Need some help...

[Schubaal]

I found a program, and am very new to C programming.... I can't make sense of the last line in the program, to change it to an equation....

int p(int n)
{
    return 1;
}

int n(int x)
{
    if (x >= 0) return n(x-5) + p(x);
    return 0;
}

int d(int x)
{
    if (x >= 0) return d(x-10) + n(x);
    return 0;
}

int q(int x)
{
    if (x >= 0) return q(x-25) + d(x);
    return 0;
}

int c(int x)
{
    if (x >= 0) return c(x-50) + q(x);
    return 0;
}

main()
{
    int i;
    
    for (i = 1; i < 101; i++)
 printf("%d: %d\n", i, c(i));
}

This is more of a math question than a C programming question, but I need to know the true significance of the line:
printf("%d: %d\n", i, c(i));

[twm]

>printf("%d: %d\n", i, c(i));
printf is a function that formats data all nice and pretty to standard output (usually the terminal). The double quoted part is the format string, it tells printf how you want things to look. In this case you want a decimal integer followed by a colon, then a space and another decimal integer, then a line break. %d means decimal signed integer, and \n means a newline. It's the same as if you hit return on the command line. The number of arguments past the format string matches the number of formats (things that start with % in the format string). The arguments will replace the formats with their value, so the first %d is replaced by i and the second %d is replaced by c(i).

It's kind of tough to get a feel for things, but once you do it all makes sense in a convoluted sort of way. But that just the way programmers are...convoluted.