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

Comparison of Android versions
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: = <

题目大意&解题思路：

　　　　1、比较两个字符串的第一个字母的大小.
　　　　2、如果两个字符串的第二个字母不同就比较接下来的三个字母的大小.
  　　　　    如果两个字符串的第二个字母相同就比较剩余的四个字母.

　　　　PS：英语好读懂题就不难。TT

Code：

 /*************************************************************************
     > File Name: shanghai_1010.cpp
     > Author: Enumz
     > Mail: 369372123@qq.com
     > Created Time: 2014年11月02日 星期日 13时43分08秒
  ************************************************************************/

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<string>
 #include<cstring>
 #include<list>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<algorithm>
 #include<cmath>
 #include<bitset>
 #include<climits>
 #define MAXN 100000
 using namespace std;
 char str1[],str2[];
 int main()
 {
     int T;
     cin>>T;
     int times=;
     while (T--)
     {
         cin>>str1>>str2;
         printf("Case %d:",times++);
         char tmp1[]={},tmp2[]={};
         tmp1[]=str1[];
         tmp2[]=str2[];
         if (strcmp(tmp1,tmp2)>)
             printf(" >");
         else if (strcmp(tmp1,tmp2)<)
             printf(" <");
         else printf(" =");
         tmp1[]=str1[];
         tmp2[]=str2[];
         if (strcmp(tmp2,tmp1)!=)
         {
             str1[]=,str2[]=;
             if (strcmp(str1+,str2+)>)
                 printf(" >\n");
             else if( strcmp(str1+,str2+)<)
                 printf(" <\n");
             else
                 printf(" =\n");
         }
         else
         {
             if (strcmp(str1+,str2+)>)
                 printf(" >\n");
             else if (strcmp(str1+,str2+)<)
                 printf(" <\n");
             else
                 printf(" =\n");
         }
     }
     return ;
 }