Look for least frequent digit in an integer C++ - c++

I have to say first there's one rule to my problem - Can't use anything other than functions, loops, arrays.
Here's what i have currently:
#include <iostream>
using namespace std;
int getLeastOccurredDigit(int);
int main()
{
int uIn;
cout << "Enter an Integer: ";
cin >> uIn;
cout << "\nThe least occurred digit in " << uIn << " is " << getLeastOccurredDigit(uIn) << endl;
return 0;
}
int getLeastOccurredDigit(int number)
{
int freqList[10] = { 0 };
int pNum = (number < 0) ? -number : number;
while (pNum != 0)
{
freqList[(pNum % 10)]++;
pNum /= 10;
}
}
I basically keep counts of all possible appearance of digits(0-9) in the array:
freqList[]
The next step is to of course compare all element in the array; ignoring 0s, and output the number that appeared the least.
I have no idea what i could do to loop through the array, compare the elements while ignoring 0s, and come out with the smallest value.

int getLeastOccurredDigit(int number)
{
int freqList[10] = { 0 };
int pNum = (number < 0) ? -number : number;
while (pNum != 0)
{
freqList[(pNum % 10)]++;
pNum /= 10;
}
int MAX_FREQUENCY = (sizeof(int)*CHAR_BIT+2)/3;
/*(sizeof(int)*CHAR_BIT+2)/3 is an upper bound on the number of octal
digits required for an int, and hence is an upper bound on the number
of decimal digits required for an int. by Martin Bonner [in comments]*/
int small=MAX_FREQUENCY,index=0;
for(int i=1;i<10;i++)
{
if(small>freqList[i]&&freqList[i]!=0)
{
index=i;
small=freqList[i];
}
}
return index;
}

Since you got answered already, I'll post alternative version, where you are not restricted just by arrays:
#include <iostream>
#include <string>
#include <map>
int LeastDigit(const int& num)
{
std::string snum = std::to_string(num);
std::map<char, size_t> occur;
std::pair<char, size_t> result;
for(auto& x : snum)
occur[x]++;
result = *occur.begin();
for(auto& x : occur)
if(x.second < result.second)
result = x;
return int(result.first-48);
}
int main()
{
int num = 112233455, result;
result = LeastDigit(num);
std::cout << "Your digit: " << result << std::endl;
return 0;
}
STL makes your life easier.

Related

Binary String Question to find maximum power of 2

We have given a binary String of length n,we can cyclically shift this string any number of times.Let X be the decimal representation of string s. Find the greatest power of 2 with which X can be divisible with, if it can be divisible with arbitrarily large power print "-1".For the result, you are required to print a single integer denoting the maximum power of 2 by which X can be divisible with.
ex:
Input:
0011
Output:
2
Explanation:We can cyclically shift the string 2 times to get "1100" which is divisible by 2^2 hence the answer is 2.
Here is my solution .. however it is giving me tle on most of the test cases and wrong answer on some of the test cases..
int highestpower(int n)
{
return (n & (~(n - 1)));
}
int findnum(string s)
{
int value = 0;
int p=0;
for(int i = s.length()-1;i>=0;i--)
{
value = value+pow(2,p)*(s[i]-'0');
p++;
}
return value;
}
int maximumPower(string s) {
int ans = 0;
for(int i=0;i<s.length();i++)
{
int num = findnum(s.substr(i)+s.substr(0,i));
ans = max(ans,highestpower(num));
}
return ans/2;
}
how can I solve this answer?Thanks..
I have some difficulty to understand the logic of your code. In practice, it failed on about all cases I have tested.
Moreover, it seems quite over-complicated. It is enough to count the number of consecutive zeros. We just have to pay attention that
this calculation must be performed in a cyclic way. For example, if s == 00100, the count number is 4, as after shifting, we get 10000. One simple way to handle this cyclicity is to concatenate the string s2 = s+s = 0010000100 and then to count the maximum number of consecutive zeros in the obtained string s2. In addition, we must pay attention that the input string is not composed of zeros only.
In the following code, I compared your code (maximumPower) with mine (maximumPower_new), on several different inputs.
Result:
0011 : 2 new: 2
0100010 : 4 new: 3
00100 : 8 new: 4
The code:
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
int highestpower(int n)
{
return (n & (~(n - 1)));
}
int findnum(const std::string& s)
{
int value = 0;
int p=0;
for(int i = s.length()-1;i>=0;i--)
{
value = value+pow(2,p)*(s[i]-'0');
p++;
}
return value;
}
int maximumPower(const std::string& s) {
int ans = 0;
for(int i = 0; i < s.length(); i++)
{
int num = findnum(s.substr(i)+s.substr(0,i));
ans = std::max(ans,highestpower(num));
}
return ans/2;
}
int maximumPower_new (const std::string& s) {
int n = s.length();
if (n == 0) return -1;
std::string s2 = s + s;
int count = 0;
int count_max = 0;
for (auto c: s2) {
if (c == '0') {
count ++;
} else {
count_max = std::max(count, count_max);
count = 0;
}
}
count_max = std::max(count, count_max);
if (count_max >= n) return -1;
else return count_max;
}
int main() {
for (std::string s: {"0011", "0100010", "00100"}) {
std::cout << s << " : " << maximumPower(s) << " new: " << maximumPower_new(s) << "\n";
}
}

How to convert an integer to an array of its digits in C++

Suppose I have the integer 1004.
