Blue Bridge Cup Basics practice hex to octal "string processing"

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 Tips the hexadecimal number is converted to a number of decimal digits, and then a binary number is converted into octal.

Idea: Convert first to binary and then into octal

Code:

 #include < stdio.h> #include <string.h> #define M 100005char S[m*4];char str[m];int ans[150005];int main () {int N;//char TEMP[6];SCANF ("%d", &n), while (n--) {scanf ("%s", str+1), memset (s, 0, sizeof (s)), int i, j, len = strlen (str+1) +1, temp; printf ("%d.. \ n ", Len); for (i = 1; i < Len; ++i) {if (Str[i] >= ' 0 ' &&str[i] <= ' 9 ') {temp = str[i]-' 0 ';} else temp = 10+str[i]-' A '; j = I*4;while (temp) {S[j] = temp%2+ ' 0 '; j--;temp/= 2;} while (J > (i-1)) {s[j--] = ' 0 ';}} printf ("%s", &s[1]), memset (ans, 0, sizeof (ANS)), int tot = 0;for (i = (len-1), i-3 >= 0; I-= 3) {ans[tot++] = (s[ i]-' 0 ') + (s[i-1]-' 0 ') *2+ (s[i-2]-' 0 ');} temp = 1;if (s[i]! = ' + ') {++tot;while (i > 0&&s[i]!= ') {ans[tot-1] + = (s[i]-' 0 ') *temp;temp *= 2;--i;}} while (tot>0&&ans[tot] = = 0)--tot;for (i = tot; I >= 0;--i) printf ("%d", Ans[i]);p UTS ("");////for (i = ())}r Eturn 0;} 

Blue Bridge Cup Basics practice hex to octal "string processing"