URI / BEE CROWD 1133 - Rest of a Division - Solution in C,C++,Python | URI - BEECROWD - BEE 1133 Solution in C,C++,Python

                                                                                             beecrowd | 1133

Rest of a Division

Adapted by Neilor Tonin, URI  Brazil

                                                                                             Timelimit: 1

URI / BEE CROWD 1133 - Rest of a Division - Solution in C,C++,Python | URI - BEECROWD - BEE 1133 Solution in C,C++,Python :

Write a program that reads two integer numbers X and Y. Print all numbers between X and Y which dividing it by 5 the rest is equal to 2 or equal to 3.

Input

The input file contains 2 any positive integers, not necessarily in ascending order.

Output

Print all numbers according to above description, always in ascending order.

Input Sample	Output Sample
10 18	12 13 17

N.B: Don't copy paste the code as same. Just try to understand it and try yourself.

  URI / BEE 1133 Solution in C :   

URI / BEECROWD Online Judge 1133 Solve  in C :

#include<stdio.h>
int main()
{
    int a,b,i;

    scanf("%d %d",&a,&b);

    if(a<b)
    {
        for(i=a+1;i<b;i++)
        {
            if(i%5==2 || i%5==3)
            {
                printf("%d\n",i);
            }
        }
    }
    else if(a>b)
    {
        for(i=b+1;i<a;i++)
        {
            if(i%5==2 || i%5==3)
            {
                printf("%d\n",i);
            }
        }
    }
    return 0;
}

  URI / BEE 1133 Solution in C++ :   

URI / BEECROWD Online Judge 1133 Solve  in C++ :

#include <iostream>
using namespace std;

int main(){
    int a,b;
    
    cin >> a;
    cin >> b;
    
    int aux;
    
    if(b < a){
         aux = a;
         a = b;
         b = aux;
    }
    
    for(int i = a+1; i < b; i++){
            if(i%5 == 2 || i%5 == 3){
                   cout << i << endl;
                   }
                   }
    return 0;
}

  URI / BEE 1133 Solution in Python :   

URI / BEECROWD Online Judge 1133 Solve  in Python :

n1=int(input())
n2=int(input())
t = n1
if (n1 > n2):
    n1 = n2
    n2 = t
while (n1 < n2):
    n1 += 1
    if (n1 % 5 == 2 or n1 % 5 == 3 and n1 != n2):
        print(n1)

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URI / BEE CROWD 1132 - Multiples of 13 - Solution in C,C++,Python | URI - BEECROWD - BEE 1132 Solution in C,C++,Python

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