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



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
FRF85B EPF21B
KTU84L KTU84M

[align=left]Sample Output[/align]

Case 1: > >
Case 2: = <

[align=left]Source[/align]
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;
}