Basic (1) Hexadecimal to octal

Basic exercises of test questions hexadecimal to octal

resource constraints

Memory limit: 512.0MB C/C++ time limit: 1.0s Java time limit: 3.0s Python time limit: 5.0s

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

Input format
　　The first line of input is a positive integer n (1<=n<=10).
　　Next n lines, each line is a string consisting of 0~9 and uppercase letters A~F, indicating 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 corresponds to the octal positive integer input.

　　[Note ]
　　The input hexadecimal number will not have a leading 0, such as 012A.
　　The output octal number must also not have leading zeros.

Sample input
　　2
　　39
　　123ABC

Sample output
　　71
　　4435274

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

The specific idea gives a conversion function, first convert the hexadecimal number into an integer, and then convert the integer into an octal string

Just look at the change function

Version 1: Using numbers as an intermediate process, too much data long long cannot be received at all.

#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;

string change(string& s)
{
	string res;
	long long num = 0;
	int step = 0;
	for (int i = s.size() - 1; i >= 0; i--)
	{
		switch (s[i])
		{
		case '1':
			num += 1 * pow(16, step++);
			break;
		case '2':
			num += 2 * pow(16, step++);
			break;
		case '3':
			num += 3 * pow(16, step++);
			break;
		case '4':
			num += 4 * pow(16, step++);
			break;
		case '5':
			num += 5 * pow(16, step++);
			break;
		case '6':
			num += 6 * pow(16, step++);
			break;
		case '7':
			num += 7 * pow(16, step++);
			break;
		case '8':
			num += 8 * pow(16, step++);
			break;
		case '9':
			num += 9 * pow(16, step++);
			break;
		case 'A':
			num += 10 * pow(16, step++);
			break;
		case 'B':
			num += 11 * pow(16, step++);
			break;
		case 'C':
			num += 12 * pow(16, step++);
			break;
		case 'D':
			num += 13 * pow(16, step++);
			break;
		case 'E':
			num += 14 * pow(16, step++);
			break;
		case 'F':
			num += 15 * pow(16, step++);
			break;
		}
	}
	int m;
	while (num)
	{
		m = num % 8;
		num /= 8;
		res += '0'+m;
        // res += to_string(m);
	}
	reverse(res.begin(), res.end());
	return res;
}

int main()
{
	int n;
	cin >> n;
	vector<string> v(n);
	vector<string> res(n);
	for (int i = 0; i < n; i++)
	{
		cin >> v[i];
		string temp = change(v[i]);
		res[i] = temp;
	}
	for (int i = 0; i < n; i++)
	{
		cout << res[i];
		cout << endl;
	}
	return 0;
}

[Solution 2] Use 16-"2-"8 base

#include <iostream>
using namespace std;

int main()
{
	int n;
	cin>>n;
	for(int i=0; i<n; i++)
	{	
		string tow;
		string sixteen;
		string eight;
		cin>>sixteen;
		for(int j=0; j<sixteen.length(); j++)
		{
			switch(sixteen[j])
			{
				case '0':tow+="0000";break;
				case '1':tow+="0001";break;
				case '2':tow+="0010";break;
				case '3':tow+="0011";break;
				case '4':tow+="0100";break;
				case '5':tow+="0101";break;
				case '6':tow+="0110";break;
				case '7':tow+="0111";break;
				case '8':tow+="1000";break;
				case '9':tow+="1001";break;
				case 'A':
				case 'a':tow+="1010";break;
				case 'B':
				case 'b':tow+="1011";break;
				case 'C':
				case 'c':tow+="1100";break;
				case 'D':
				case 'd':tow+="1101";break;
				case 'E':
				case 'e':tow+="1110";break;
				case 'F':
				case 'f':tow+="1111";break;
			}
		}
		if(tow.length()%3==1)
			tow="00"+tow;
		if(tow.length()%3==2)
			tow="0"+tow;
		if(!(tow[0]=='0'&&tow[1]=='0'&&tow[2]=='0'))
		{
			char temp;
			temp = (tow[0]-'0')*4+(tow[1]-'0')*2+tow[2];
			eight += temp;
		}
		for(int j=3; j<tow.length(); j+=3)
		{
			eight+=(tow[j]-'0')*4+(tow[j+1]-'0')*2+tow[j+2];
		}
		cout<<eight<<endl;
	}
	return 0;
}