# 1501091239-Blue Bridge cup-basic practice hex to octal

Source: Internet
Author: User

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. Thinking of solving problems converting 16 into octal, you can now convert the 16 binary to binary, because a 16 binary can be represented by a binary four bit , and then the binary is converted to octal, because The three-digit binary can be represented by an octal . Example: Hex A1Binary 1010 0001 octal 010 000 001 (if the binary number is not 3 divisible, add 0 to the left)so 16-->2, you can follow, 2-->8 when you need to reverse, so enough to add 0 on the leftCode
`#include <stdio.h> #include <math.h> #include <string.h>int temp[410000];char jz16[110000];int Answer[410000];int SwitchA (char a) {if (a>= ' 0 ' &&a<= ' 9 ')    return a-' 0 '; else    return a-' a ' +10;} int main () {int n;int len;int now;int i,j,k;int sum;scanf ("%d", &n), while (n--) {scanf ("%s", jz16); Len=strlen (JZ16); for (i=0;i<len;i++) {Now=switcha (Jz16[i]), for (j= (i+1) *4-1;j>=i*4;j--) {temp[j]=now%2;now/=2;}} A hexadecimal becomes 4 binary followed by memset (answer,0,sizeof (answer)); for (i=len*4-1,sum=0;i>=0;i-=3,sum++)//Three binary into an octal inverse For     (j=i,k=0;j>i-3&&j>=0;j--, k++)        answer[sum]+=temp[j]* (int) pow (2,k), for (i=sum;; i--)    if (answer[i]!=0)        break;//Note Do not output the preceding redundant 0 for (i=i;i>=0;i--)    printf ("%d", Answer[i]);p rintf ("\ n");} return 0;}`

1501091239-Blue Bridge cup-basic practice hex to octal

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