I have an assignment to make a program that should convert a number from it's integer value to a binary value. For some reason my array is always filled with zeroes and won't add "1"'s from my if statements. I know there are probably solutions to this assignment on internet but I would like to understand what is problem with my code. Any help is appreciated.
Here is what I tried:
#include <iostream>
/*Write a code that will enable input of one real number in order to write out it's binary equivalent.*/
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
while (number > 0) {
int i = 0;
if ((number / 10) % 2 == 0) {
binaryNumber[i] = 0;
}
if ((number / 10) % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 10;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
You need to remove number/10 in both the if statements. Instead, just use number. you need the last digit every time to get the ith bit.
Moreover, you need to just half the number in every iteration rather than doing it /10.
// Updated Code
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
int i = 0;
while (number > 0) {
if (number % 2 == 0) {
binaryNumber[i] = 0;
}
if (number % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 2;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
The first thing is the variable 'i' in the while loop. Consider it more precisely: every time you iterate over it, 'i' is recreated again and assigned the value of zero. It's the basics of the language itself.
The most relevant mistake is logic of your program. Each iteration we must take the remainder of division by 2, and then divide our number by 2.
The correct code is:
#include <iostream>
int main()
{
int x = 8;
bool repr[32]{};
int p = 0;
while(x)
{
repr[p] = x % 2;
++p;
x /= 2;
}
for(int i = 31; i >= 0; --i)
std::cout << repr[i];
return 0;
}
... is always filled with zeroes ... I would like to understand what is problem with my code
int i = 0; must be before the while, having it inside you only set the index 0 of the array in your loop because i always values 0.
But there are several other problems in your code :
using int binaryNumber[32] you suppose your int are on 32bits. Do not use 32 but sizeof(int)*CHAR_BIT, and the same for your last loop in case you want to also write 0 on the left of the first 1
you look at the value of (number / 10) % 2, you must look at the value of number % 2
it is useless to do the test then its reverse, just use else, or better remove the two ifs and just do binaryNumber[i] = number & 1;
number = number / 10; is the right way when you want to produce the value in decimal, in binary you have to divide by 2
in for (int i = 31; i >= 0; i--) { except for numbers needing 32 bits you will write useless 0 on the left, why not using the value of i from the while ?
There are some logical errors in your code.
You have taken (number/10) % 2, instead, you have to take (number %2 ) as you want the remainder.
Instead of taking i = 31, you should use this logic so you can print the following binary in reverse order:
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
Here is the code to convert an integer to its binary equivalent:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
// function to convert integer to binary
void DecBinary(int n)
{
// Array to store binary number
int BinaryNumb[32];
int i = 0;
while (n > 0)
{
// Storing remainder in array
BinaryNumb[i] = n % 2;
n = n / 2;
i++;
}
// Printing array in reverse order
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
}
// Main Program
int main()
{
int testcase;
//Loop is optional
for(int i = 0; i < testcase; i++)
{
cin >> n;
DecToBinary(n);
}
return 0;
}
I have to create a program which calculates the factorial of any number, the problem is if I input any number above 20 it just returns that number. What in my else if statement could be causing this and is there a better way to solve this? ( this function is called in main and works if num <= 20)
void factorial() {
//User input for number
long long num;
std::cout << "Input any positive integer to find its factorial: ";
std::cin >> num;
unsigned long long numFact = 1;
if (num <= 20) {
while (num > 0) {
numFact = numFact * num;
num = num - 1;
}
std::cout << numFact;
}
else if (num > 20) {
std::vector<int> multFactorial;
//stores num as seperate elements in vector multFactorial
while (num > 0) {
int remain = num % 10;
num = num / 10;
multFactorial.insert(multFactorial.begin(), remain);
}
std::vector<int> answer;
std::vector<int> answerFinal;
//Manually multiplies elements in multFactorial
//Then adds new vectors created by multiplying to get final answer
//Repeats until factorial is solved
//Ex: 21 * 20; 0 * 1 and 0 * 2 stored as {0 , 0}
//2*1 and 2*2 stored as {4, 2, 0}
//Vectors will be addes to get {4, 2, 0} and then that will be multiplied
by 19 until num = 1
while (num > 1) {
for (int i = multFactorial.size() - 1; i >= 0; i--) {
int remain1 = ((num - 1) % 10) * multFactorial[i];
answer.insert(answer.begin(), remain1);
int remain2 = (((num - 1) / 10) * multFactorial[i]);
answerFinal.insert(answerFinal.begin(), remain2);
}
answerFinal.insert(answerFinal.begin(), 0);
//Adds vectors to get final value seperate as digits
for (int i = multFactorial.size() - 1; i >= 0; i--) {
multFactorial[i] = answer[i] + answerFinal[i];
}
num = num - 1;
}
//Prints what should be the factorial of the number input
for (size_t i = 0; i < multFactorial.size(); i++) {
std::cout << multFactorial[i];
}
}
}
Factorials of large numbers results in huge numbers. This can be accommodated in languages like C, C++ etc by putting the results into arbitrary length strings.
