/*
;First homework exercise:
;Based on the definitions below of count-change, cc, and next, 
;draw a tree showing the tree recursive calls of cc 
;in execution of (count-change 26).
;What would the difference be for the call (count-change 27)?

; submit this homework on paper.
*/

long count_change(int n);
long cc(int n, int largest_coin);
int next(int c);

// memoization system
#include <map>
using namespace std;
typedef pair<int, int> args;
map<args, long> M;
long setmemo(int n, int coin, long ans)
{ M[args(n, coin)] = ans; return ans; }
bool getmemo(int n, int coin, long& ans)
{ return ans = M[args(n, coin)]; }

/*
(define (count-change n)
 ; return the number of different ways to make change of n cents.
  (cc n 50))
*/

long count_change(int n)
{ return cc(n, 50); }

/*
(define (cc n largest-coin)
 ; return the number of different ways to make change of n cents 
 ; using no coins larger in amount than largest-coin.
 (cond ((zero? n)          1)
       ((= largest-coin 1) 1)
       ((< n largest-coin) (cc n (next largest-coin)))
       (else               (+ (cc n 
				  (next largest-coin))
		              (cc (- n largest-coin) 
				  largest-coin))) ))
*/
long cc(int n, int largest_coin)
{	long ans;
	if (n == 0) return 1;
	if (largest_coin == 1) return 1;
	if (0 != getmemo(n, largest_coin, ans)) return ans;
	else
	if (n < largest_coin) return setmemo(n, largest_coin, cc(n, next(largest_coin)));
	else 
		return setmemo(n, largest_coin, 
			cc(n, next(largest_coin)) // no use of largest
		
			+ 
		       cc(n-largest_coin, largest_coin) // at least one use of largest
		       );
}

/*
(define (next coin)
  (cond ((= coin 50) 25)
        ((= coin 25) 10)
        ((= coin 10) 5)
        ((= coin 5) 1)
	(else (error "next: unrecognized coin" coin)) ))
*/
#include <iostream>

int next(int c)
{
	switch(c) {
		case 50: return 25;
		case 25: return 10;
		case 10: return 5;
		case 5: return 1;
		default: cerr << "next: no next coin for " << c << endl;
		}
	return -1; // error
}

int main()
{
	cout << 5 << " " << count_change(5) << endl;
	cout << 6 << " " << count_change(6) << endl;
	cout << 10 << " " << count_change(10) << endl;
	cout << 23 << " " << count_change(23) << endl;
	cout << 230 << " " << count_change(230) << endl;
	cout << 2304 << " " << count_change(2304) << endl;
	cout << 23045 << " " << count_change(23045) << endl;
	cout << 230456 << " " << count_change(230456) << endl;
}