I want to store this in the array A with the following pattern:
A[0]=1
A[1]=0
A[2]=0
A[3]=4
How can I get the value at that index ?
How can I do this in C++?
You get the last index of a number by using modulo 10 and then remove that value by dividing the number by 10.
So assume you do this:
1004 % 10 = 4
1004 / 10 = 100
Then repeat that for each digit
Using c++ static memory:
int originalNumber = 1004;
int digitArray[10] = {0};
int idx = 0;
while (originalNumber > 0)
{
int digit = n % 10;
originalNumber /= 10;
digitArray[idx] = digit;
++idx;
}
// Reverse the order of the array
std::reverse(std::begin(digitArray), std::begin(digitArray)+(idx-1));
I'm not sure if it's the most efficient way of doing this but it definitely works.
You can enter each digit in the number to the array but from the end of the array to the beginning.
For example, there is an array with the size of 4, so you get the last digit of the number like this: num % 10, and push the digit to the third index of the array.
Code example:
#define SIZE 4
int* numToArray(int num)
{
int* arr = new int[SIZE]; // assuming you already know the number of digits in the number
for(int i = SIZE-1; i >= 0; i++)
{
arr[i] = num % 10; // Enters the last digit to the array
num /= 10; // Gets rid of the last digit in the number
}
return arr;
}
Instead of ordinary integer array, I suggest you using std::vector instead.
Then, you can have something like the following:
#include <iostream>
#include <vector>
int main() {
int number = 1004;
std::vector<int> digits;
while (number != 0) {
digits.insert(digits.begin(), number % 10);
number /= 10;
}
for (auto const i : digits) {
std::cout << i << " "; // 1 0 0 4
}
// or by index
std::cout << std::endl;
std::cout << "[0]" << digits[0] << std::endl; // 1
std::cout << "[1]" << digits[1] << std::endl; // 0
std::cout << "[2]" << digits[2] << std::endl; // 0
std::cout << "[3]" << digits[3] << std::endl; // 4
return 0;
}
Demo
Adding to all the existing answers I'd like to propose a more elegant, if probably less efficient approach utilizing the many wonders of the standard library.
#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
int main()
{
auto number = 1004;
auto asString = std::to_string(number); //
std::vector<int> digits(asString.length());
std::transform(asString.begin(), asString.end(), digits.begin(), [](char c){return c -'0';});
for(auto d : digits)
{
std::cout << d << ' ';
}
}

Sum of factoriais to be equal a given number

I'm trying to solve the following problem:
What is the smallest number of factoriais summed that are needed to be equal an given number a? (1 ≤ a ≤ 10^5)
Example:
Input: 10, Output: 3. (10 = 3! + 2! + 2!)
Input: 25, Output: 2. (25 = 4! + 1!)
My code:
#include<bits/stdc++.h>
using namespace std;
int a;
int rec(int vet){
int count = 0;
a = a - vet;
if(a >= vet){
count++;
rec(vet);
}
count++;
return count;
}
int main(){
int vet[8] = {1}, count = 0;
cin >> a;
for(int i = 2; i <= 8; i++){
vet[i-1] = vet[i-2]*i;
}
for(int i = 7; i >= 0; i--){
if(a < vet[i]){
continue;
}
count += rec(vet[i]);
}
cout << count << endl;
}
My logic:
1°: a max is equal to 100000, so the maximum fatorial we have to
compare is 8!;
2°: I take a factioral that is equal or nearest small to a,
subtract the factorial from it and count++; If after the subtraction,
a still bigger then my factorial, I do the same step recursively.
This code pass on the base cases, but I got a wrong answer. I wasn't capable to find what case it didn't pass, so I'm here.
Can you find where am I wrong? Or if my solution is not good and I should try another approach.
Thanks for the help!
The problem is easily solved by a recursive approach.
Here is checked code:
#include <iostream>
using namespace std;
int factorial(int n) {
return n<=1 ? 1 : n * factorial(n-1);
}
int MinFact(int number)
{
static int num_of_facts;
int a = 1;
if (number)
{
while(factorial(a+1)<=number)a++;
cout << a << "!" << endl;
num_of_facts++;
MinFact((number-factorial(a)));
}
return num_of_facts;
}
int main()
{
int num;
cout << "Enter number" << endl;
cin >> num;
num = MinFact(num);
cout << "Number of factorials: " << num;
return 0;
}
As I mentioned in the comment, the issue is with the rec function. Due to rec being local, the count is not being incremented correctly.
