By using generating functions, you can get a closed-form constant-time solution to the problem. Graham, Knuth, and Patashnik’s Concrete Mathematics is the book for this, and contains a fairly extensive discussion of the problem. Essentially you define a polynomial where the nth coefficient is the number of ways of making change for n dollars.
Pages 4-5 of the writeup show how you can use Mathematica (or any other convenient computer algebra system) to compute the answer for 10^10^6 dollars in a couple seconds in three lines of code.
(And this was long enough ago that that’s a couple of seconds on a 75Mhz Pentium...)
#!/usr/bin/python
# find the number of ways to reach a total with the given number of combinations
cents = 200
denominations = [25, 10, 5, 1]
names = {25: "quarter(s)", 10: "dime(s)", 5 : "nickel(s)", 1 : "pennies"}
def count_combs(left, i, comb, add):
if add: comb.append(add)
if left == 0 or (i+1) == len(denominations):
if (i+1) == len(denominations) and left > 0:
if left % denominations[i]:
return 0
comb.append( (left/denominations[i], demoninations[i]) )
i += 1
while i < len(denominations):
comb.append( (0, denominations[i]) )
i += 1
print(" ".join("%d %s" % (n,names[c]) for (n,c) in comb))
return 1
cur = denominations[i]
return sum(count_combs(left-x*cur, i+1, comb[:], (x,cur)) for x in range(0, int(left/cur)+1))
count_combs(cents, 0, [], None)
I would favor a recursive solution. You have some list of denominations, if the smallest one can evenly divide any remaining currency amount, this should work fine.
Basically, you move from largest to smallest denominations. Recursively,
You have a current total to fill, and a largest denomination (with more than 1 left). If there is only 1 denomination left, there is only one way to fill the total. You can use 0 to k copies of your current denomination such that k * cur denomination <= total.
For 0 to k, call the function with the modified total and new largest denomination.
Add up the results from 0 to k. That's how many ways you can fill your total from the current denomination on down. Return this number.
Here's my python version of your stated problem, for 200 cents. I get 1463 ways. This version prints all the combinations and the final count total.
#!/usr/bin/python
# find the number of ways to reach a total with the given number of combinations
cents = 200
denominations = [25, 10, 5, 1]
names = {25: "quarter(s)", 10: "dime(s)", 5 : "nickel(s)", 1 : "pennies"}
def count_combs(left, i, comb, add):
if add: comb.append(add)
if left == 0 or (i+1) == len(denominations):
if (i+1) == len(denominations) and left > 0:
if left % denominations[i]:
return 0
comb.append( (left/denominations[i], demoninations[i]) )
i += 1
while i < len(denominations):
comb.append( (0, denominations[i]) )
i += 1
print(" ".join("%d %s" % (n,names[c]) for (n,c) in comb))
return 1
cur = denominations[i]
return sum(count_combs(left-x*cur, i+1, comb[:], (x,cur)) for x in range(0, int(left/cur)+1))
count_combs(cents, 0, [], None)
def countChange(money: Int, coins: List[Int]): Int = {
def loop(money: Int, lcoins: List[Int], count: Int): Int = {
// if there are no more coins or if we run out of money ... return 0
if ( lcoins.isEmpty || money < 0) 0
else{
if (money == 0 ) count + 1
/* if the recursive subtraction leads to 0 money left - a prefect division hence return count +1 */
else
/* keep iterating ... sum over money and the rest of the coins and money - the first item and the full set of coins left*/
loop(money, lcoins.tail,count) + loop(money - lcoins.head,lcoins, count)
}
}
val x = loop(money, coins, 0)
Console println x
x
}
Scala function :
def countChange(money: Int, coins: List[Int]): Int = {
def loop(money: Int, lcoins: List[Int], count: Int): Int = {
