Translation of number into strings [closed] - c++

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I would like to know whether there is a way in which we can print the number alphabetically i.e
123 should be printed as one two three.
The only condition is that we should not reverse the number and we should not use array.
I only know these two ways:
"Reverse the number", that is, taking the last digit and cutting it off. For each cut-off digit, one can use an array to look up the correct string.
using switch and a lot of cases
Any ideas?

for hundreds place:
int hundreds = my_num / 100 //Needs "/", NOT "%"
if(hundreds == 0)
cout << "zero";
else if(hundreds == 1)
cout << "one";
//repeat for 2-9
This process could be tweaked to do the other digits as well. It is also worth mentioning that the if/else block a) could be done with a switch/case if preferred, and b) could pretty easily be made into a separate function to avoid having to repeat the block of code over and over, I just wrote out as much as I did for clarity's sake. Note that this assumes the number you're "translating" is an integer. With integers the "/" operator will return the full quotient WITHOUT the remainder, e.g. 123 / 100 = 1, not 1.23

Not necessarily the easiest route, but you can make a function, say DigitToWord which will take a digit 0, 1, 2, ...etc to its word with a switch statement. Then I recommend using a for loop over the number, continuously dividing by 10 and taking the mod for the loop:
int num; //my number i want to print
int div = pow(10, (int)log10(num)); //find the largest power of 10 smaller than num
while(num > 0) {
int remainder = num%div;
int digit = num/div;
DigitToWord();
num = remainder;
}

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Given a string output it in a specific way using recursion? [closed]

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Given a string we have to output the string in a special way.
• If the string consists of one char, we output that char normally.
• Otherwise, we divide the string into two equal parts (if the number of letters in the substr is odd, the second part of the substr will be one letter longer than the first), and we output the first part twice and then the second part (according to the same rules).
For example, let's assume that we want to output the string YOGURT. We divide that string into two equal parts: YOG and URT.
How will we output the substr YOG? Again, it will be divided into two parts - Y and OG. The substr Y we output normally (but in the output of the substr YOG we will do it twice), and the substr OG we output as OOG. So the substr YOG we output as YYOOG.
Analogously, the substr URT is going to give the output UURRT. So the string YOGURT is going to be output as YYOOGYYOOGUURRT.
Length of the string can at max be 10000.
Now I tried using a non recursion way to solve this problem but it was way to slow so I have come to an conclusion I have to do this with recursion. And since I don't have that much experience with recursion I would really need some help.
This is very naturally implemented with recursion like so:
void print(std::string_view s) {
if (s.size() <= 1) std::cout << s;
else {
auto m = s.size() / 2;
print(s.substr(0, m));
print(s.substr(0, m));
print(s.substr(m));
}
}

Want to see if a one digit of a 2 digit number is divisible by the other digit [closed]

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The problem I am having is how to say if(//code=Whole) I dont know how to tell the program if the number equals whole number because if its a number like 39 9 divided 3 equals a whole number if a number is not divisible it will go to the decimals which is how I am willing to solve the problem by saying if digit 1 or 2 divisible by the other digit is whole then cout<<Divisible
I tried searching up didnt work heres my code
#include <iostream>
using namespace std;
int main()
{
int x,z,n;
cin>>x;
x%10=z;
x/10=n;
if(z/n==int)
{
cout<<"YES";
}
else if(n/z==int)
{
cout<<"YES"
}
else
{
cout<<"NO";
}
}
I dont know how to say if it equals whole So i dont know what to put there plus there might be something wrong in the program other than that.
This is your code.
int main()
{
int x,z,n;
cin>>x;
x%10=z;
x/10=n;
if(z/n==int)
{
cout<<"YES";
}
else if(n/z==int)
{
cout<<"YES"
}
else
{
cout<<"NO";
}
}
A few things. First, please use whitespace to make your code readable. It's going to safe you from a wide variety of bugs over the years.
Next, this is bad code:
x%10=z;
x/10=n;
So is this:
if(z/n==int)
The first few lines of your main method should be like this:
int x;
cin >> x;
int z = x % 10;
int n = x / 10;
Note that I've done two things. First, I moved where the variables are declared to as late as possible so that they aren't hanging around, uninitialized for a while. This is just good practice. Define them and assign them at the same time.
Next, the way you wrote that code doesn't make sense. An assignment statement can't be written backwards the way you did it.
In your head, you could say, "z becomes x % 10" if you have to.
This line won't compile:
if(z/n==int)
If you want to see if z divides evenly by n:
if ( z % n == 0 )
You have the same problem in the else-if a few lines later.

