Consistently comparing digits symmetrically to its middle digit. If first number is bigger than the last , first is wining and I have to display it else I display last and that keep until I reach middle digit(this is if I have odd number of digits), if digit don't have anything to be compared with it wins automatically.
For example number is 13257 the answer is 7 5 2.
Another one 583241 the answer is 5 8 3.
For now I am only trying to catch when number of digits is odd. And got stuck.. This is my code. The problem is that this code don't display any numbers, but it compares them in the if statement(I checked while debugging).
#include <iostream>
using namespace std;
int countDigit(int n) {
int count = 0;
while (n != 0) {
count++;
n /= 10;
}
return count;
}
int main() {
int n;
cin >> n;
int middle;
int count = countDigit(n);
if (count % 2 == 0) {
cout<<"No mid digit exsist!!";
}
else {
int lastDigit = n % 10;
middle = (count + 1) / 2;
for (int i = 0; i < middle; i++) {
for (int j = lastDigit; j<middle; j--) {
if (i > j) {
cout << i <<' ';
}
else {
cout << j;
}
}
}
}
return 0;
}
An easier approach towards this, in my opinion, would be using strings. You can check the size of the string. If there are even number of characters, you can just compare the first half characters, with the last half. If there are odd numbers, then do the same just print the middle character.
Here's what I'd do for odd number of digits:
string n;
cin>>n;
int i,j;
for(i=0,j=n.size()-1;i<n.size()/2,j>=(n.size()+1)/2;i++,j--)
{
if(n[i]>n[j]) cout<<n[i]<<" ";
else cout<<n[j]<<" ";
}
cout<<n[n.size()/2]<<endl;
We analyze the requirements and then come up with a design.
If we have a number, consisting of digits, we want to compare "left" values with "right" values. So, start somehow at the left and the right index of digits in a number.
Look at this number: 123456789
Index: 012345678
Length: 9
in C and C++ indices start with 0.
So, what will we do?
Compare index 0 with index 8
Compare index 1 with index 7
Compare index 2 with index 6
Compare index 3 with index 5
Compare index 4 with index 4
So, the index from the left is running up and the index from the right is running down.
We continue as long as the left index is less than or equal the right index. All this can be done in a for or while loop.
It does not matter, wether the number of digits is odd or even.
Of course we also do need functions that return the length of a number and a digit of the number at a given position. But I see that you know already how to write these functions. So, I will not explain it further here.
I show you 3 different examples.
Ultra simple and very verbose. Very inefficient, because we do not have arrays.
Still simple, but more compressed. Very inefficient, because we do not have arrays.
C++ solution, not allowed in your case
Verbose
#include <iostream>
// Get the length of a number
unsigned int length(unsigned long long number) {
unsigned int length = 0;
while (number != 0) {
number /= 10;
++length;
}
return length;
}
// Get a digit at a given index of a number
unsigned int digitAt(unsigned int index, unsigned long long number) {
index = length(number) - index - 1;
unsigned int result = 0;
unsigned int count = 0;
while ((number != 0) && (count <= index)) {
result = number % 10;
number /= 10;
++count;
}
return result;
}
// Test
int main() {
unsigned long long number;
if (std::cin >> number) {
unsigned int indexLeft = 0;
unsigned int indexRight = length(number) - 1;
while (indexLeft <= indexRight) {
if (digitAt(indexLeft, number) > digitAt(indexRight, number)) {
std::cout << digitAt(indexLeft, number);
}
else {
std::cout << digitAt(indexRight, number);
}
++indexLeft;
--indexRight;
}
}
}
Compressed
#include <iostream>
// Get the length of a number
size_t length(unsigned long long number) {
size_t length{};
for (; number; number /= 10) ++length;
return length;
}
// Get a digit at a given index of a number
unsigned int digitAt(size_t index, unsigned long long number) {
index = length(number) - index - 1;
unsigned int result{}, count{};
for (; number and count <= index; ++count, number /= 10)
result = number % 10;
return result;
}
// Test
int main() {
if (unsigned long long number; std::cin >> number) {
// Iterate from left and right at the same time
for (size_t indexLeft{}, indexRight{ length(number) - 1 }; indexLeft <= indexRight; ++indexLeft, --indexRight)
std::cout << ((digitAt(indexLeft,number) > digitAt(indexRight, number)) ? digitAt(indexLeft, number) : digitAt(indexRight, number));
}
}
More modern C++
#include <iostream>
#include <string>
#include <algorithm>
#include <cctype>
int main() {
if (std::string numberAsString{}; std::getline(std::cin, numberAsString) and not numberAsString.empty() and
std::all_of(numberAsString.begin(), numberAsString.end(), std::isdigit)) {
for (size_t indexLeft{}, indexRight{ numberAsString.length() - 1 }; indexLeft <= indexRight; ++indexLeft, --indexRight)
std::cout << ((numberAsString[indexLeft] > numberAsString[indexRight]) ? numberAsString[indexLeft] : numberAsString[indexRight]);
}
}
You are trying to do something confusing with nested for-cycles. This is obviously wrong, because there is nothing “quadratic” (with respect to the number of digits) in the entire task. Also, your code doesn’t seem to contain anything that would determine the highest-order digit.