A simple solution would be to replace the rec function as follows
int rec(int vec) {
int count = a / vec;
a = a % vec;
return count;
}
Edit : for a failing case try 18. The solution will be 3 but you will get 2.
I guess you can figure out how this logic works. If not you could do it with a loop.

Palindrome testing function

This function seems to be working for numbers with an odd number of digits but not for every number with an even number of digits (e.g. it returns true for 2662 but not for 906609). I've been trying to figure it out for the last 20-30 minutes but I haven't found out why.
#include <math.h>
int digits(int n)
{
return log10(n)+1;
}
bool ispalindrome(int n)
{
int c=digits(n);
for(int i=0; i<c/2; i++){
int a=pow(10,i),b=pow(10,c-i-1);
if( int(n/a) %10 != int(n/b) %10 ) return false;
}
return true;
}
#include <iostream>
#include <cstdlib>
int main(int, char**argv)
{
while (*++argv)
std::cout << *argv
<< (ispalindrome(std::atoi(*argv)) ? " is a palindrome." : " is not a palindrome.")
<< std::endl;
}
I can't reproduce, but I think it'll be easier to compare int as a string == reversedString:
bool isPalindrome(const int& n)
{
std::stringstream ssN;
ssN << n;
auto numberAsString = ssN.str();
auto reversed = numberAsString;
std::reverse(numberAsString.begin(), numberAsString.end());
return numberAsString == reversed;
}
Please, check the code: main.cpp
I can't reproduce your results - I get the expected answer for both your test cases.
BTW, you can avoid the need for <math.h> by simply reversing the digits and then comparing with the original:
bool ispalindrome(int n)
{
int a = 0;
for (int r = n; r; r /= 10)
a = a*10 + r%10;
return a == n;
}

C++ - Decimal to binary converting

I wrote a 'simple' (it took me 30 minutes) program that converts decimal number to binary. I am SURE that there's a lot simpler way so can you show me?
Here's the code:
#include <iostream>
#include <stdlib.h>
using namespace std;
int a1, a2, remainder;
int tab = 0;
int maxtab = 0;
int table[0];
int main()
{
system("clear");
cout << "Enter a decimal number: ";
cin >> a1;
a2 = a1; //we need our number for later on so we save it in another variable
while (a1!=0) //dividing by two until we hit 0
{
remainder = a1%2; //getting a remainder - decimal number(1 or 0)
a1 = a1/2; //dividing our number by two
maxtab++; //+1 to max elements of the table
}
maxtab--; //-1 to max elements of the table (when dividing finishes it adds 1 additional elemnt that we don't want and it's equal to 0)
a1 = a2; //we must do calculations one more time so we're gatting back our original number
table[0] = table[maxtab]; //we set the number of elements in our table to maxtab (we don't get 10's of 0's)
while (a1!=0) //same calculations 2nd time but adding every 1 or 0 (remainder) to separate element in table
{
remainder = a1%2; //getting a remainder
a1 = a1/2; //dividing by 2
table[tab] = remainder; //adding 0 or 1 to an element
tab++; //tab (element count) increases by 1 so next remainder is saved in another element
}
tab--; //same as with maxtab--
cout << "Your binary number: ";
while (tab>=0) //until we get to the 0 (1st) element of the table
{
cout << table[tab] << " "; //write the value of an element (0 or 1)
tab--; //decreasing by 1 so we show 0's and 1's FROM THE BACK (correct way)
}
cout << endl;
return 0;
}
By the way it's complicated but I tried my best.
edit - Here is the solution I ended up using:
std::string toBinary(int n)
{
std::string r;
while(n!=0) {r=(n%2==0 ?"0":"1")+r; n/=2;}
return r;
}
std::bitset has a .to_string() method that returns a std::string holding a text representation in binary, with leading-zero padding.
Choose the width of the bitset as needed for your data, e.g. std::bitset<32> to get 32-character strings from 32-bit integers.
#include <iostream>
#include <bitset>
int main()
{
std::string binary = std::bitset<8>(128).to_string(); //to binary
std::cout<<binary<<"\n";
unsigned long decimal = std::bitset<8>(binary).to_ulong();
std::cout<<decimal<<"\n";
return 0;
}
EDIT: Please do not edit my answer for Octal and Hexadecimal. The OP specifically asked for Decimal To Binary.
The following is a recursive function which takes a positive integer and prints its binary digits to the console.
Alex suggested, for efficiency, you may want to remove printf() and store the result in memory... depending on storage method result may be reversed.
/**
* Takes a unsigned integer, converts it into binary and prints it to the console.
* #param n the number to convert and print
*/
void convertToBinary(unsigned int n)
{
if (n / 2 != 0) {
convertToBinary(n / 2);
}
printf("%d", n % 2);
}
Credits to UoA ENGGEN 131
*Note: The benefit of using an unsigned int is that it can't be negative.
You can use std::bitset to convert a number to its binary format.