Recursive function, geometric progression [closed]

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I've just started learning C++ and I came across a simple problem but couldn't solve. Help me please,
In geometric progression, b1=2, r=3.
Create recursive function that outputs initial n sequences of progression. n should be input.
Assuming b1 is the initial number and r is the multiplier (ratio?), that would give 2, 6, 18, 54, ....
The pseudo-code for such a beast would be:
def geometric (limit, current, multiplier):
if limit <= 0: return
print(current)
geometric(limit - 1, current * multiplier, multiplier)
def main():
b1 = 2
r = 3
n = input("How many terms? ")
geometric(n, b1, r)
The reason it's pseudo-code is that I tend to prefer for educational questions since people learn a lot more by converting it rather than just being given an answer.
So, in terms of how the recursive function works, you pass it the number of terms remaining along with the current term and the multiplier.
Provided it hasn't reached its base case (number of terms remaining is zero), it prints the current term then recursively calls itself, adjusting the arguments so as to approach the base case.
Once it has reached the base case, it will return (from all recursive levels).
Turning that into C++ is an exercise I'll leave for the reader, but I can provide something similar so you will see the approach taken.
Let's say you wanted to sum all the numbers between 1 and a given number. It has the same features as any recursive solution, including the one posed in your question:
an operation that can be defined in terms of a simpler operation; and
the base case.
In C++, a complete program to do that would be something like:
#include <iostream>
void recurse(int number, int sum) {
// Base case for printing and returning.
if (number <= 0) {
std::cout << "Sum is " << sum << '\n';
return;
}
// Recurse to next level.
recurse(number - 1, sum + number);
}
int main() {
int myNum;
std::cout << "Enter your number: ";
std::cin >> myNum;
recurse(myNum, 0);
}
Some sample runs:
Enter your number: 3
Sum is 6
Enter your number: 4
Sum is 10
Enter your number: 5
Sum is 15
Enter your number: 6
Sum is 21
Enter your number: 100
Sum is 5050
The explanatory text on your specific question, along with the C++ code showing how to do it for a different question, should hopefully be enough for you to work it out.

How to start on the maths challenge "find the sum of the multiples of X under Y"? [closed]

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I'm new in C++ and I have been thinking about this.
There is a question that asks:
Find the sum of the multiples of 2 under 547.
How am I supposed to code the program to keep on multiplying the user input and adding them at the same time until they reach the number under 547, then stop the process and count the result?
Find the sum of the multiples of 2 under 547
Lets break this down...
A "Multiple of two" is a number that can be divided by two evenly. Meaning there is no remainder. Look at this link for more info.
Ok we know what a "Multiple of two" is now lets look at the sum...
We want to add up "Multiples of two" under 547. Meaning we will count up to 547 by two and add them together.
so 2 + 4 + 6 + .... + 544 + 546 = A big number
Like #Mureinik posted...we start by defining and initializing something to hold our sum (adding up all the counting by two) in.
long sum = 0;
Then we need to count up to 547 by twos.
for(int i = 0; i < 547; i += 2){
}
This is a for loop. It will execute what is inside the { } until the condition is false. It defines i as the integer of 0. The condition is i < 547. It will also increment i by two each time through the loop. That is what the i += 2 part does. (i += 2 is the same thing as i = i + 2)
So now we have something to hold our summation in (sum) and we have a means of counting by two (the for loop). All we have to do is add up the numbers.
sum += i;
This will take care of that for you. Like you hopefully have guessed sum += i is short hand for sum = sum + i;. i will always be a "Multiple of two" because we are always adding 2 to i. Once i gets to 548 it will fall out of the loop and stop adding up the sum.
Like others have commented it will not hurt to look up some tutorials on the web. I know that it can be over whelming you just have to stick with it.
Hope this clears up some stuff for you. Code On.
You just need to iterate over the multiples and sum them:
long sum = 0;
for (int i = 0; i < 547; i +=2) {
sum += i;
}

Separating number input into units [closed]

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I need to separate users input into units and store them in an array.
e.g if user enters 6547. Array will store {6,5,4,7} Using C++ on Linux
I would appreciate if you can help me with pseudocode or explain an algorithm.
I'm a beginner so please restrain from advising advanced function (and explain its use if you do) as we have studied basics so far
N.B| If such question has already been answered and I skipped it in search, please do point me to it.
The math for isolating right most digit:
digit = digit % 10;
The math for shifting a number right by one digit:
new_number = old_number / 10;
Every letter and number can be represented as a text character. For example, '5' is a character representing the single decimal digit 5.
The math for converting a textual digit (character) to numeric:
digit = char_digit - '0';
Example:
digit = '9' - '0';
The math for converting a numeric digit to a textual digit (character):
char_digit = digit + '0';
Example:
char_digit = 5 + '0';
Your problem basically breaks into few parts, which you need to figure out:
how to read one character from input
how to convert one character to the digit it represents
how to store it in the array
Please, try to explain if you have problem with some particular point from the list above or there is a problem somewhere else.
Suppose Variable input_string holds the number entered by the user & you want to store it in an array named 'a'...Here's a C snippet.. you can easily convert it into C++ code..
I would recommend taking the input as string rather than int so that you can directly insert the digits extracted from the end...(else you can start storing the integer from beginning and then reverse the array)
scanf("%s",&input_string)
size = strlen(input_string)-1
input = atoi(input_string)
while (input/10>0)
{
i=input%10;
input=input/10;
a[size]=i;
size--;
}
Hope that helps!
Here's a C++11 solution:
std::string input;
std::cin >> input;
int num = std::stoi(input);
std::vector<int> v_int;
for (unsigned int i = 0; i < input.size(); i++)
{
v_int.push_back(num % 10);
num /= 10;
}
// To get the numbers in the original order
std::sort(v_int.rbegin(), v_int.rend());
for (unsigned int i = 0; i < v_int.size(); i++) {
std::cout << v_int[i] << std::endl;
}
If you want it in a c-style array, do this:
int* dynamic_array = new int[v_int.size()];
std::copy(v_int.begin(), v_int.end(), dynamic_array);
delete dynamic_array;