2014上海全国邀请赛 j 题！！

Comparison of Android versions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

2FRF85B EPF21BKTU84L KTU84M

 

Sample Output

Case 1: > >Case 2: = <

 

看完题目我才发现英语是多么的重要！！！！！！

以后好好学英语~~~~(>_<)~~~~ 

题意：比较版本发行的大小和时间的大小；

思路：比较版本时就比较第一个字母就ok了；

比较时间时还要判断是否是in the same code branch （就是看第二个字母是否相同）；之前看的时候差点晕死了:-(

相同就多比较最后一个字母，不相同就不比较最后一个字母；

很奇怪是吧，好像这场比赛题目都这么奇怪:-O

AC代码（写的比较挫。。）：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int main(){char str1[10], str2[10], ans1, ans2;int T;scanf("%d", &T);for(int k=1; k<=T; k++){int a, b;char s1[10], s2[10];scanf("%s %s", str1, str2);if(str1[0]>str2[0]) ans1='>';else if(str1[0]==str2[0]) ans1='=';else if(str1[0]<str2[0]) ans1='<';sscanf(str1+3, "%d", &a); sscanf(str2+3, "%d", &b);if(str1[2]>str2[2]) ans2='>';else if(str1[2]<str2[2]) ans2='<';else if(str1[2]==str2[2]) {if(a>b)ans2='>';else if(a<b)ans2='<';else if(a==b && str1[1]!=str2[1]) ans2='=';else if(a==b && str1[1]==str2[1]){if(str1[5]>str2[5]) ans2='>';else if(str1[5]<str2[5]) ans2='<';else if(str1[5]==str2[5]) ans2='=';}}printf("Case %d: %c %c\n", k, ans1, ans2);}return 0;}