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