Java Application Simple recursion

After graduation on how to learn the algorithm, still in school when learning data structure, and now basically back to the teacher, but the teacher tuition fees did not return me ...

Scenario: Like given a number, count how many numbers he consists of, such as: 36 now has 10, 5, 1, then the best handsome 3 10, a 5, a 1 composition (the default priority to take maximum). Because we need to develop a business on this side of the ticket, and sometimes do not want a count, let us write a. But the number of his invoice is not fixed, endless, so every time to judge, if: 36 now the base is still 10, 5, 1 but only let you take 1 10, 2 5, a number of 1, then the combination should be: a 10, 2 5, 16 1, probably this meaning.

Add two lists, one to hold the cardinality (from large to small) and the corresponding number of the base to hold.

List List =NewArrayList (); List.add (10); List.add (100); List.add (1); List.add (20); List.add (50); List.add (5);        Collections.sort (list, Collections.reverseorder ()); List List2=NewArrayList (); List2.add (1);//100 corresponding number (hereinafter)List2.add (4);// -List2.add (1);// -List2.add (20);//TenList2.add (50);//5List2.add (1000);//1

Here's the recursion. No difficulty. The parameters are: the total number of inputs, the initial subscript of the set, the base set, the number of cardinality sets

 Public voidCintTotle,intflag, List List, List list2) {        //System.out.println ("current Total" + Totle + ", current quota" + list.get (flag));        intsum = 0;  for(inti = 0; I < list.size (); i++) {sum+= (int) List.get (i) * (int) List2.get (i); }        if(Sum <totle) {System.out.println ("Invoice total is not enough, please add invoice!!!!" "); } Else {            if(Totle% (int) list.get (flag) = = 0) {                ints = totle/(int) list.get (flag); //System.out.println (S + "");                if(S > (int) {list2.get (flag)) {System.out.println ("Need" + list.get (flag) + "yuan Invoice" + S + "Zhang, currently only" + list2.get (flag) + "Zhang, all use! "); Totle-= (int) list2.get (flag) * (int) list.get (flag);                    System.out.println (Totle); C (Totle,++flag, list, list2); } Else{System.out.println ("Need" + list.get (flag) + "yuan Invoice" + S + "Zhang"); }            } Else {                ints = totle/(int) list.get (flag); if(S > (int) {list2.get (flag)) {System.out.println ("Need" + list.get (flag) + "yuan Invoice" + S + "Zhang, currently only" + list2.get (flag) + "Zhang, all use! "); Totle-= (int) list2.get (flag) * (int) list.get (flag);                    System.out.println (Totle); C (Totle,++flag, list, list2); } Else{totle= Totle% (int) list.get (flag); System.out.println ("Need" + list.get (flag) + "yuan Invoice" + S + "Zhang");                    System.out.println (Totle); C (Totle,++flag, list, list2); }            }        }    }

Assume:

C (365, 0, list, list2);

Output

If the input exceeds all of the total, then the following:

Very simple, record, but also remind yourself, the algorithm is quite important ... Do not have to follow slowly study!!!

Java Application Simple recursion