Blue Bridge Cup-basic exercise-16 binary to octal (is timed out!!!) )

Basic practice hex to octal time limit: 1.0s memory limit: 512.0MBThe problem describes the given n hexadecimal positive integers, outputting their corresponding octal numbers. Enter the first behavior of the input format as a positive integer n (1<=n<=10).
Next n rows, each line a string of 0~9, uppercase letters A~F, representing the hexadecimal positive integer to be converted, each hexadecimal number is not more than 100000. The output format outputs n rows, and each behavior enters a corresponding octal positive integer. Note that the hexadecimal number entered does not have a leading 0, such as 012A.
The octal number of the output cannot have a leading 0. Sample Input 2
39
123ABC Sample Output 71
4435274 indicates that the hexadecimal number is converted to a binary number, and then converted to octal by a number of digits.

Brother I tried, or overtime, I did not think! Blue Bridge, the fucking data is so sick! Do you have a download to see his data? My algorithm is the same as the C + + algorithm, but I am timed out! C + + will pass! Java is still not suitable for a problem!

Import Java.io.*;import java.util.*;p ublic class main{public static void Main (string[] args) {//TODO auto-generated Metho D stubscanner input = new Scanner (system.in); int n=input.nextint (); Input.nextline (); for (int i=0;i<n;i++) {String STR1 = Input.nextline ();                              String str2= ""; Used to string the concatenation of char a[] = Str1.tochararray (); Record the string just entered into the character array for (int j=0;j<str1.length (); j + +)//16 binary conversion to binary {switch (A[j]) {case ' 0 ': str2+= "0000"; Case ' 1 ': str2+= "0001"; Break;case ' 2 ': str2+= "0010"; Break;case ' 3 ': str2+= "0011"; Break;case ' 4 ': str2+= "0100"; Case ' 5 ': str2+= "0101"; Break;case ' 6 ': str2+= "0110"; Break;case ' 7 ': str2+= "0111"; Break;case ' 8 ': str2+= "n"; break; Case ' 9 ': str2+= "1001", Break;case ' A ': str2+= "1010"; Break;case ' B ': str2+= "1011"; Break;case ' C ': str2+= "1100"; break; Case ' D ': str2+= "1101"; Break;case ' E ': str2+= "1110"; Break;case ' F ': str2+= "1111"; break;default:break;}}             System.out.println (STR2); Number of digits before the test correction//correction digit int arry[] = new Int[10000001];if (Str2.length ()%3==1) {str2= "xx" +STR2;} if (Str2.length ()%3==2) {str2= "0" +STR2;}              System.out.println (STR2);     Test the corrected number of digits in the case of//system.out.println (Str2.length ());       Test the corrected number of digits!char b[] = Str2.tochararray ();                             After correcting the number of digits, the original A-character array cannot be used!, and replaced by the B-character array int j=0; Record 8 binary digits for (int k=0;k<=str2.length () -3;k+=3) {arry[j]= (b[k]-' 0 ') *4+ (b[k+1]-' 0 ') *2+ (b[k+2]-' 0 '); j + +;} for (int k=0;k<j;++k) {if (k==0&&arry[k]==0) continue; System.out.print (Arry[k]);} System.out.println ();}}}

C++:AC Code!

