错误,说明PDO已经安装,如果说明对象不存在,则修改PHP配置文件php.ini,取消php_pdo_yourssqlserverhere. ...

  5. 乐字节-Java8核心特性实战之Lambda表达式

    大家好,小乐又来给大家分享Java8核心特性了,上一篇文章是<乐字节|Java8核心实战-接口默认方法>,这次就来讲Java8核心特征之Lambda表达式. Java8 引入Lambda表 ...

  6. java使用FileUtils文件操作神器

    前言: 在工作当中我们往往遇到很多文件的操作,我们也习惯写一些自己定义的工具类来简化文件操作,其实apache的commons的FileUtils类就是这样一个工具类,使用它能大大的简化我们对文件的操 ...

  7. asp.net 字符串过滤

    /// <summary> /// 去除HTML标记 /// </summary> /// <param name="Htmlstring">包 ...

  8. Centos7 docker nginx容器搭建

    一.安装docker http://www.cnblogs.com/WJ--NET/p/8553807.html 二.创建Dockerfile #创建文件夹 mkdir centos_nginx cd ...

  9. Java 解决一些ACM中大数问题

    大数中算术运算结果的首选标度 运算 结果的首选标度 加 max(addend.scale(), augend.scale()) 减 max(minuend.scale(), subtrahend.sc ...

  10. 我的新书《计算机图形学基础(OpenGL版)》

    我的新书<计算机图形学基础(OpenGL版)>今年6月份在清华大学出版社出版了!新书与原在机械工业出版社出的<计算机图形学>相比,主要有以下不同: 1.加重OpenGL的内容, ...