Here is an algorithm for that - similar to yours.
https://www.geeksforgeeks.org/factorial-large-number/
Best advice is to check your code against this.
Use a debugger if you have one and step through the code line by line.
If not print out intermediate results and compare with expected.
EDIT: As per review comment, the code at above ref is similar to below- just in case link is broken in future.
// C++ program to compute factorial of big numbers
#include<iostream>
using namespace std;
// Maximum number of digits in output
#define MAX 100 // change to whatever value you need
int multiply(int x, int res[], int res_size);
// Calculate factorial of large number
void factorial(int n)
{
int res[MAX];
// Initialize result
res[0] = 1;
int res_size = 1;
// Apply factorial formula
for (int x=2; x<=n; x++)
res_size = multiply(x, res, res_size);
// print out the result
cout << "Factorial is \n";
for (int i=res_size-1; i>=0; i--)
cout << res[i];
}
// Multiplies x with the number represented by res[].
// res_size is size of res[] or number of digits in the
// number represented by res[].
int multiply(int x, int res[], int res_size)
{
int carry = 0; // Initialize carry
// One by one multiply n with individual digits of res[]
for (int i=0; i<res_size; i++)
{
int prod = res[i] * x + carry;
// Store last digit of 'prod' in res[]
res[i] = prod % 10;
// Put rest in carry
carry = prod/10;
}
// Put carry in res and increase result size
while (carry)
{
res[res_size] = carry%10;
carry = carry/10;
res_size++;
}
return res_size;
}
// Main program
int main()
{
//put code here to read a number
factorial(50); // take 50 for example
return 0;
}
So my problem is bigger but I just do not know what to do with my code. I can do what I want if I use an array works just fine but we are not using arrays yet so I have no idea how to do it. So I have to take user input as a string validate that the string is 16 characters long, all of them are digits, and most importantly I have to multiply every other or even character by 2. Then if it is a double digit add the two digit (ex. 10 1+0). Oh by the way I do not know why but every time I do i%2 == 0 I get the odd numbers. Is it because i is unsigned?
for(unsigned i = 1; i < card.length(); i++){
if (i % 2 == 1){
}
else {
}
}
return sum;
}
You could use an array of strings where each string contains a number.
Go through them checking for 2 conditions:
Double the number if it is even (i.e., i % 2 == 0)
Add the digits if the number has 2 digits (i.e., string's length is 2)
Code:
#include <iostream>
#include <iterator> using namespace std;
int TOTAL_CARDS = 16;
void printCards(string msg, string *array) {
cout<<msg<<endl;
for(int i = 0; i < TOTAL_CARDS; i++) {
cout<<"array["<<i<<"]="<<array[i]<<endl;
}
cout<<"\n"<<endl; }
int main() {
string cards[TOTAL_CARDS];
// hardcoded numbers 0 up to TOTAL_CARDS for demo purposes
for(int i = 0; i < TOTAL_CARDS; i++) {
cards[i] = to_string(i);
}
printCards("Before:", cards);
for (unsigned i = 1; i < TOTAL_CARDS; i++){
// double if even
if (i % 2 == 0){
cards[i] = to_string(stoi(cards[i]) * 2);
}
// add digits if double digit number
if (cards[i].length() == 2) {
// get each digit
string currentNum = cards[i];
int firstDigit = currentNum[0] - '0'; // char - '0' gives int
int secondDigit = currentNum[1] - '0';
// do sum and put in array
int sum = firstDigit + secondDigit;
cards[i] = to_string(sum);
}
}
printCards("After:", cards); }
Output:
Before:
array[0]=0
array[1]=1
array[2]=2
array[3]=3
array[4]=4
array[5]=5
array[6]=6
array[7]=7
array[8]=8
array[9]=9
array[10]=10
array[11]=11
array[12]=12
array[13]=13
array[14]=14
array[15]=15
After:
array[0]=0
array[1]=1
array[2]=4
array[3]=3
array[4]=8
array[5]=5
array[6]=3
array[7]=7
array[8]=7
array[9]=9
array[10]=2
array[11]=2
array[12]=6
array[13]=4
array[14]=10
array[15]=6
If you wanted to get user input for the numbers:
// get user to enter numbers
cout<<"Please enter "<<TOTAL_CARDS<<" numbers: "<<endl;
for(int i = 0; i < TOTAL_CARDS; i++) {
cin>>cards[i];
}
I found the answer to it. I first needed to create char variable named num. Convert the char to an int using chnum and then multiply.
