Blue Bridge Cup: Hexadecimal to octal (16-2-8: C version and C++ version)

Blue Bridge Cup Practice System Submission Link

Exam questions basic practice hexadecimal to octal

Problem description
　　Given n hexadecimal positive integers, output their corresponding octal numbers.

Input format
　　The first line of input is a positive integer n (1<=n<=10).
　　In the next n lines, each line contains a string consisting of 0 _{9 and uppercase letters A} F, which represents the hexadecimal positive integer to be converted, and the length of each hexadecimal number does not exceed 100000.

Output format
　　output n lines, each line input corresponding octal positive integer.

[Note] The
　　entered hexadecimal number will not have leading 0, such as 012A.
　　The output octal number cannot have a leading 0 either.

Sample input
2
39
123ABC

Sample output
71
4435274

[Prompt]
　　First convert the hexadecimal number into a certain hexadecimal number, and then convert the certain hexadecimal number into an octal number.

Problem-solving ideas:

Idea : Convert hexadecimal to binary, and then from binary to octal

Rule:
Hexadecimal to binary: 1 hexadecimal to 4 binary
binary to octal: 3 binary to 1 octal

Example:
hexadecimal number 12
hexadecimal to binary 0001 0010
two to eight 0 00 010 010
(counting three digits from right to left, not enough three digits to be filled with 0)
Result: 22 (leading 0 is not output)

Code C version:

//该题不能用strcat函数，会超时的 
#include<stdio.h>
#include<string.h>
#define N 100010
char p[16][5]={"0000","0001","0010","0011",
			   "0100","0101","0110","0111",
			   "1000","1001","1010","1011",
			   "1100","1101","1110","1111"};//代表[0,15];
char s[N];//输入的16进制 
char a[N*4];  //转化后的二进制 
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,'\0',sizeof(a));
		a[0]='$';
		a[1]='$';
    	//a[2]='\0';
		int cnt=2;
		int i,j,k,m;
		scanf("%s",s);
		int ls=strlen(s); 
		for(i=0;i<ls;i++)
		{
			if(s[i]>='0'&&s[i]<='9')
			{
				m=s[i]-48;
				for(j=0;j<4;j++)
					a[cnt++]=p[m][j];
				//a[cnt]='\0';
			}
			if(s[i]>='A'&&s[i]<='F')
			{
				m=s[i]-55;
				for(j=0;j<4;j++)
					a[cnt++]=p[m][j];
				//a[cnt]='\0';
			}
		}
		//二进制的真实长度是cnt-2 
		if((cnt-2)%3==0)
			k=2;
		if((cnt-2)%3==1)
		{
			k=0;
			a[0]='0',a[1]='0';//补两个0 
		}
		if((cnt-2)%3==2)
		{
			k=1;
			a[1]='0';
		}
		int flag=0,y;
		for(;k<cnt;k+=3)
		{
			y=4*(a[k]-'0')+2*(a[k+1]-'0')+(a[k+2]-'0');//3位2进制化为1为8进制 
			if(y!=0)
				flag=1;
			if(flag==1)
				printf("%d",y);	
		}
		printf("\n");
	}
	return 0;
}

Code C++ version:

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int i,j,n,num,len,p;
	string s1,s2;
	cin>>n;
	while(n--)
	{
		cin>>s1;
		s2="";
		for (i=0; i<s1.length(); i++)
		{
			switch(s1[i])
			{
				case '0':
					s2+="0000";
					break;
				case '1':
					s2+="0001";
					break;
				case '2':
					s2+="0010";
					break;
				case '3':
					s2+="0011";
					break;
				case '4':
					s2+="0100";
					break;
				case '5':
					s2+="0101";
					break;
				case '6':
					s2+="0110";
					break;
				case '7':
					s2+="0111";
					break;
				case '8':
					s2+="1000";
					break;
				case '9':
					s2+="1001";
					break;
				case 'A':
					s2+="1010";
					break;
				case 'B':
					s2+="1011";
					break;
				case 'C':
					s2+="1100";
					break;
				case 'D':
					s2+="1101";
					break;
				case 'E':
					s2+="1110";
					break;
				case 'F':
					s2+="1111";
					break;
			}
		}
		len=s2.length();
		if (len%3==1)//三个二进制代表一个八进制
			s2="00"+s2;
		else if (len%3==2)
			s2="0"+s2;
		int flag=0;
		for (i=0; i<=s2.length()-3; i+=3)
		{
			int num=4*(s2[i]-'0')+2*(s2[i+1]-'0')+1*(s2[i+2]-'0');
			//二进制每三位算出一个8进制数 
			if (num)
				flag=1;//防止前面三个前导零？ 
			if (flag)
				cout<<num;
		}
		cout<<endl;
	}
	return 0;
}