I would suggest that you start with something very simple: string’ify the number and then iterate over the digits in the string. This is obviously neither elegant nor particularly fast, but it will be a working solution to start with and you can improve it later.
BTW, the sooner you get out of the bad habit of using namespace std; the better. It is an antipattern, please avoid it.
Side note: There is no need to treat odd and even numbers of digits differently. Just let the algorithm compare the middle digit (if it exists) against itself and select it; no big deal. It is a tiny efficiency drawback in exchange for a big code simplicity benefit.
#include <cstdint>
#include <iostream>
#include <string>
using std::size_t;
using std::uint64_t;
uint64_t extract_digits(uint64_t source) {
const std::string digits{std::to_string(source)};
auto i = digits.begin();
auto j = digits.rbegin();
const auto iend = i + (digits.size() + 1) / 2;
uint64_t result{0};
for (; i < iend; ++i, ++j) {
result *= 10;
result += (*i > *j ? *i : *j) - '0';
}
return result;
}
int main() {
uint64_t n;
std::cin >> n;
std::cout << extract_digits(n) << std::endl;
}
If the task disallows the use of strings and arrays, you could try using pure arithmetics by constructing a “digit-inverted” version of the number and then iterating over both numbers using division and modulo. This will (still) have obvious limitations that stem from the data type size, some numbers cannot be inverted properly etc. (Use GNU MP for unlimited integers.)
#include <cstdint>
#include <iostream>
using std::size_t;
using std::uint64_t;
uint64_t extract_digits(uint64_t source) {
uint64_t inverted{0};
size_t count{0};
for (uint64_t div = source; div; div /= 10) {
inverted *= 10;
inverted += div % 10;
++count;
}
count += 1;
count /= 2;
uint64_t result{0};
if (count) for(;;) {
const uint64_t a{source % 10}, b{inverted % 10};
result *= 10;
result += a > b ? a : b;
if (!--count) break;
source /= 10;
inverted /= 10;
}
return result;
}
int main() {
uint64_t n;
std::cin >> n;
std::cout << extract_digits(n) << std::endl;
}
Last but not least, I would strongly suggest that you ask questions after you have something buildable and runnable. Having homework solved by someone else defeats the homework’s purpose.
One of our assignments in working with C++ in 1st year programming was to write a function that can check if a number is palindromic or not (e.g 16461). I'd upload my code but I don't even know where to begin when it comes to extracting digits from an integer or checking the number of digits in an integer. Any help or hints would be appreciated!
There are many ways to solve this. I like most the solution that builds the mirror number and checks whether it is identical to the original (even though, it is arguably not the most efficient way). The code should be something like:
bool isPalindrom(int n) {
int original = n;
int mirror = 0;
while (n) {
mirror = mirror * 10 + n % 10;
n /= 10;
}
return mirror == original;
}
You can use modulo arithmetic (% operator) to extract individual digits. An alternative would be to get the string representation of your number and work with that.
Hints:
"Number of digits" is a tricky thing to define, since you can always add 0's on the left-hand side and still have the same number. Read carefully and think carefully about how you want to define this.
The digits of an integer are associated with powers of 10 (recall 123 = 1*100 + 2*10 + 3*1), so to extract digits you need to be extracting powers of 10. The key operations are remainder and truncated division. What happens if you do (123 % 10)? How about (123 / 10)? How about ((123 / 10) % 10)?
Best convert the integer into a string first. Testing a string if it is a palindrome is much easier.
#include <sstream>
#include <iostream>
#include <string>
bool isPalindrome(int value)
{
// convert integer into text
std::ostringstream oss;
oss << value;
std::string valueText = oss.str();
if (valueText.size()%2==0) {
return false;
}
for (int i = 0; i < (valueText.size()/2); ++i) {
if (valueText[i]!=valueText[valueText.size()-i-1]) {
return false;
}
}
return true;
}
int main()
{
for (int i = 0; i < 100000; ++i) {
if (isPalindrome(i)) {
std::cout << i << std::endl;
}
}
return 0;
}
First convert the integer into a std::string:
std::ostringstream oss;
oss << value;
std::string valueText = oss.str();
Now check if the string has a odd number of digits:
if (valueText.size()%2==0) {
return false;
}
If the string has a odd number of digits, test if the digits match:
for (int i = 0; i < (valueText.size()/2); ++i) {
if (valueText[i]!=valueText[valueText.size()-i-1]) {
return false;
}
}
Here's a solution that converts the integer to a C-style string and go from there.
