50 50.1
Answer: 501

10 20
Answer: 6

0.05 5.02
Answer: 20

0.05 5
Answer: 21

99.98
99.99
Answer: 5001

25 50
Answer: 3

13 14.1
Answer: 15

92 93.4
Answer: 13

30.2 50.01
Answer: 2

0.01 0.02
Answer: 5001

45.55 45.56
Answer: 90

49.09 50.01
Answer: 2

0.01 99.99
Answer: 2

0.29 0.3
Answer: 334

60 70
Answer: 3

Edited by author 08.02.2023 14:48

With e=0.0000001 I have WA#14

read(p,q);
  for i:=2 to 10000000 do begin
    a:=i/100*p;
    b:=i/100*q;
    a1:=trunc(a);
    b1:=trunc(b);
    if (b1-a1)>=0.99 then begin
      write(i);
      halt;
    end;
  end;
why WA ?

Input in the task looks like:
13 14.1

Not like:
13
14.1

So change input in your program to
p,q=map(float, input().split())

Hello!
Seems there is a duplication in english statement of the problem:
>> We know that there are more than P% conductors and less than Q% conductors of all citizens of Ekaterinburg.

>>By percentage, we know that there are more than P% conductors and less than Q% conductors of all Russian citizens in this city

получил АС, долго искал где долбанный баг, т.к. ТЕСТ 22 выдавал ошибку по времени, если у вас такая же проблема и вы используете преобразование double в int то в 22 тесте точность P и Q 3 знака, поэтому ТЛ и выдает, вероятно на вход подаются какие нибудь числа 22.333 и 22.334
удачи

code like:

size_t out = 0;
printf ("%zu\n", out);

result - > wa#1 (((

if i use

printf ("%d\n", out);

all  test accepted (( Why?
https://wandbox.org/ & gcc 7.1  - The first version works correctly (((

less than can't be equal!

Можно просто перебирать ответ наименьшего возможного,  двойки,  пока не найдем нужное значение.

Для данного количества жителей можно за log этого количества по времени найти требуемое количество кондукторов.

#include <bits/stdc++.h>

using namespace std;
float p,q,t1,t2,t3,l1,l2;
int main()
{
    cin>>p>>q;
    for(int i=1; i<=1000000; i++){
        t1=(p*(float)i)/(float)100;
        t2=(q*(float)i)/(float)100;
        l1=(int)t1;
        l2=(int)t2;
        if(l1==t1||l2==t2||l1!=l2){
            if(l1!=t1) l1=l1+1;
            for(int j=l1; j<=l2; j++){
                t3=((float)j/(float)i)*(float)100;
                if(p-0.000001<t3&&t3<q+0.000001){
                    cout<<i;
                    return 0;
                }
            }
        }
    }
}

The follwing is my pgm. Could some give a test case not satisfied here?
var
s1,i:integer;
f:boolean;
a2,b2,a1,b1,a,b:real;
begin

 f:=false;
 read(a);
 read(b);
 a1:=a/100;
 b1:=b/100;
 if b<>0 then
 i:=trunc(1/b1);
 a2:=a1*(i);
 b2:=b1*(i);
 while not f do
    begin
   if (((trunc(b2)<>trunc(a2)))and (frac(b2)<>0)   then f:=true;
   a2:=a2+a1;
   b2:=b2+b1;
   i:=i+1;
 end;
 i:=i-1;
 writeln(i);

end.

1)Not use double format;
2)Use int64;

Submitted following solution, but WA on 1st test. I tried very different inputs and works fine actually.

def get_input_number
        gets.chomp.to_f
end

def calculate_conductors(low_percent, high_percent)
        (2..10000).each do |citizens|
                konduktors = (low_percent*citizens).ceil
                if (low_percent*citizens < konduktors) && (konduktors < high_percent*citizens)
                        return citizens
                end
        end
end

def prepare_input_and_go
        lower_border = get_input_number/100
        upper_border = get_input_number/100
        puts calculate_conductors(lower_border, upper_border)
end

prepare_input_and_go

// 严格按照题目来
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;

const int p = 10000;
double xs, ys;
int x, y, ans;

int main(void)
{
    scanf ("%lf%lf", &xs, &ys);
    ans = 1, x = y = 0;
    // 严格按照题目来 (more than ... ; less than ... )
    ys -= 1E-10, xs += 1E-10;
    while (x == y)
          ans++, x = int (ans * xs) / 100, y = int (ans * (ys)) / 100;
    printf ("%d", ans);
    return 0;
}

import java.util.Scanner;

public class _1011 {
    public static void main(String[] args) throws Exception {
        Scanner s = new Scanner(System.in);
        double p, q;
        p = Double.parseDouble(s.nextLine());
        q = Double.parseDouble(s.nextLine());
        p /= 100;
        q /= 100;
        answer(p, q, 1, 1, 0);
    }

    public static void answer(double p, double q, double m, double mp, long n) {
        long t = (long) (m / q) + 1;
        if (p * t < mp) {
            System.out.println(n + t);
        } else {
            answer(p, q, 1 - q * t + m, 1 - p * t + mp, n + t);
        }
    }
}

Why I have Runtime error? I use C#. Code is here.
using System;
using System.Globalization;

class EntryPoint
{
    static void Main()
    {
        CultureInfo ci = new CultureInfo("en-US");
        double p = Convert.ToDouble(Console.ReadLine(),ci);
        double q = Convert.ToDouble(Console.ReadLine(),ci);
        int x = 1;
        while (true)
        {
            if ((int)(q * x / 100) - (int)(p * x / 100) > 0)
                break;
            x++;
        }
        Console.WriteLine(x);
    }
}

#include<iostream>

using namespace std;

int main()
{
    double p,q;
    cin>>p;
    cin.get();
    cin>>q;
    long tmp_q = 1,tmp_p = 0;
    for(int i = 1;i < size_t(-1); ++i)
    {
    tmp_p = p * i * 100;
    tmp_q = q * i * 100;
    long res =  (int)(tmp_p/10000);
    ++res;
    if(tmp_q % 10000 == 0)
    {
        if(res < (int)(tmp_q/10000))
        {
            cout<<i<<endl;
            break;
        }
    }
    else
    {
        if(res <= (int)(tmp_q/10000))
        {
        cout<<i<<endl;
        break;
        }
    }
    }
}

I do not see error here, but didn't pass the 17'th test
var p,q: extended;
    a,b: longint;
BEGIN
    read(p,q); p:=p/100; q:=q/100;
    a:=1;
    while (true) do begin
       b:=1;
     while (a/b>=q) do inc(b);
     if (a/b>p) then begin
          write(b);
          halt(0);
       end else inc(a);
    end;
END.

The key here is "more than p, less than q", not "more than or equal to p, less than or equal to q"