for(unsigned i = 0; i < card.length(); i++){
if (i % 2 == 1){
num = card.at(i);
chnum = (num -'0');
add = chnum * 2;
if(add >= 10){
char ho = (add + '0');
string str(1,ho);
for (unsigned j = 0; j < str.length();j++){
char digi = str.at(j);
int chub = (digi - '0');
cout << digi;
//add = (chub) + (chub);
}
}
sum += add;
}
I'm trying to print the numbers from 1 to N in lexicographic order, but I get a failed output. for the following input 100, I get the 100, but its shifted and it doesn't match with the expected output, there is a bug in my code but I can not retrace it.
class Solution {
public:
vector<int> lexicalOrder(int n) {
vector<int> result;
for(int i = 1; i <= 9; i ++){
int j = 1;
while( j <= n){
for(int m = 0; m < j ; ++ m){
if(m + j * i <= n){
result.push_back(m+j*i);
}
}
j *= 10;
}
}
return result;
}
};
Input:
100
Output:
[1,10,11,12,13,14,15,16,17,18,19,100,2,20,21,22,23,24,25,26,27,28,29,3,30,31,32,33,34,35,36,37,38,39,4,40,41,42,43,44,45,46,47,48,49,5,50,51,52,53,54,55,56,57,58,59,6,60,61,62,63,64,65,66,67,68,69,7,70,71,72,73,74,75,76,77,78,79,8,80,81,82,83,84,85,86,87,88,89,9,90,91,92,93,94,95,96,97,98,99]
Expected:
[1,10,100,11,12,13,14,15,16,17,18,19,2,20,21,22,23,24,25,26,27,28,29,3,30,31,32,33,34,35,36,37,38,39,4,40,41,42,43,44,45,46,47
Think about when i=1,j=10 what will happen in
for(int m = 0; m < j ; ++ m){
if(m + j * i <= n){
result.push_back(m+j*i);
}
}
Yes,result will push_back 10(0+10*1),11(1+10*1),12(2+10*1)..
Here is a solution:
#include <iostream>
#include <vector>
#include <string>
std::vector<int> fun(int n)
{
std::vector<std::string> result;
for (int i = 1; i <= n; ++i) {
result.push_back(std::to_string(i));
}
std::sort(result.begin(),result.end());
std::vector<int> ret;
for (auto i : result) {
ret.push_back(std::stoi(i));
}
return ret;
}
int main(int argc, char *argv[])
{
std::vector<int> result = fun(100);
for (auto i : result) {
std::cout << i << ",";
}
std::cout << std::endl;
return 0;
}
You are looping through all 2 digit numbers starting with 1 before outputting the first 3 digit number, so your approach won't work.
One way to do this is to output the digits in base 11, padded out with leading spaces to the maximum number of digits, in this case 3. Output 0 as a space, 1 as 0, 2 as 1 etc. Reject any numbers that have any non-trailing spaces in this representation, or are greater than n when interpreted as a base 10 number. It should be possible to jump past multiple rejects at once, but that's an unnecessary optimization. Keep a count of the numbers you have output and stop when it reaches n. This will give you a lexicographical ordering in base 10.
Example implementation that uses O(1) space, where you don't have to generate and sort all the numbers up front before you can output the first one:
void oneToNLexicographical(int n)
{
if(n < 1) return;
// count max digits
int digits = 1, m = n, max_digit11 = 1, max_digit10 = 1;
while(m >= 10)
{
m /= 10; digits++; max_digit11 *= 11; max_digit10 *= 10;
}
int count = 0;
bool found_n = false;
// count up starting from max_digit * 2 (first valid value with no leading spaces)
for(int i = max_digit11 * 2; ; i++)
{
int val = 0, trailing_spaces = 0;
int place_val11 = max_digit11, place_val10 = max_digit10;
// bool valid_spaces = true;
for(int d = 0; d < digits; d++)
{
int base11digit = (i / place_val11) % 11;
if(base11digit == 0)
{
trailing_spaces++;
val /= 10;
}
else
{
// if we got a non-space after a space, it's invalid
// if(trailing_spaces > 0)
// {
// valid_spaces = false;
// break; // trailing spaces only
// }
val += (base11digit - 1) * place_val10;
}
place_val11 /= 11;
place_val10 /= 10;
}
// if(valid_spaces && (val <= n))
{
cout << val << ", ";
count++;
}
if(val == n)
{
found_n = true;
i += 10 - (i % 11); // skip to next number with one trailing space
}
// skip past invalid numbers:
// if there are multiple trailing spaces then the next run of numbers will have spaces in the middle - invalid
if(trailing_spaces > 1)
i += (int)pow(11, trailing_spaces - 1) - 1;
// if we have already output the max number, then all remaining numbers
// with the max number of digits will be greater than n
else if(found_n && (trailing_spaces == 1))
i += 10;
if(count == n)
break;
}
}
This skips past all invalid numbers, so it's not necessary to test valid_spaces before outputting each.