// if there are no more coins or if we run out of money ... return 0
if ( lcoins.isEmpty || money < 0) 0
else{
if (money == 0 ) count + 1
/* if the recursive subtraction leads to 0 money left - a prefect division hence return count +1 */
else
/* keep iterating ... sum over money and the rest of the coins and money - the first item and the full set of coins left*/
loop(money, lcoins.tail,count) + loop(money - lcoins.head,lcoins, count)
}
}
val x = loop(money, coins, 0)
Console println x
x
}
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
if (argc != 2)
{
printf("usage: change amount-in-cents\n");
return 1;
}
int total = atoi(argv[1]);
printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);
int combos = 0;
for (int q = 0; q <= total / 25; q++)
{
int total_less_q = total - q * 25;
for (int d = 0; d <= total_less_q / 10; d++)
{
int total_less_q_d = total_less_q - d * 10;
for (int n = 0; n <= total_less_q_d / 5; n++)
{
int p = total_less_q_d - n * 5;
printf("%d\t%d\t%d\t%d\n", q, d, n, p);
combos++;
}
}
}
printf("%d combinations\n", combos);
return 0;
}
但我对计算组合数量的子问题非常感兴趣。 我怀疑它有一个封闭式方程。
Here's some absolutely straightforward C++ code to solve the problem which did ask for all the combinations to be shown.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
if (argc != 2)
{
printf("usage: change amount-in-cents\n");
return 1;
}
int total = atoi(argv[1]);
printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);
int combos = 0;
for (int q = 0; q <= total / 25; q++)
{
int total_less_q = total - q * 25;
for (int d = 0; d <= total_less_q / 10; d++)
{
int total_less_q_d = total_less_q - d * 10;
for (int n = 0; n <= total_less_q_d / 5; n++)
{
int p = total_less_q_d - n * 5;
printf("%d\t%d\t%d\t%d\n", q, d, n, p);
combos++;
}
}
}
printf("%d combinations\n", combos);
return 0;
}
But I'm quite intrigued about the sub problem of just calculating the number of combinations. I suspect there's a closed-form equation for it.
/* Q: Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars),
find the number of combinations of coins that make up the dollar value.
There are only penny, nickel, dime, and quarter.
(quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */
/* A:
Reference: http://andrew.neitsch.ca/publications/m496pres1.nb.pdf
f(n, k): number of ways of making change for n cents, using only the first
k+1 types of coins.
+- 0, n < 0 || k < 0
f(n, k) = |- 1, n == 0
+- f(n, k-1) + f(n-C[k], k), else
*/
#include <iostream>
#include <vector>
using namespace std;
int C[] = {1, 5, 10, 25};
// Recursive: very slow, O(2^n)
int f(int n, int k)
{
if (n < 0 || k < 0)
return 0;
if (n == 0)
return 1;
return f(n, k-1) + f(n-C[k], k);
}
// Non-recursive: fast, but still O(nk)
int f_NonRec(int n, int k)
{
vector<vector<int> > table(n+1, vector<int>(k+1, 1));
for (int i = 0; i <= n; ++i)
{
for (int j = 0; j <= k; ++j)
{
if (i < 0 || j < 0) // Impossible, for illustration purpose
{
table[i][j] = 0;
}
else if (i == 0 || j == 0) // Very Important
{
table[i][j] = 1;
}
else
{
// The recursion. Be careful with the vector boundary
table[i][j] = table[i][j-1] +
(i < C[j] ? 0 : table[i-C[j]][j]);
}
}
}
return table[n][k];
}
int main()
{
cout << f(100, 3) << ", " << f_NonRec(100, 3) << endl;
cout << f(200, 3) << ", " << f_NonRec(200, 3) << endl;
cout << f(1000, 3) << ", " << f_NonRec(1000, 3) << endl;
return 0;
}
The sub problem is a typical Dynamic Programming problem.
/* Q: Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars),
find the number of combinations of coins that make up the dollar value.
There are only penny, nickel, dime, and quarter.
(quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */
/* A:
Reference: http://andrew.neitsch.ca/publications/m496pres1.nb.pdf
f(n, k): number of ways of making change for n cents, using only the first
k+1 types of coins.
