hdoj-5099-Comparison of Android versions

Comparison of Android versions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1172 Accepted Submission(s): 470

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: = <

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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#include<stdio.h> 
#include<string.h>
int main(){
	int tt,ncas=0;
	scanf("%d",&tt);
	while(tt--){
		ncas++;
		char s[10],t[10];
	
		scanf("%s%s",s,t);	
		int len1=strlen(s),len2=strlen(t);
		printf("Case %d: ",ncas);

		if(s[0]==t[0]){
			printf("= ");
		}
		else{
			if(s[0]>t[0]) printf("> ");
			else printf("< ");
		}
		if(s[1]!=t[1]){
			s[len1-1]='\0';
			t[len2-1]='\0';
		}
		int temp=strcmp(s+2,t+2);
		if(temp>0) printf(">");
		else if(temp<0) printf("<");
		else printf("=");
		printf("\n");
	}
	return 0;
}

#include<stdio.h> 
#include<string.h>
int f(char a[]){
    int i,res=0;
    for(i=3;;++i){
      if(a[i]>='0'&&a[i]<='9') res=res*10+a[i]-'0';
      else return res;
    }
}
int main(){
    int tt,ncas=0;
    scanf("%d",&tt);
    while(tt--){
        ncas++;
        char s[10],t[10];
        
        scanf("%s%s",s,t);
		int len1=strlen(s),len2=strlen(t);
        printf("Case %d: ",ncas);
        if(s[0]==t[0]){
            printf("= ");
        }
        else{
            if(s[0]>t[0]) printf("> ");
            else printf("< ");
        }
        if(s[2]>t[2]) printf(">");
        else if(s[2]<t[2]) printf("<");
        else{
		    int x=f(s),y=f(t);
        	if(x==y){
            if(s[1]==t[1]){
                if(s[len1-1]==t[len2-1]) printf("=");
                else if(s[len1-1]>t[len2-1]) printf(">");
                else printf("<");
            }
            else printf("=");
	        }
	        else if(x>y) printf(">");
	        else printf("<");
        }
        printf("\n");
    }
    return 0;
}