The inner loop can be removed by doing the base11 -> base 10 conversion using differences, making the algorithm O(N) - the inner while loop tends towards a constant:
int val = max_digit10;
for(int i = max_digit11 * 2; ; i++)
{
int trailing_spaces = 0, pow11 = 1, pow10 = 1;
int j = i;
while((j % 11) == 0)
{
trailing_spaces++;
pow11 *= 11;
pow10 *= 10;
j /= 11;
}
int output_val = val / pow10;
if(output_val <= n)
{
cout << output_val << ", ";
count++;
}
if(output_val == n)
found_n = true;
if(trailing_spaces > 1)
{
i += (pow11 / 11) - 1;
}
else if(found_n && (trailing_spaces == 1))
{
i += 10;
val += 10;
}
else if(trailing_spaces == 0)
val++;
if(count == n)
break;
}
Demonstration
The alternative, simpler approach is just to generate N strings from the numbers and sort them.
Maybe more general solution?
#include <vector>
#include <algorithm>
using namespace std;
// returns true is i1 < i2 according to lexical order
bool lexicalLess(int i1, int i2)
{
int base1 = 1;
int base2 = 1;
for (int c = i1/10; c > 0; c/=10) base1 *= 10;
for (int c = i2/10; c > 0; c/=10) base2 *= 10;
while (base1 > 0 && base2 > 0) {
int d1 = i1 / base1;
int d2 = i2 / base2;
if (d1 != d2) return (d1 < d2);
i1 %= base1;
i2 %= base2;
base1 /= 10;
base2 /= 10;
}
return (base1 < base2);
}
vector<int> lexicalOrder(int n) {
vector<int> result;
for (int i = 1; i <= n; ++i) result.push_back(i);
sort(result.begin(), result.end(), lexicalLess);
return result;
}
The other idea for lexicalLess(...) is to convert integers to string before comparision:
#include <vector>
#include <algorithm>
#include <string>
#include <boost/lexical_cast.hpp>
using namespace std;
// returns true is i1 < i2 according to lexical order
bool lexicalLess(int i1, int i2)
{
string s1 = boost::lexical_cast<string>(i1);
string s2 = boost::lexical_cast<string>(i2);
return (s1 , s2);
}
You need Boost to run the second version.
An easy one to implement is to convert numbers to string, them sort the array of strings with std::sort in algorithm header, that sorts strings in lexicographical order, then again turn numbers to integer
Make a vector of integers you want to sort lexicographically, name it numbers.
Make an other vector and populate it strings of numbers in the first vector. name it strs.
Sort strs array.4. Convert strings of strs vector to integers and put it in vectors
List item
#include <cstdlib>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
string int_to_string(int x){
string ret;
while(x > 0){
ret.push_back('0' + x % 10);
x /= 10;
}
reverse(ret.begin(), ret.end());
return ret;
}
int main(){
vector<int> ints;
ints.push_back(1);
ints.push_back(2);
ints.push_back(100);
vector<string> strs;
for(int i = 0; i < ints.size(); i++){
strs.push_back(int_to_string((ints[i])));
}
sort(strs.begin(), strs.end());
vector<int> sorted_ints;
for(int i = 0; i < strs.size(); i++){
sorted_ints.push_back(atoi(strs[i].c_str()));
}
for(int i = 0; i < sorted_ints.size(); i++){
cout<<sorted_ints[i]<<endl;
}
}
As the numbers are unique from 1 to n, you can use a set of size n and insert all of them into it and then print them out.
set will automatically keep them sorted in lexicographical order if you store the numbers as a string.
Here is the code, short and simple:
void lexicographicalOrder(int n){
set<string> ans;
for(int i = 1; i <= n; i++)
ans.insert(to_string(i));
for(auto ele : ans)
cout <<ele <<"\n";
}