#include <iostream> #include <string>using namespace Std;int arr[10000001];int main () {int n,len_str,i,j;    String str,str2;    cin>>n;        while (n--) {cin>>str;        Len_str=str.length ();                Str2= ""; 16 binary conversion to binary for (i=0;i<len_str;++i) {switch (Str[i]) {case ' 0 ': str2+= "            0000 "; break;            Case ' 1 ': str2+= "0001";            Case ' 2 ': str2+= "0010";            Case ' 3 ': str2+= "0011";            Case ' 4 ': str2+= "0100";            Case ' 5 ': str2+= "0101";            Case ' 6 ': str2+= "0110";            Case ' 7 ': str2+= "0111";            Case ' 8 ': str2+= "n";            Case ' 9 ': str2+= "1001";            Case ' A ': str2+= "1010";            Case ' B ': str2+= "1011";            Case ' C ': str2+= "1100";            Case ' D ': str2+= "1101";            Case ' E ': str2+= "1110";       Case ' F ': str2+= "1111";     Default:break;                }}//fixed number of digits if (len_str%3==1) str2= "xx" +STR2;                        else if (len_str%3==2) str2= "0" +STR2;        Len_str=str2.length ();        binary conversion octal j=0;            for (i=0;i<=len_str-2;i+=3) {arr[j]= (str2[i]-' 0 ') *4+ (str2[i+1]-' 0 ') *2+ (str2[i+2]-' 0 ');        ++j;            } for (I=0;i<j;++i) {if (i==0 && arr[i]==0) continue;        cout<<arr[i];            } cout<<endl; } return 0;}

A classmate of the Java AC Code: (Must learn!) )//Attached author URL: http://blog.csdn.net/u010887744/article/details/44889037

Import Java.io.bufferedreader;import java.io.ioexception;import Java.io.inputstreamreader;import Java.util.Scanner ;p Ublic class Main {public static void Main (string[] args) throws IOException {BufferedReader in = new Buffere        Dreader (New InputStreamReader (system.in));        int n = integer.parseint (In.readline ());        String a[] = new String[n];        for (int i = 0; i < n; i++) {A[i] = In.readline ();            } for (int i = 0; i < n; i++) {char[] temp = A[i].tochararray ();            StringBuffer s2 = new StringBuffer ();            2 int k = temp.length; for (int j = 0; J < K; J + +) {switch (Temp[j]) {case ' 0 ': S2.append ("                    0000 ");                Break                    Case ' 1 ': S2.append ("0001");                Break                    Case ' 2 ': S2.append ("0010");        Break        Case ' 3 ': S2.append ("0011");                Break                    Case ' 4 ': S2.append ("0100");                Break                    Case ' 5 ': S2.append ("0101");                Break                    Case ' 6 ': S2.append ("0110");                Break                    Case ' 7 ': S2.append ("0111");                Break                    Case ' 8 ': S2.append ("1000");                Break                    Case ' 9 ': S2.append ("1001");                Break                    Case ' A ': S2.append ("1010");                Break                    Case ' B ': s2.append ("1011");                Break                    Case ' C ': S2.append ("1100");                Break                    Case ' D ': S2.append ("1101");                Break    Case ' E ':                S2.append ("1110");                Break                    Case ' F ': s2.append ("1111");                Break            }}//2 to 8 stringbuffer s3 = new StringBuffer ();            int m = 0;                if (4 * k% 3 = = 1) {s3.append (s2.substring (0, 1));            M + = 1; } else if (4 * k% 3 = = 2) {switch (s2.substring (0, 2)) {case "S3":                    Append ("1");                Break                    Case "Ten": S3.append ("2");                Break                    Case "One": S3.append ("3");                Break                Default:break;            } m + = 2; } for (int j = m; J < 4 * k;)                    {Switch (S2.substring (J, J + 3)) {case ': S3.append ("0"); BreAk                    Case "001": S3.append ("1");                Break                    Case "010": S3.append ("2");                Break                    Case "011": S3.append ("3");                Break                    Case "S3.append": ("4");                Break                    Case "101": S3.append ("5");                Break                    Case "S3.append": ("6");                Break                    Case "111": S3.append ("7");                Break            } j + = 3; }//Delete 0//Use Delete (old is 0) or charAt if (s3.length () = = 2 && s3.charat            (0) = = ' 0 ') {//0-->00-->delete 00-->notany System.out.println (s3.substring (1));                } else {int q = 0;            while (S3.charat (q) = = ' 0 ') {        q++;                } String s = s3.tostring ();            System.out.println (s3.substring (q)); }        }    }}

Blue Bridge Cup-basic exercise-16 binary to octal (is timed out!!!) )