AW>      problem.  The program is to input a value (like $1. $2. $5..)
AW>      and spit out a table showing all the different cominations of coins
AW>      that could be used to make up that value.  So for $0.10 the table
    This sounds like a problem that recursion can solve.
    First, let's put all the coins in an array.
    double coins[] = {0.01, 0.05, 0.10, 0.25};  //normal coins only
      // or maybe just int coins[] = {1,5,10,25};
    Now, we can systematically access a coin as an enumerated types.
    
    Now, lets write only part of the problem. At each step we wish to
    try to see if adding a coin will make or break the target dollar
    amount.
    void allCoins(double target, double tally) {
      if (tally < target) {
         for (int i = 0; i<4; i++) {
           if (tally + coins[i] <= target) {
             allCoins(target, tally + coins[i]); 
           }
         }
      }
    }
    The first run it will have 100 allCoins calls, each using a 
    penny. Next, it will have 95 pennies and a nickel, 90 pennies
    and a dime, and 75 pennies and a quarter. Etc. Etc.
    The REAL problem will be keeping track of which coins are 
    'all ready in the tally'.
    I think a Stack would be the best answer.
    So
    Stack s;
    void allCoins(double target, double tally) {
      if (tally < target) {
         for (int i = 0; i<4; i++) {
           if (tally + coins[i] <= target) {****
             s.push( coins[i] );
             allCoins(target, tally + coins[i]);
             s.pop();
           }
         }
      } else if (tally == target) {
        // write the contents of the stack
        // assuming that Stack overloads the << operator
        cout << s;
      }
    }
    There you have it! A nice an somewhat simple solution.
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