Comparison of Android versions

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: = <

题解：读懂题意，主要是第二个字母的判断。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define mem(a,b) memset(a,b,sizeof(a))
#define For(a,b) for(int a=0;a<b;a++)
using namespace std;
const int maxn =  1e3+500;
const int INF = 0x3f3f3f3f;
const int inf = 0x3f;
typedef long long ll;
const ll mod = 1e9+7;
char c[2][maxn];
int juge(char a,char b) {
    int flag = 0;
    if(a>b)
        return 1;
    if(a<b)
        return -1;
    return 0;
    }
int main() {
    int t;
    int flag1,flag2;
    cin >> t;
    For(q,t) {
        cin >>c[0]>>c[1];
        flag1 = juge(c[0][0],c[1][0]);
        flag2 = juge(c[0][1],c[1][1]);
        if(flag2 != 0) {
            flag2 = juge(c[0][2],c[1][2]);
            if(flag2 == 0)
                flag2 = juge(c[0][3],c[1][3]);
            if(flag2 == 0)
                flag2 = juge(c[0][4],c[1][4]);
            }
        else{
            if(flag2 == 0)
                flag2 = juge(c[0][2],c[1][2]);
            if(flag2 == 0)
                flag2 = juge(c[0][3],c[1][3]);
            if(flag2 == 0)
                flag2 = juge(c[0][4],c[1][4]);
            if(flag2 == 0)
                flag2 = juge(c[0][5],c[1][5]);
        }
        cout <<"Case "<<q+1<<": ";
        if(flag1 == 0)
            cout <<"= ";
        else if(flag1 < 0)
            cout <<"< ";
        else
            cout <<"> ";
        if(flag2 == 0)
            cout <<"=";
        else if(flag2 <0)
            cout <<"<";
        else
            cout <<">";
        cout << endl;
        }
    return 0;
    }