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
bool isPalindrome(int n) {
char s[256];
sprintf(s, "%d", n);
char *p = s, *q = &s[strlen(s) - 1];
while (*p == *q && p++ < q--)
;
return (p == q) || (*p == *q);
}
int main() {
int n = 1644451;
cout << isPalindrome(n) << endl;
return 0;
}
If performance is not an issue, and if you can do c++11, an easy to read solution :
template<class T>
bool isPalindrome(T i)
{
auto s = std::to_string(i);
auto s2 = s;
std::reverse(s.begin(), s.end());
return s == s2;
}
which is of course much slower than manually going through the digits, but imho is very readable...
call it with:
std::cout << isPalindrome<int>(12321);
std::cout << isPalindrome<int>(1232);
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');}
i have to reverse the position of integer like this
input = 12345
output = 54321
i made this but it gives wrong output e.g 5432
#include <iostream>
using namespace std;
int main(){
int num,i=10;
cin>>num;
do{
cout<< (num%i)/ (i/10);
i *=10;
}while(num/i!=0);
return 0;
}
Here is a solution
int num = 12345;
int new_num = 0;
while(num > 0)
{
new_num = new_num*10 + (num % 10);
num = num/10;
}
cout << new_num << endl;
Your loop terminates too early. Change
}while(num/i!=0);
to
}while((num*10)/i!=0);
to get one more iteration, and your code will work.
If you try it once as an example, you'll see your error.
Input: 12
first loop:
out: 12%10 = 2 / 1 = 2
i = 100
test: 12/100 = 0 (as an integer)
aborts one too early.
One solution could be testing
(num % i) != num
Just as one of many solutions.
Well, remember that integer division always rounds down (or is it toward zero?) in C. So what would num / i be if num < 10 and i = 10?
replace your while statement
with
while (i<10*num)
If I were doing it, I'd (probably) start by creating the new value as an int, and then print out that value. I think this should simplify the code a bit. As pseudocode, it'd look something like:
output = 0;
while (input !=0)
output *= 10
output += input % 10
input /= 10
}
print output
The other obvious possibility would be to convert to a string first, then print the string out in reverse:
std::stringstream buffer;
buffer << input;
cout << std::string(buffer.str().rbegin(), buffer.str().rend());
int _tmain(int argc, _TCHAR* argv[])
{
int x = 1234;
int out = 0;
while (x != 0)
{
int Res = x % (10 );
x /= 10;
out *= 10;
out += Res;
}
cout << out;
}
This is a coding assignment for my college course. This assignment comes just after a discussion on Operator Overloading in C++. Although it doesn't make it clear if Overloading should be used for the assignment or not.
The following code works for a two-digit number only.
#include<iostream>
using namespace std;
int main() {
int n;
cin >> n;
cout << (n%10) << (n/10);
return 0;
}
int a,b,c,d=0;
cout<<"plz enter the number"<<endl;
cin>>a;
b=a;
do
{
c=a%10;
d=(d*10)+c;
a=a/10;
}
while(a!=0);
cout<<"The reverse of the number"<<d<<endl;
if(b==d)
{
cout<<"The entered number is palindom"<<endl;
}
else
{
cout<<"The entered number is not palindom"<<endl;
}
}
template <typename T>
T reverse(T n, size_t nBits = sizeof(T) * 8)
{
T reverse = 0;
auto mask = 1;
for (auto i = 0; i < nBits; ++i)
{
if (n & mask)
{
reverse |= (1 << (nBits - i - 1));
}
mask <<= 1;
}
return reverse;
}
This will reverse bits in any signed or unsigned integer (short, byte, int, long ...). You can provide additional parameter nBits to frame the bits while reversing.
i. e.
7 in 8 bit = 00000111 -> 11100000
7 in 4 bit = 0111 -> 1110
public class TestDS {
public static void main(String[] args) {
System.out.println(recursiveReverse(234));
System.out.println(recursiveReverse(234 ,0));
}
public static int reverse(int number){
int reversedNumber = 0;
int temp = 0;
while(number > 0){
//use modulus operator to strip off the last digit
temp = number%10;
//create the reversed number
reversedNumber = reversedNumber * 10 + temp;
number = number/10;
}
return reversedNumber;
}
private static int reversenumber =0;
public static int recursiveReverse(int number){
if(number <= 0){
return reversenumber;
}
reversenumber = reversenumber*10+(number%10);
number =number/10;
return recursiveReverse(number);
}
public static int recursiveReverse(int number , int reversenumber){
if(number <= 0){
return reversenumber;
}
reversenumber = reversenumber*10+(number%10);
number =number/10;
return recursiveReverse(number,reversenumber);
}
}
I have done this simply but this is applicable upto 5 digit numbers but hope it helps
#include<iostream>
using namespace std;
void main()
{
int a,b,c,d,e,f,g,h,i,j;
cin>>a;
b=a%10;
c=a/10;
d=c%10;
e=a/100;
f=e%10;
g=a/1000;
h=g%10;
i=a/10000;
j=i%10;
cout<<b<<d<<f<<h<<j;
}`