2 New To have a Clue

Original - cpp Code #include <stdio.h> int main(void) { int i; for(i = 0; i < 25; i += 5) printf(" %2d %2d %2d %2d %2d \n", i, i+1, i+2, i+3, i+4); return 0; }

#include <stdio.h>
 
int main(void)
{
    int i;
    for(i = 0; i < 25; i += 5)
        printf(" %2d %2d %2d %2d %2d \n", i, i+1, i+2, i+3, i+4);
 
    return 0;
} 

Original - cpp Code int fac(int x) { if(x == 1) return 1; else return x * fac(x - 1); }

int fac(int x)
{
    if(x == 1)
        return 1;
    else
        return x * fac(x - 1);
} 

Original - cpp Code #include <stdio.h> #include <stdlib.h> // global variable to store the set we're currently working on int amounts[6]; // recursive function, takes a value, a bill to start at, and the number of options so far. int calculate_options(int value, int denom, int counter) { // set possible denominations static int vals[] = {1, 2, 5, 10, 20, 50}; // at the 1 dollar bill yet? if(denom > 0) // no, for all the possible amounts the current bill fits in the current value, get all possible sub-amounts. { for(amounts[denom] = 0; value - amounts[denom] * vals[denom] >= 0; amounts[denom]++) counter = calculate_options((value - amounts[denom] * vals[denom]), (denom - 1), counter); } else // yes, output current set of values { // another option completed, increment counter counter++; // print the current set of bills printf(" %2d * %5d * %6d * %3d * %4d * %3d * %3d \n", counter, amounts[5], amounts[4], amounts[3], amounts[2], amounts[1], value); } return counter; } // main, takes arguments to read the command line int main(int argc, char **argv) { // declarations int bill, counter; // see if there's a value on command line if(argc == 1) // no, default to 100 bill = 100; else // yes, read value bill = atoi(argv[1]); // print table header printf(" No * fifty * twenty * ten * five * two * one \n"); printf("**********************************************\n"); // call function counter = calculate_options(bill, 5, 0); // print number of possibilities printf("There are %d different ways to break up %d dollars\n", counter, bill); // exit return 0; }

#include <stdio.h>
#include <stdlib.h>
 
// global variable to store the set we're currently working on
int amounts[6];
 
// recursive function, takes a value, a bill to start at, and the number of options so far.
int calculate_options(int value, int denom, int counter)
{
    // set possible denominations
    static int vals[] = {1, 2, 5, 10, 20, 50};
    
    // at the 1 dollar bill yet?
    if(denom > 0)      // no, for all the possible amounts the current bill fits in the current value, get all possible sub-amounts.
    {
        for(amounts[denom] = 0; value - amounts[denom] * vals[denom] >= 0; amounts[denom]++)
            counter = calculate_options((value - amounts[denom] * vals[denom]), (denom - 1), counter);
    }
    else                // yes, output current set of values
    {
        // another option completed, increment counter
        counter++;
        // print the current set of bills
        printf(" %2d * %5d * %6d * %3d * %4d * %3d * %3d \n", counter, amounts[5], amounts[4], amounts[3], amounts[2], amounts[1], value);
    }
    return counter;
}
 
// main, takes arguments to read the command line
int main(int argc, char **argv)
{
    // declarations
    int bill, counter;
 
    // see if there's a value on command line
    if(argc == 1)   // no, default to 100
        bill = 100;
    else            // yes, read value
        bill = atoi(argv[1]);
 
    // print table header
    printf(" No * fifty * twenty * ten * five * two * one \n");
    printf("**********************************************\n");
 
    // call function
    counter = calculate_options(bill, 5, 0);
 
    // print number of possibilities
    printf("There are %d different ways to break up %d dollars\n", counter, bill);
 
    // exit
    return 0;
}