pattern_10_10.c

/*

    pattern_10_10.c

    https://www.facebook.com/photo/?fbid=1368408023607660&set=gm.5173435132677435

    Task pattern:

n = 4 , number of rows.

                  1
              10 10
        100 100 100
1000 1000 1000 1000

In the first row we have those 1 (10 ^ 0)
In the second row we have two 10 (10 ^ 1)
In the third row we have three 100 (10 ^ 2)
. . .
And in the nth row we have n 10 ^ n

The problem is that the numbers are aligned with the right margin and
therefore for each line we have to calculate how many blank characters
precede the beginning of the number printing.

n = 4 , number of rows.

------------------1     ,  row = 0 has 18 blanks in front of the first digit
--------------10 10     ,  row = 1 has 14 blanks in front of the first digit
--------100 100 100     ,  row = 2 has  8 blanks in front of the first digit
1000 1000 1000 1000     ,  row = 3 has  0 blanks in front of the first digit

The length of the longest row is n*n+n-1 .

In each row the numbers take up (row+1)*(row+1)+row places.
The rest up to the length of the longest row are blank characters.

So, number of blank characters in each row is
blanks = n*n+n-1 - ( (row+1)*(row+1)+row ) .


    You can find all my C programs at Dragan Milicev's pastebin:

    https://pastebin.com/u/dmilicev

*/

#include <stdio.h>
#include <math.h>   // due to the function of pow()

// implementation of function pow() only for integers
int my_pow_for_integers(int base, int exponent){
    int result = 1;
    for (exponent; exponent>0; exponent--)
        result = result * base;
    return result;
}

// print a pattern along the left edge
void print_pattern1(int n){
    int i, j;

    for(i=0;i<n;i++){       // i count rows
        for(j=0;j<=i;j++)   // j counts 10^i
            printf("%.f ", pow(10,i) );
        printf("\n\n");
    }
}

// print a pattern aligned along the right edge
void print_pattern2(int n){
    int i, row, blanks;

    for(row=0;row<n;row++){                 // print rows, row count rows

        blanks = n*n+n-1 - ( (row+1)*(row+1)+row ); // calculate how many blanks

        for(i=0;i<blanks;i++)               // print blanks, i counts how many
            printf("-");

        for(i=0;i<=row;i++)                 // print the numbers 10^row, i counts how many
            printf("%.f ", pow(10,row) );

        printf("\n\n");
    }
}


int main(void){
    int n=4;                // n is number of rows

    print_pattern1(n);      // print a pattern along the left edge

    printf("\n\n");

    print_pattern2(n);      // print pattern aligned along the right edge

//  printf("\n 10 ^ %d = %d\n", n, my_pow_for_integers(10,n) );

    printf("\n\n");
    return 0;
} // main()