HDU 5099 Comparison of Android versions（字符串）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5099

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: = <

 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

 

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PS:

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，

如果第二个字母相同就比较剩余的四个字母！

代码如下：

#include <cstdio>#include <cstring>const int maxn = 17;int main(){    int t;    char a[maxn], b[maxn];    char aa[maxn], bb[maxn];    int cas = 0;    scanf("%d",&t);    getchar();    while(t--)    {        scanf("%s",a);        scanf("%s",b);        printf("Case %d: ",++cas);        if(a[0] > b[0])        {            printf("> ");        }        else if(a[0] == b[0])        {            printf("= ");        }        else if(a[0] < b[0])        {            printf("< ");        }        if(a[1] == b[1])        {            aa[0] = a[2];            aa[1] = a[3];            aa[2] = a[4];            aa[3] = a[5];            aa[4] = '\0';            bb[0] = b[2];            bb[1] = b[3];            bb[2] = b[4];            bb[3] = b[5];            bb[4] = '\0';            if(strcmp(aa,bb) < 0)            {                printf("<");            }            else if(strcmp(aa,bb) == 0)            {                printf("=");            }            else if(strcmp(aa,bb) > 0)            {                printf(">");            }        }        else        {            aa[0] = a[2];            aa[1] = a[3];            aa[2] = a[4];            aa[3] = '\0';            //aa[3] = a[5];            bb[0] = b[2];            bb[1] = b[3];            bb[2] = b[4];            bb[3] = '\0';            //bb[3] = b[5];            if(strcmp(aa,bb) < 0)            {                printf("<");            }            else if(strcmp(aa,bb) == 0)            {                printf("=");            }            else if(strcmp(aa,bb) > 0)            {                printf(">");            }        }        printf("\n");    }    return 0;}