+- 0, n < 0 || k < 0
f(n, k) = |- 1, n == 0
+- f(n, k-1) + f(n-C[k], k), else
*/
#include <iostream>
#include <vector>
using namespace std;
int C[] = {1, 5, 10, 25};
// Recursive: very slow, O(2^n)
int f(int n, int k)
{
if (n < 0 || k < 0)
return 0;
if (n == 0)
return 1;
return f(n, k-1) + f(n-C[k], k);
}
// Non-recursive: fast, but still O(nk)
int f_NonRec(int n, int k)
{
vector<vector<int> > table(n+1, vector<int>(k+1, 1));
for (int i = 0; i <= n; ++i)
{
for (int j = 0; j <= k; ++j)
{
if (i < 0 || j < 0) // Impossible, for illustration purpose
{
table[i][j] = 0;
}
else if (i == 0 || j == 0) // Very Important
{
table[i][j] = 1;
}
else
{
// The recursion. Be careful with the vector boundary
table[i][j] = table[i][j-1] +
(i < C[j] ? 0 : table[i-C[j]][j]);
}
}
}
return table[n][k];
}
int main()
{
cout << f(100, 3) << ", " << f_NonRec(100, 3) << endl;
cout << f(200, 3) << ", " << f_NonRec(200, 3) << endl;
cout << f(1000, 3) << ", " << f_NonRec(1000, 3) << endl;
return 0;
}
public class RepresentCents {
public static int sum(int n) {
int count = 0;
for (int i = 0; i <= n / 25; i++) {
for (int j = 0; j <= n / 10; j++) {
for (int k = 0; k <= n / 5; k++) {
for (int l = 0; l <= n; l++) {
int v = i * 25 + j * 10 + k * 5 + l;
if (v == n) {
count++;
} else if (v > n) {
break;
}
}
}
}
}
return count;
}
public static void main(String[] args) {
System.out.println(sum(100));
}
}
The code is using Java to solve this problem and it also works... This method may not be a good idea because of too many loops, but it's really a straight forward way.
public class RepresentCents {
public static int sum(int n) {
int count = 0;
for (int i = 0; i <= n / 25; i++) {
for (int j = 0; j <= n / 10; j++) {
for (int k = 0; k <= n / 5; k++) {
for (int l = 0; l <= n; l++) {
int v = i * 25 + j * 10 + k * 5 + l;
if (v == n) {
count++;
} else if (v > n) {
break;
}
}
}
}
}
return count;
}
public static void main(String[] args) {
System.out.println(sum(100));
}
}
$denoms = [1,5,10,25]
def all_combs(sum,last)
return 1 if sum == 0
return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom|
total+all_combs(sum-denom,denom)}
end
这会运行得很慢，因为它不会被记住，但你明白了。
semi-hack to get around the unique combination problem - force descending order:
$denoms = [1,5,10,25]
def all_combs(sum,last)
return 1 if sum == 0
return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom|
total+all_combs(sum-denom,denom)}
end
This will run slow since it won't be memoized, but you get the idea.
def crossprod (list1, list2):
output = 0
for i in range(0,len(list1)):
output += list1[i]*list2[i]
return output
def breakit(target, coins):
coinslimit = [(target / coins[i]) for i in range(0,len(coins))]
count = 0
temp = []
for i in range(0,len(coins)):
temp.append([j for j in range(0,coinslimit[i]+1)])
r=[[]]
for x in temp:
t = []
for y in x:
for i in r:
t.append(i+[y])
r = t
for targets in r:
if crossprod(targets, coins) == target:
print targets
count +=1
return count
if __name__ == "__main__":
coins = [25,10,5,1]
target = 78
print breakit(target, coins)
This is my answer in Python. It does not use recursion:
def crossprod (list1, list2):
output = 0
for i in range(0,len(list1)):
output += list1[i]*list2[i]
return output
def breakit(target, coins):
coinslimit = [(target / coins[i]) for i in range(0,len(coins))]
count = 0
temp = []
for i in range(0,len(coins)):
temp.append([j for j in range(0,coinslimit[i]+1)])
r=[[]]
for x in temp:
t = []
for y in x:
for i in r:
t.append(i+[y])
r = t
for targets in r:
if crossprod(targets, coins) == target:
print targets
count +=1
return count
if __name__ == "__main__":
coins = [25,10,5,1]
target = 78
print breakit(target, coins)
Both: iterate through all denominations from high to low, take one of denomination, subtract from requried total, then recurse on remainder (constraining avilable denominations to be equal or lower to current iteration value.)
