While Loop

An ascending integer occurs when each digit is greater than any digit that precedes it. An example is
478. Write a program that finds the number of ascending integers between 200 and 300.

I know how to do basic while loops but cannot figure out how to separate a number and determine if it is an ascending integer.

C++.

Answered by Coder100 [earned 5 cycles]

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So sorry about that, logic mistakes :(


#include <iostream>
#include <vector>

int main()
{
  unsigned int hundred = 2, ten = 0, one = 0;
  std::vector<int> nums;
  
  while (hundred < 3)
  {
    while (ten < 10)
    {
      while (one < 10)
      {
        if (hundred < ten && ten < one)
        {
          nums.push_back(100 * hundred + 10 * ten + one);
        }
        one ++;
      }

      one = 0;
      ten++;
    }

    ten = 0;
    hundred ++;
  }

  for (int i : nums)
  {
    std::cout << i << " ";
  }

  std::cout << std::endl;
  return 0;
}

If you need I can explain the code to you, but it should be pretty straightforward (I think)

Thank you very much.

np :)

here is better solution


#include <iostream>

int d1 = 2;
int d2 = d1 + 1;

int main(){
  while(d2<10) {
  int d3 = d2 + 1;
  while (d3<10) {
    int number = d1*100+d2*10+d3;
    std::cout<<number;
    std::endl;
    d3++;
  }
  d2++;
  }
}

pls do std::endl

idk cpp meanie

xd i mean its just user preference

fixed

but the thing is std::endl is better because some people use \r\n so you could have just gotten garbled text jk jk most os now do that convert automatically

ok at edit your answer so it is the better version >:)

its not rlly user preference cus realTronsi didnt use use namespace std so it wouldnt work if he/she didnt use std::endl

yes you do if you dont put mark it as standard library

After thinking about this problem, I got to this solution.


#include <iostream>
#include <vector>

int main()
{
  unsigned int hundred = 2, ten = 0, one = 0;
  std::vector<int> nums;
  
  while (hundred < 3)
  {
    while (ten < 10)
    {
      while (one < 10)
      {
        one ++;
        if (hundred < ten && ten < one)
        {
          nums.push_back(100 * hundred + 10 * ten + one);
        }
      }
      
      one = 0;
      ten++;
    }

    ten = 0;
    hundred ++;
  }

  for (int i : nums)
  {
    std::cout << i << " ";
  }

  std::cout << std::endl;
  return 0;
}

thank god for 0!

Hmm there's an extraneous 300 in there, how strnage

he asked for number of, not the exact numbers (and your answer also doesn't work) but literally just 6+5+4+3+2+1 (21) is your answer

basic permutation as such:

23_ ,4-9: 6 possibilities
24_ ,5-9: 5 possibilities,

etc.

Your script gave these numbers:


234 235 236 237 238 239 240 245 246 247 248 249 250 256 257 258 259 260 267 268 269 270 278 279 280 289 290 300

some such as 280 aren't ascending, and you also missed a few

num.size() iirc

yeah unfortunately i did a bad plz read the marked one

ok good. Just that your answer is questionable.


//I'm not gonnna write actual code since I don't know much cpp but this is the actual logic:

digit2 = digit1 + 1
while digit2 < 10
  digit3 = digit2 + 1
  while digit3 < 10
     cout << digit1*100 + digit2*10 + digit3
     digit3++
  digit2++

(not actual code since idk much cpp but there is no point in looping over every single number.)

hey does it not work?

it stores all the integers, and if I took the time to study permutations, it might be more elegant.

yours work but it's a lazy solution lol

Something like:
3*2 * 8! * 7!
should suffice but it doesn't i don't think

did you see my code answer though? If you put that in cpp syntax it should work, I could put it in js syntax to prove a point

indeed plz helpz makez betterz

OH RIGHT i should probably skip a few numbers yes yes


let d1 = 2
let d2 = d1+1

while(d2<10){
  let d3 = d2+1;
  while(d3<10){
    console.log(d1*100+d2*10+d3)
    d3++
  }
  d2++
}

This extremely short chunk of code in js works, (just an extra 9 at the end that can easily be solved)

hmm true, but whats with the 9 anyways

idk its nodejs problem, try putting a console.log() after the while loops and the 9 disappears

ill convert it into cpp

c++ god


int d1 = 2;
int d2 = d1 + 1;

do {
  int d3 = d3 + 1;
  do {
    std::cout<<d1*100+d2*10+d3<<
    d3++
  } while (d3<10);
  d2++
} while(d2<10);

there might be syntax errors since im not the best cpp


#include <iostream>

int d1 = 2;
int d2 = d1 + 1;

while(d2<10) {
  int d3 = d3 + 1;
  while (d3<10) {
    std::cout<<d1*100+d2*10+d3<<
    d3++;
  }
  d2++;
}

theres an error unqualifed-id idk how to fix but anyways theres code

hol' up why is there an extra <<

thats an old version dont look at it

what lang?

C++.

alright!

please read the question thanks

btw your example won't work endl isn't defined

actually read the question smh smh