Use the following code snippet:
std::string binary = std::bitset<8>(n).to_string();
I found this on stackoverflow itself. I am attaching the link.
A pretty straight forward solution to print binary:
#include <iostream>
using namespace std;
int main()
{
int num,arr[64];
cin>>num;
int i=0,r;
while(num!=0)
{
r = num%2;
arr[i++] = r;
num /= 2;
}
for(int j=i-1;j>=0;j--){
cout<<arr[j];
}
}
Non recursive solution:
#include <iostream>
#include<string>
std::string toBinary(int n)
{
std::string r;
while(n!=0) {r=(n%2==0 ?"0":"1")+r; n/=2;}
return r;
}
int main()
{
std::string i= toBinary(10);
std::cout<<i;
}
Recursive solution:
#include <iostream>
#include<string>
std::string r="";
std::string toBinary(int n)
{
r=(n%2==0 ?"0":"1")+r;
if (n / 2 != 0) {
toBinary(n / 2);
}
return r;
}
int main()
{
std::string i=toBinary(10);
std::cout<<i;
}
An int variable is not in decimal, it's in binary. What you're looking for is a binary string representation of the number, which you can get by applying a mask that filters individual bits, and then printing them:
for( int i = sizeof(value)*CHAR_BIT-1; i>=0; --i)
cout << value & (1 << i) ? '1' : '0';
That's the solution if your question is algorithmic. If not, you should use the std::bitset class to handle this for you:
bitset< sizeof(value)*CHAR_BIT > bits( value );
cout << bits.to_string();
Here are two approaches. The one is similar to your approach
#include <iostream>
#include <string>
#include <limits>
#include <algorithm>
int main()
{
while ( true )
{
std::cout << "Enter a non-negative number (0-exit): ";
unsigned long long x = 0;
std::cin >> x;
if ( !x ) break;
const unsigned long long base = 2;
std::string s;
s.reserve( std::numeric_limits<unsigned long long>::digits );
do { s.push_back( x % base + '0' ); } while ( x /= base );
std::cout << std::string( s.rbegin(), s.rend() ) << std::endl;
}
}
and the other uses std::bitset as others suggested.
#include <iostream>
#include <string>
#include <bitset>
#include <limits>
int main()
{
while ( true )
{
std::cout << "Enter a non-negative number (0-exit): ";
unsigned long long x = 0;
std::cin >> x;
if ( !x ) break;
std::string s =
std::bitset<std::numeric_limits<unsigned long long>::digits>( x ).to_string();
std::string::size_type n = s.find( '1' );
std::cout << s.substr( n ) << std::endl;
}
}
The conversion from natural number to a binary string:
string toBinary(int n) {
if (n==0) return "0";
else if (n==1) return "1";
else if (n%2 == 0) return toBinary(n/2) + "0";
else if (n%2 != 0) return toBinary(n/2) + "1";
}
For this , In C++ you can use itoa() function .This function convert any Decimal integer to binary, decimal , hexadecimal and octal number.
#include<bits/stdc++.h>
using namespace std;
int main(){
int a;
char res[1000];
cin>>a;
itoa(a,res,10);
cout<<"Decimal- "<<res<<endl;
itoa(a,res,2);
cout<<"Binary- "<<res<<endl;
itoa(a,res,16);
cout<<"Hexadecimal- "<<res<<endl;
itoa(a,res,8);
cout<<"Octal- "<<res<<endl;return 0;
}
However, it is only supported by specific compilers.
You can see also: itoa - C++ Reference
Here is modern variant that can be used for ints of different sizes.
#include <type_traits>
#include <bitset>
template<typename T>
std::enable_if_t<std::is_integral_v<T>,std::string>
encode_binary(T i){
return std::bitset<sizeof(T) * 8>(i).to_string();
}
Your solution needs a modification. The final string should be reversed before returning:
std::reverse(r.begin(), r.end());
return r;
DECIMAL TO BINARY NO ARRAYS USED *made by Oya:
I'm still a beginner, so this code will only use loops and variables xD...
Hope you like it. This can probably be made simpler than it is...
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
int main()
{
int i;
int expoentes; //the sequence > pow(2,i) or 2^i
int decimal;
int extra; //this will be used to add some 0s between the 1s
int x = 1;
cout << "\nThis program converts natural numbers into binary code\nPlease enter a Natural number:";
cout << "\n\nWARNING: Only works until ~1.073 millions\n";
cout << " To exit, enter a negative number\n\n";
while(decimal >= 0){
cout << "\n----- // -----\n\n";
cin >> decimal;
cout << "\n";
if(decimal == 0){
cout << "0";
}
while(decimal >= 1){
i = 0;
expoentes = 1;
while(decimal >= expoentes){
i++;
expoentes = pow(2,i);
}
x = 1;
cout << "1";
decimal -= pow(2,i-x);
extra = pow(2,i-1-x);
while(decimal < extra){
cout << "0";
x++;
extra = pow(2,i-1-x);
}
}
}
return 0;
}
here a simple converter by using std::string as container. it allows a negative value.
