# Blue Bridge Cup Basics practice hex to octal

Source: Internet
Author: User

Basic practice hex to octal time limit: 1.0s memory limit: 512.0MB Problem Description
Given n hexadecimal positive integers, output their corresponding octal numbers.

Input Format
The first behavior of the input is 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.

output Format
Outputs n rows, each of which enters a corresponding octal positive integer.

"Note
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

the prompt
The hexadecimal number is converted to a number of decimal digits, and then a binary number is converted into octal. The hexadecimal number is converted to a binary number, and then the binary number is converted to octal. I just started to define a char character array that has been timed out, see the string defined by string on the net does not time out, depressed timeout code
`#include <stdio.h> #include <string.h> #include <string> #include <math.h> #include < Algorithm> #define LL Long long#define PI atan (1.0) *4#define DD doublea#define MAX 100100#define mod 10007using Namespac    E Std;char a[max];int p[max*10];int main () {int n,m,j,i,len,sum;    scanf ("%d", &n);    while (n--) {scanf ("%s", &a);    Char s[max*10]= "";    Len=strlen (a);        for (i=0;i<len;i++)//conversion to binary {switch (A[i]) {case ' 0 ': strcat (S, "0000");        Case ' 1 ': strcat (S, "0001");        Case ' 2 ': strcat (S, "0010");        Case ' 3 ': strcat (S, "0011");        Case ' 4 ': strcat (S, "0100");        Case ' 5 ': strcat (S, "0101");        Case ' 6 ': strcat (S, "0110");        Case ' 7 ': strcat (S, "0111");        Case ' 8 ': strcat (S, "n");        Case ' 9 ': strcat (S, "1001");        Case ' A ': strcat (S, "1010");        Case ' B ': strcat (S, "1011");        Case ' C ': strcat (S, "1100"); CaSe ' D ': strcat (S, "1101");        Case ' E ': strcat (S, "1110");        Case ' F ': strcat (S, "1111");        Default:break;    }}//puts (s);    Strrev (s);    int Len1=strlen (s);    if (len1%3==1)//Because the binary to octal, is a three-bit binary into an octal {//So to complement, in front of 0 so that the length can be divisible by 3 strcat (S, "00");    len1+=2;    } else if (len1%3==2) {strcat (S, "0");    Len1+=1;    } int op=1;    for (i=0;i<len1-2;i+=3)//binary turn octal {p[op++]= (s[i]-' 0 ') *1+ (s[i+1]-' 0 ') *2+ (s[i+2]-' 0 ');    } int flag=1;    for (i=op;i>0;i--) {if (p[i]!=0) flag=0;    if (P[i]==0&&flag) continue;    printf ("%d", p[i]);    } printf ("\ n");  }return 0;}`

AC Code:

`#include <stdio.h> #include <string.h> #include <string> #include <math.h> #include < Algorithm> #define LL Long long#define PI atan (1.0) *4#define DD doublea#define MAX 100100#define mod 10007using Namespac    E std;int p[max*10];int Main () {int n,m,j,i,len,sum;    scanf ("%d", &n);    while (n--) {string a,s; A.resize (MAX);    Requires pre-allocated space scanf ("%s", &a[0]);    S= "";        Len=a.length ();//String length for (i=0;i<len;i++) {switch (A[i]) {case ' 0 ': s+= "0000";        Case ' 1 ': s+= "0001";        Case ' 2 ': s+= "0010";        Case ' 3 ': s+= "0011";        Case ' 4 ': s+= "0100";        Case ' 5 ': s+= "0101";        Case ' 6 ': s+= "0110";        Case ' 7 ': s+= "0111";        Case ' 8 ': s+= "n";        Case ' 9 ': s+= "1001";        Case ' A ': s+= "1010";        Case ' B ': s+= "1011";        Case ' C ': s+= "1100";        Case ' D ': s+= "1101";        Case ' E ': s+= "1110"; Case ' F ':s+= "1111";        Default:break;    }}//puts (s);    Strrev (s);    int Len1=s.length ();  if (len1%3==1) {s= "xx" +s;    00+s is to add 00 to the string before s+00 is to add 00 to the string after len1+=2;    } else if (len1%3==2) {s= "0" +s;    Len1+=1;    } int op=1;    for (i=0;i<len1-2;i+=3) {p[op++]= (s[i]-' 0 ') *4+ (s[i+1]-' 0 ') *2+ (s[i+2]-' 0 ') * *;    } int flag=1;    for (i=0;i<op;i++) {if (p[i]!=0) flag=0;    if (P[i]==0&&flag) continue;    printf ("%d", p[i]);    } printf ("\ n");  }return 0;}`

Blue Bridge Cup Basics practice hex to octal

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