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：
1 /************************************************************************* 2 > File Name: shanghai_1010.cpp 3 > Author: Enumz 4 > Mail: 369372123@qq.com 5 > Created Time: 2014年11月02日 星期日 13时43分08秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<cstdio> 10 #include<cstdlib> 11 #include<string> 12 #include<cstring> 13 #include<list> 14 #include<queue> 15 #include<stack> 16 #include<map> 17 #include<set> 18 #include<algorithm> 19 #include<cmath> 20 #include<bitset> 21 #include<climits> 22 #define MAXN 100000 23 using namespace std; 24 char str1[100],str2[100]; 25 int main() 26 { 27 int T; 28 cin>>T; 29 int times=1; 30 while (T--) 31 { 32 cin>>str1>>str2; 33 printf("Case %d:",times++); 34 char tmp1[4]={0},tmp2[4]={0}; 35 tmp1[0]=str1[0]; 36 tmp2[0]=str2[0]; 37 if (strcmp(tmp1,tmp2)>0) 38 printf(" >"); 39 else if (strcmp(tmp1,tmp2)<0) 40 printf(" <"); 41 else printf(" ="); 42 tmp1[0]=str1[1]; 43 tmp2[0]=str2[1]; 44 if (strcmp(tmp2,tmp1)!=0) 45 { 46 str1[5]=0,str2[5]=0; 47 if (strcmp(str1+2,str2+2)>0) 48 printf(" > "); 49 else if( strcmp(str1+2,str2+2)<0) 50 printf(" < "); 51 else 52 printf(" = "); 53 } 54 else 55 { 56 if (strcmp(str1+2,str2+2)>0) 57 printf(" > "); 58 else if (strcmp(str1+2,str2+2)<0) 59 printf(" < "); 60 else 61 printf(" = "); 62 } 63 } 64 return 0; 65 }