#include <iostream>
#include <string>
#include <limits>
int main()
{
int x = -14;
int n = std::numeric_limits<int>::digits - 1;
std::string s;
s.reserve(n + 1);
do
s.push_back(((x >> n) & 1) + '0');
while(--n > -1);
std::cout << s << '\n';
}
This is a more simple program than ever
//Program to convert Decimal into Binary
#include<iostream>
using namespace std;
int main()
{
long int dec;
int rem,i,j,bin[100],count=-1;
again:
cout<<"ENTER THE DECIMAL NUMBER:- ";
cin>>dec;//input of Decimal
if(dec<0)
{
cout<<"PLEASE ENTER A POSITIVE DECIMAL";
goto again;
}
else
{
cout<<"\nIT's BINARY FORM IS:- ";
for(i=0;dec!=0;i++)//making array of binary, but reversed
{
rem=dec%2;
bin[i]=rem;
dec=dec/2;
count++;
}
for(j=count;j>=0;j--)//reversed binary is printed in correct order
{
cout<<bin[j];
}
}
return 0;
}
There is in fact a very simple way to do so. What we do is using a recursive function which is given the number (int) in the parameter. It is pretty easy to understand. You can add other conditions/variations too. Here is the code:
int binary(int num)
{
int rem;
if (num <= 1)
{
cout << num;
return num;
}
rem = num % 2;
binary(num / 2);
cout << rem;
return rem;
}
// function to convert decimal to binary
void decToBinary(int n)
{
// array to store binary number
int binaryNum[1000];
// counter for binary array
int i = 0;
while (n > 0) {
// storing remainder in binary array
binaryNum[i] = n % 2;
n = n / 2;
i++;
}
// printing binary array in reverse order
for (int j = i - 1; j >= 0; j--)
cout << binaryNum[j];
}
refer :-
https://www.geeksforgeeks.org/program-decimal-binary-conversion/
or
using function :-
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;cin>>n;
cout<<bitset<8>(n).to_string()<<endl;
}
or
using left shift
#include<bits/stdc++.h>
using namespace std;
int main()
{
// here n is the number of bit representation we want
int n;cin>>n;
// num is a number whose binary representation we want
int num;
cin>>num;
for(int i=n-1;i>=0;i--)
{
if( num & ( 1 << i ) ) cout<<1;
else cout<<0;
}
}
#include <iostream>
#include <bitset>
#define bits(x) (std::string( \
std::bitset<8>(x).to_string<char,std::string::traits_type, std::string::allocator_type>() ).c_str() )
int main() {
std::cout << bits( -86 >> 1 ) << ": " << (-86 >> 1) << std::endl;
return 0;
}
Okay.. I might be a bit new to C++, but I feel the above examples don't quite get the job done right.
Here's my take on this situation.
char* DecimalToBinary(unsigned __int64 value, int bit_precision)
{
int length = (bit_precision + 7) >> 3 << 3;
static char* binary = new char[1 + length];
int begin = length - bit_precision;
unsigned __int64 bit_value = 1;
for (int n = length; --n >= begin; )
{
binary[n] = 48 | ((value & bit_value) == bit_value);
bit_value <<= 1;
}
for (int n = begin; --n >= 0; )
binary[n] = 48;
binary[length] = 0;
return binary;
}
#value = The Value we are checking.
#bit_precision = The highest left most bit to check for.
#Length = The Maximum Byte Block Size. E.g. 7 = 1 Byte and 9 = 2 Byte, but we represent this in form of bits so 1 Byte = 8 Bits.
#binary = just some dumb name I gave to call the array of chars we are setting. We set this to static so it won't be recreated with every call. For simply getting a result and display it then this works good, but if let's say you wanted to display multiple results on a UI they would all show up as the last result. This can be fixed by removing static, but make sure you delete [] the results when you are done with it.
#begin = This is the lowest index that we are checking. Everything beyond this point is ignored. Or as shown in 2nd loop set to 0.
#first loop - Here we set the value to 48 and basically add a 0 or 1 to 48 based on the bool value of (value & bit_value) == bit_value. If this is true the char is set to 49. If this is false the char is set to 48. Then we shift the bit_value or basically multiply it by 2.
#second loop - Here we set all the indexes we ignored to 48 or '0'.
SOME EXAMPLE OUTPUTS!!!
int main()
{
int val = -1;
std::cout << DecimalToBinary(val, 1) << '\n';
std::cout << DecimalToBinary(val, 3) << '\n';
std::cout << DecimalToBinary(val, 7) << '\n';
std::cout << DecimalToBinary(val, 33) << '\n';
std::cout << DecimalToBinary(val, 64) << '\n';
std::cout << "\nPress any key to continue. . .";
std::cin.ignore();
return 0;
}
00000001 //Value = 2^1 - 1
00000111 //Value = 2^3 - 1.
01111111 //Value = 2^7 - 1.
0000000111111111111111111111111111111111 //Value = 2^33 - 1.
