Problem to remove space after line break

Question

#include <stdio.h>

int main (){

    int x,y,i;

    scanf("%d",&x);
    scanf("%d",&y);

    if ((x<y)&&(x>1)&&(x<20)&&(y<10000)){
        for (i=1;i<=x;i++){
            printf("%d ",i);
                }
            printf("\n");
       for (i=(x+1);i<=y;i++){
           printf("%d ",i);
       }
   }
 return 0;
}

objective of the programme:

1st: The program reads two numbers X and Y (X minor Y). Next shows a sequence from 1 to Y, passing to the next line to each X numbers.

2º: Each sequence shall be printed on one line, with a blank space between each number.

3º: The entry contains two integers X (greater than 1 and less than 20) and Y (greater than X and less than 100000) respectively.

4º:OBS: The program does not need to have an answer for cases where the program does not run, and also does not need user interaction.

***Example demonstrating my program error:

By removing that last space before the break of each line the problem of the question will be solved, I just can’t find a way to do it.

Answer 1

Treat the first one as an exception that doesn’t have a space, the rest will all go within the normal.

#include <stdio.h>

int main () {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    if (x < y && x > 1 && x < 20 && y < 10000) {
         printf("%d", 1);
        for (int i = 2; i <= x; i++) printf(" %d", i);
        printf("\n%d", x + 1);
        for (int i = x + 2; i <= y; i++) printf(" %d", i);
    }
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

You can do the reverse and treat the latter as an exception, but I don’t like it.

Answer 2

Your program displays the sequence of 1 a X, however, the question statement says nothing about displaying this sequence! Which of the two is correct?

To identify values of Y that are multiples of X just check if the rest of the division (or module) of Y for X is equal to zero, check it out:

if( y % x == 0 )
    printf( "Y eh multiplo de X!\n" );

Following his reasoning and the statement of the question, follows a commented solution:

#include <stdio.h>

int main( void )
{
    int i,x,y;

    /* 1) O programa le dois numeros X e Y (X menor Y)... */
    printf("X: ");
    scanf( "%d", &x );

    printf("Y: ");
    scanf( "%d", &y );

    /*
        3) A entrada contem dois numeros inteiros X (maior que 1 e menor que 20) e
           Y (maior que X e menor que 100000) respectivamente.
    */
    if( (x<y) && (x>1) && (x < 20) && (y < 10000) )
    {
        /* 1) A seguir mostra uma sequencia de 1 a Y... */
        for ( i = 1; i <= y; i++ )
        {
            printf( "%d", i );

            /* 1) Passando para a proxima linha a cada X numeros... */
            /* ...e passando pra ultima linha na ultima iteracao    */
            if( (i % x == 0) || (i == y) )
                printf("\n");
            else
                printf(" "); /* Pulo do gato! */
        }
    }
    else
    {
        /* 4) OBS: O programa nao precisar ter resposta para os casos em que o programa nao roda */
        printf("Entrada invalida!\n");
    }

    return 0;
}

Testing (X=19 and Y=60):

X: 19
Y: 60
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38
39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57
58 59 60

Testing (X=3 and Y=17):

X: 3
Y: 17
1 2 3
4 5 6
7 8 9
10 11 12
13 14 15
16 17

Testing (X=4 and Y=15):

X: 4
Y: 15
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15

Testing (X=2 and Y=8):

X: 2
Y: 8
1 2
3 4
5 6
7 8