Blue Bridge Cup - binary conversion exercise

Based on practice Decimal to hexadecimal

Problem description

Hexadecimal number is a kind of integer expression that is often used in program design . It has 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F common 16 Symbols , They represent decimal numbers respectively 0 to 15. The hexadecimal counting method is full 16 Into the 1, So decimal numbers 16 In hexadecimal it's 10, And decimal 17 In hexadecimal it's 11, And so on , Decimal 30 In hexadecimal it's 1E.
　　 Give a nonnegative integer , Express it in hexadecimal form .

Input format

The input contains a non negative integer a, Represents the number to convert .0<=a<=2147483647

Output format

The output of this integer is 16 Hexadecimal said

The sample input

30

Sample output

1E

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
using namespace std;
map<int,string>types;
int main()
{
    
	types[0] = "0";types[1] = "1";types[2] = "2";types[3] = "3";
	types[4] = "4";types[5] = "5";types[6] = "6";types[7] = "7";
	types[8] = "8";types[9] = "9";types[10] = "A";types[11] = "B";
	types[12] = "C";types[13] = "D";types[14] = "E";types[15] = "F";
	string s16;
	long long n;//n It's a decimal number , To turn into 16 Hexadecimal number 
	cin>>n;
	// If n yes 0, Direct output 0 Exit procedure 
	if(n == 0){
    
		cout<<0<<endl;
		return 0;
	}
	// If n Less than 0, Output the minus sign first , Then deal with it as a positive number 
	if(n < 0){
    
		cout<<"-";
		n = -n;
	}
	// except 16 Remainder 
	while(n){
    
		s16.insert(0,types[n%16]);// The remainder is inserted in the first place 
		n = n/16;
	}
	cout<<s16<<endl;
	return 0;
}

Based on practice Hexadecimal to decimal

Problem description

Enter a number from the keyboard that does not exceed 8 A string of positive hexadecimal digits of , Convert it to a positive decimal number and output .
　　 notes ： In hexadecimal numbers 10~15 Use capital letters respectively A、B、C、D、E、F Express .

The sample input

FFFF

Sample output

65535

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<map>
using namespace std;
map<char,int>types;
int main()
{
    
	types['0'] = 0;types['1'] = 1;types['2'] = 2;types['3'] = 3;
	types['4'] = 4;types['5'] = 5;types['6'] = 6;types['7'] = 7;
	types['8'] = 8;types['9'] = 9;types['A'] = 10;types['B'] = 11;
	types['C'] = 12;types['D'] = 13;types['E'] = 14;types['F'] = 15;
	string s16;//s16 yes 16 String form of hexadecimal number ,
	long long n = 0;//n Is the decimal number converted , The initial value is set to 0
	cin>>s16;
	// If 0, Direct output 0 And quit the program 
	if(s16 == "0"){
    
		cout<<0<<endl;
		return 0;
	}
	// The hexadecimal input specified in the title is a positive number 
	for(int i=0;i<s16.length();i++){
    
		n *= 16;// Position right   Or rewrite it into  n= n << 4;
		n += types[s16[i]];// Add mantissa 

	}
	cout<<n<<endl;
	return 0;
}

Based on practice Hexadecimal to octal

Problem description

Given n Six 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 That's ok , Each line is made up of 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F Composed string , Represents the hexadecimal positive integer to convert , The length of each hexadecimal number does not exceed 100000.

Output format

Output n That's ok , Enter the corresponding octal positive integer for each line .

【 Be careful 】
　　 The hexadecimal number you enter will not have a leader 0, such as 012A.
　　 The output octal number also can't have leading 0.

The sample input

2
39
123ABC

Their thinking

The idea is to 16 Replace the hexadecimal number with a hexadecimal number , Then install and replace it with octal numbers .

Since the length of each hexadecimal number does not exceed 100000, Obviously not long long Deposit , It must all be in the form of strings .

that , You can replace hexadecimal numbers with binary numbers first , Then change from binary to octal .
Each hexadecimal digit corresponds to 4 Bit binary number , Every time 3 Bit binary number corresponds to one bit octal number .
among ：

1. Each hexadecimal 0-F Corresponding to binary 0000-1111
2. Each octal 0-7 Corresponding to binary 000-111
3. use map Store the mapping relationship between strings , For example, hexadecimal corresponds to binary "0"–>“0000”,“A”–>“1010”, Binary corresponds to octal "010"–>“2”
4. Converted 0-1 character string ( Binary string ) If the length is not 3 Multiple , Insert a leading... At the beginning of its string 0 Until the length is 3 Multiple
5. The last octal string installed and replaced should clear the leading 0, Until the first place is not the leader 0 Then output

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<map>
using namespace std;
// Hexadecimal to octal 
// Method ： Turn to binary first , Then convert to hexadecimal 
string sixteento2(string s16)
{
    
	string s2;
	map<char,string>types;
	types['0'] = "0000";types['1'] = "0001";types['2'] = "0010";types['3'] = "0011";
	types['4'] = "0100";types['5'] = "0101";types['6'] = "0110";types['7'] = "0111";
	types['8'] = "1000";types['9'] = "1001";types['A'] = "1010";types['B'] = "1011";
	types['C'] = "1100";types['D'] = "1101";types['E'] = "1110";types['F'] = "1111";
	for(int i=0;i<s16.length();i++){
    
		;;
		s2.append(types[s16[i]]);
	}
	return s2;
}
//2 Turn into the system 8 Base number 
string doubleto8(string s2)
{
    
	map<string,string>types;
	types["000"] = "0";types["001"] = "1";types["010"] = "2";types["011"] = "3";
	types["100"] = "4";types["101"] = "5";types["110"] = "6";types["111"] = "7";
	string s8;
	// If the binary string length is not 3 Multiple , Make up operation 
	if(s2.length()%3 == 1)
		s2.insert(0,"00");
	else if(s2.length()%3 == 2)
		s2.insert(0,"0");
	for(int i=0;i<=s2.length()-3;i=i+3){
    
		s8.append(types[string(s2,i,3)]);
	}
	// Remove the possible leading 0
	while(s8[0] == '0')
		s8 = string(s8,1,s8.length()-1);
	return s8;
}
int main()
{
    
	int m;// Every conversion m Time 
	cin>>m;
	for(int i=1;i<=m;i++){
    
		string s;
		cin>>s;
		string sout = doubleto8(sixteento2(s));
		cout<<sout<<endl;
	}
	return 0;
}