1111111111111111111111111111111111111111111111111111111111111111 //Value = 2^64 - 1.
SPEED TESTS
Original Question's Answer: "Method: toBinary(int);"
Executions: 10,000 , Total Time (Milli): 4701.15 , Average Time (Nanoseconds): 470114
My Version: "Method: DecimalToBinary(int, int);"
//Using 64 Bit Precision.
Executions: 10,000,000 , Total Time (Milli): 3386 , Average Time (Nanoseconds): 338
//Using 1 Bit Precision.
Executions: 10,000,000, Total Time (Milli): 634, Average Time (Nanoseconds): 63
Below is simple C code that converts binary to decimal and back again. I wrote it long ago for a project in which the target was an embedded processor and the development tools had a stdlib that was way too big for the firmware ROM.
This is generic C code that does not use any library, nor does it use division or the remainder (%) operator (which is slow on some embedded processors), nor does it use any floating point, nor does it use any table lookup nor emulate any BCD arithmetic. What it does make use of is the type long long, more specifically unsigned long long (or uint64_t), so if your embedded processor (and the C compiler that goes with it) cannot do 64-bit integer arithmetic, this code is not for your application. Otherwise, I think this is production quality C code (maybe after changing long to int32_t and unsigned long long to uint64_t). I have run this overnight to test it for every 2³² signed integer values and there is no error in conversion in either direction.
We had a C compiler/linker that could generate executables and we needed to do what we could do without any stdlib (which was a pig). So no printf() nor scanf(). Not even an sprintf() nor sscanf(). But we still had a user interface and had to convert base-10 numbers into binary and back. (We also made up our own malloc()-like utility also and our own transcendental math functions too.)
So this was how I did it (the main program and calls to stdlib were there for testing this thing on my mac, not for the embedded code). Also, because some older dev systems don't recognize "int64_t" and "uint64_t" and similar types, the types long long and unsigned long long are used and assumed to be the same. And long is assumed to be 32 bits. I guess I could have typedefed it.
// returns an error code, 0 if no error,
// -1 if too big, -2 for other formatting errors
int decimal_to_binary(char *dec, long *bin)
{
int i = 0;
int past_leading_space = 0;
while (i <= 64 && !past_leading_space) // first get past leading spaces
{
if (dec[i] == ' ')
{
i++;
}
else
{
past_leading_space = 1;
}
}
if (!past_leading_space)
{
return -2; // 64 leading spaces does not a number make
}
// at this point the only legitimate remaining
// chars are decimal digits or a leading plus or minus sign
int negative = 0;
if (dec[i] == '-')
{
negative = 1;
i++;
}
else if (dec[i] == '+')
{
i++; // do nothing but go on to next char
}
// now the only legitimate chars are decimal digits
if (dec[i] == '\0')
{
return -2; // there needs to be at least one good
} // digit before terminating string
unsigned long abs_bin = 0;
while (i <= 64 && dec[i] != '\0')
{
if ( dec[i] >= '0' && dec[i] <= '9' )
{
if (abs_bin > 214748364)
{
return -1; // this is going to be too big
}
abs_bin *= 10; // previous value gets bumped to the left one digit...
abs_bin += (unsigned long)(dec[i] - '0'); // ... and a new digit appended to the right
i++;
}
else
{
return -2; // not a legit digit in text string
}
}
if (dec[i] != '\0')
{
return -2; // not terminated string in 64 chars
}
if (negative)
{
if (abs_bin > 2147483648)
{
return -1; // too big
}
*bin = -(long)abs_bin;
}
else
{
if (abs_bin > 2147483647)
{
return -1; // too big
}
*bin = (long)abs_bin;
}
return 0;
}
void binary_to_decimal(char *dec, long bin)
{
unsigned long long acc; // 64-bit unsigned integer
if (bin < 0)
{
*(dec++) = '-'; // leading minus sign
bin = -bin; // make bin value positive
}
acc = 989312855LL*(unsigned long)bin; // very nearly 0.2303423488 * 2^32
acc += 0x00000000FFFFFFFFLL; // we need to round up
acc >>= 32;
acc += 57646075LL*(unsigned long)bin;
// (2^59)/(10^10) = 57646075.2303423488 = 57646075 + (989312854.979825)/(2^32)
int past_leading_zeros = 0;
for (int i=9; i>=0; i--) // maximum number of digits is 10
{
acc <<= 1;
acc += (acc<<2); // an efficient way to multiply a long long by 10
// acc *= 10;
unsigned int digit = (unsigned int)(acc >> 59); // the digit we want is in bits 59 - 62
if (digit > 0)
{
past_leading_zeros = 1;
}
if (past_leading_zeros)
{
*(dec++) = '0' + digit;
}
acc &= 0x07FFFFFFFFFFFFFFLL; // mask off this digit and go on to the next digit
}
if (!past_leading_zeros) // if all digits are zero ...
{
*(dec++) = '0'; // ... put in at least one zero digit
}
*dec = '\0'; // terminate string
}
#if 1
#include <stdlib.h>
#include <stdio.h>
int main (int argc, const char* argv[])
{
char dec[64];
long bin, result1, result2;
unsigned long num_errors;
long long long_long_bin;
num_errors = 0;
for (long_long_bin=-2147483648LL; long_long_bin<=2147483647LL; long_long_bin++)
{
bin = (long)long_long_bin;
if ((bin&0x00FFFFFFL) == 0)
{
printf("bin = %ld \n", bin); // this is to tell us that things are moving along
}
binary_to_decimal(dec, bin);
decimal_to_binary(dec, &result1);
sscanf(dec, "%ld", &result2); // decimal_to_binary() should do the same as this sscanf()
if (bin != result1 || bin != result2)
{
num_errors++;
printf("bin = %ld, result1 = %ld, result2 = %ld, num_errors = %ld, dec = %s \n",
bin, result1, result2, num_errors, dec);
}
}
printf("num_errors = %ld \n", num_errors);
return 0;
}
#else
#include <stdlib.h>
#include <stdio.h>
int main (int argc, const char* argv[])
{
char dec[64];
long bin;
printf("bin = ");
scanf("%ld", &bin);
while (bin != 0)
{
binary_to_decimal(dec, bin);
printf("dec = %s \n", dec);
printf("bin = ");
scanf("%ld", &bin);
}
return 0;
}
#endif
My way of converting decimal to binary in C++. But since we are using mod, this function will work in case of hexadecimal or octal also. You can also specify bits. This function keeps calculating the lowest significant bit and place it on the end of the string. If you are not so similar to this method than you can vist: https://www.wikihow.com/Convert-from-Decimal-to-Binary
#include <bits/stdc++.h>
using namespace std;
string itob(int bits, int n) {
int count;
char str[bits + 1]; // +1 to append NULL character.
str[bits] = '\0'; // The NULL character in a character array flags the end
// of the string, not appending it may cause problems.
count = bits - 1; // If the length of a string is n, than the index of the
// last character of the string will be n - 1. Cause the
// index is 0 based not 1 based. Try yourself.
do {
if (n % 2)
str[count] = '1';
else
str[count] = '0';
n /= 2;
count--;
} while (n > 0);
while (count > -1) {
str[count] = '0';
count--;
}
return str;
}
int main() {
cout << itob(1, 0) << endl; // 0 in 1 bit binary.
cout << itob(2, 1) << endl; // 1 in 2 bit binary.
cout << itob(3, 2) << endl; // 2 in 3 bit binary.
cout << itob(4, 4) << endl; // 4 in 4 bit binary.
cout << itob(5, 15) << endl; // 15 in 5 bit binary.
cout << itob(6, 30) << endl; // 30 in 6 bit binary.
cout << itob(7, 61) << endl; // 61 in 7 bit binary.
cout << itob(8, 127) << endl; // 127 in 8 bit binary.
return 0;
}
The Output:
0
01
010
0100
01111
011110
0111101
01111111
Since you asked for a simple way, I am sharing this answer, after 8 years
Here is the expression!
Is it not interesting when there is no if condition, and we can get 0 or 1 with just a simple expression?
Well yes, NO if, NO long division
Here is what each variable means
Note: variable is the orange highlighted ones
Number: 0-infinity (a value to be converted to binary)
binary holder: 1 / 2 / 4 / 8 / 16 / 32 / ... (Place of binary needed, just like tens, hundreds)
Result: 0 or 1
If you want to make binary holder from 1 / 2 / 4 / 8 / 16 /... to 1 / 2 / 3 / 4 / 5/...
then use this expression
The procedure is simple for the second expression
First, the number variable is always, your number needed, and its stable.
Second the binary holder variable needs to be changed ,in a for loop, by +1 for the second image, x2 for the first image
I don't know c++ a lot ,here is a js code,for your understanding
function FindBinary(Number) {
var x,i,BinaryValue = "",binaryHolder = 1;
for (i = 1; Math.pow(2, i) <= Number; i++) {}//for trimming, you can even remove this and set i to 7,see the result
for (x = 1; x <= i; x++) {
var Algorithm = ((Number - (Number % binaryHolder)) / binaryHolder) % 2;//Main algorithm
BinaryValue = Algorithm + BinaryValue;
binaryHolder += binaryHolder;
}
return BinaryValue;
}
console.log(FindBinary(17));//your number
more ever, I think language doesn't matters a lot for algorithm questions
You want to do something like:
cout << "Enter a decimal number: ";
cin >> a1;
cout << setbase(2);
cout << a1
#include "stdafx.h"
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int main() {
// Initialize Variables
double x;
int xOct;
int xHex;
//Initialize a variable that stores the order if the numbers in binary/sexagesimal base
vector<int> rem;
//Get Demical value
cout << "Number (demical base): ";
cin >> x;
//Set the variables
xOct = x;
xHex = x;
//Get the binary value
for (int i = 0; x >= 1; i++) {
rem.push_back(abs(remainder(x, 2)));
x = floor(x / 2);
}
//Print binary value
cout << "Binary: ";
int n = rem.size();
while (n > 0) {
n--;
cout << rem[n];
} cout << endl;
//Print octal base
cout << oct << "Octal: " << xOct << endl;
//Print hexademical base
cout << hex << "Hexademical: " << xHex << endl;
system("pause");
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int a,b;
cin>>a;
for(int i=31;i>=0;i--)
{
b=(a>>i)&1;
cout<<b;
}
}
HOPE YOU LIKE THIS SIMPLE CODE OF CONVERSION FROM DECIMAL TO BINARY
#include<iostream>
using namespace std;
int main()
{
int input,rem,res,count=0,i=0;
cout<<"Input number: ";
cin>>input;`enter code here`
int num=input;
while(input > 0)
{
input=input/2;
count++;
}
int arr[count];
while(num > 0)
{
arr[i]=num%2;
num=num/2;
i++;
}
for(int i=count-1 ; i>=0 ; i--)
{
cout<<" " << arr[i]<<" ";
}
return 0;
}
#include <iostream>
// x is our number to test
// pow is a power of 2 (e.g. 128, 64, 32, etc...)
int printandDecrementBit(int x, int pow)
{
// Test whether our x is greater than some power of 2 and print the bit
if (x >= pow)
{
std::cout << "1";
// If x is greater than our power of 2, subtract the power of 2
return x - pow;
}
else
{
std::cout << "0";
return x;
}
}
int main()
{
std::cout << "Enter an integer between 0 and 255: ";
int x;
std::cin >> x;
x = printandDecrementBit(x, 128);
x = printandDecrementBit(x, 64);
x = printandDecrementBit(x, 32);
x = printandDecrementBit(x, 16);
std::cout << " ";
x = printandDecrementBit(x, 8);
x = printandDecrementBit(x, 4);
x = printandDecrementBit(x, 2);
x = printandDecrementBit(x, 1);
return 0;
}
this is a simple way to get the binary form of an int. credit to learncpp.com. im sure this could be used in different ways to get to the same point.
In this approach, the decimal will be converted to the respective binary number in the string formate. The string return type is chosen since it can handle more range of input values.
class Solution {
public:
string ConvertToBinary(int num)
{
vector<int> bin;
string op;
for (int i = 0; num > 0; i++)
{
bin.push_back(num % 2);
num /= 2;
}
reverse(bin.begin(), bin.end());
for (size_t i = 0; i < bin.size(); ++i)
{
op += to_string(bin[i]);
}
return op;
}
};
using bitmask and bitwise and .
string int2bin(int n){
string x;
for(int i=0;i<32;i++){
if(n&1) {x+='1';}
else {x+='0';}
n>>=1;
}
reverse(x.begin(),x.end());
return x;
}
You Could use std::bitset:
#include <bits/stdc++.h>
int main()
{
std::string binary = std::bitset<(int)ceil(log2(10))>(10).to_string(); // decimal number is 10
std::cout << binary << std::endl; // 1010
return 0;
}
SOLUTION 1
Shortest function. Recursive. No headers required.
size_t bin(int i) {return i<2?i:10*bin(i/2)+i%2;}
The simplicity of this function comes at the cost of some limitations. It returns correct values only for arguments between 0 and 1048575 (2 to the power of how many digits the largest unsigned int has, -1). I used the following program to test it:
#include <iostream> // std::cout, std::cin
#include <climits> // ULLONG_MAX
#include <math.h> // pow()
int main()
{
size_t bin(int);
int digits(size_t);
int i = digits(ULLONG_MAX); // maximum digits of the return value of bin()
int iMax = pow(2.0,i)-1; // maximum value of a valid argument of bin()
while(true) {
std::cout << "Decimal: ";
std::cin >> i;
if (i<0 or i>iMax) {
std::cout << "\nB Integer out of range, 12:1";
return 0;
}
std::cout << "Binary: " << bin(i) << "\n\n";
}
return 0;
}
size_t bin(int i) {return i<2?i:10*bin(i/2)+i%2;}
int digits(size_t i) {return i<10?1:digits(i/10)+1;}
SOLUTION 2
Short. Recursive. Some headers required.
std::string bin(size_t i){return !i?"0":i==1?"1":bin(i/2)+(i%2?'1':'0');}
This function can return the binary representation of the largest integers as a string. I used the following program to test it:
#include <string> // std::string
#include <iostream> // std::cout, std::cin
int main()
{
std::string s, bin(size_t);
size_t i, x;
std::cout << "Enter exit code: "; // Used to exit the program.
std::cin >> x;
while(i!=x) {
std::cout << "\nDecimal: ";
std::cin >> i;
std::cout << "Binary: " << bin(i) << "\n";
}
return 0;
}
std::string bin(size_t i){return !i?"0":i==1?"1":bin(i/2)+(i%2?'1':'0');}