If the currency system allows it, a simple greedy algorithm that takes as many of each coin as possible, starting with the highest value currency.
Otherwise, dynamic programming is required to find an optimal solution quickly since this problem is essentially the knapsack problem.
For example, if a currency system has the coins: {13, 8, 1}, the greedy solution would make change for 24 as {13, 8, 1, 1, 1}, but the true optimal solution is {8, 8, 8}
Edit: I thought we were making change optimally, not listing all the ways to make change for a dollar. My recent interview asked how to make change so I jumped ahead before finishing to read the question.
function denomination(coins, original_amount){
var original_amount = original_amount;
var original_best = [ ];
for(var i=0;i<coins.length; i++){
var amount = original_amount;
var best = [ ];
var tempBest = [ ]
while(coins[i]<=amount){
amount = amount - coins[i];
best.push(coins[i]);
}
if(amount>0 && coins.length>1){
tempBest = denomination(coins.slice(0,i).concat(coins.slice(i+1,coins.length)), amount);
//best = best.concat(denomination(coins.splice(i,1), amount));
}
if(tempBest.length!=0 || (best.length!=0 && amount==0)){
best = best.concat(tempBest);
if(original_best.length==0 ){
original_best = best
}else if(original_best.length > best.length ){
original_best = best;
}
}
}
return original_best;
}
denomination( [1,10,3,9] , 19 );
这是一个 javascript 解决方案并使用递归。
I know this is a very old question. I was searching through the proper answer and couldn't find anything that is simple and satisfactory. Took me some time but was able to jot down something.
function denomination(coins, original_amount){
var original_amount = original_amount;
var original_best = [ ];
for(var i=0;i<coins.length; i++){
var amount = original_amount;
var best = [ ];
var tempBest = [ ]
while(coins[i]<=amount){
amount = amount - coins[i];
best.push(coins[i]);
}
if(amount>0 && coins.length>1){
tempBest = denomination(coins.slice(0,i).concat(coins.slice(i+1,coins.length)), amount);
//best = best.concat(denomination(coins.splice(i,1), amount));
}
if(tempBest.length!=0 || (best.length!=0 && amount==0)){
best = best.concat(tempBest);
if(original_best.length==0 ){
original_best = best
}else if(original_best.length > best.length ){
original_best = best;
}
}
}
return original_best;
}
denomination( [1,10,3,9] , 19 );
def countChange(money: Int, coins: List[Int]): Int = {
money match {
case 0 => 1
case x if x < 0 => 0
case x if x >= 1 && coins.isEmpty => 0
case _ => countChange(money, coins.tail) + countChange(money - coins.head, coins)
}
}
In Scala Programming language i would do it like this:
def countChange(money: Int, coins: List[Int]): Int = {
money match {
case 0 => 1
case x if x < 0 => 0
case x if x >= 1 && coins.isEmpty => 0
case _ => countChange(money, coins.tail) + countChange(money - coins.head, coins)
}
}
This is a simple recursive algorithm that takes a bill, then takes a smaller bill recursively until it reaches the sum, it then takes another bill of same denomination, and recurses again. See sample output below for illustration.
var bills = new int[] { 100, 50, 20, 10, 5, 1 };
void PrintAllWaysToMakeChange(int sumSoFar, int minBill, string changeSoFar)
{
for (int i = minBill; i < bills.Length; i++)
{
var change = changeSoFar;
var sum = sumSoFar;
while (sum > 0)
{
if (!string.IsNullOrEmpty(change)) change += " + ";
change += bills[i];
sum -= bills[i];
if (sum > 0)
{
PrintAllWaysToMakeChange(sum, i + 1, change);
}
}
if (sum == 0)
{
Console.WriteLine(change);
}
}
}
PrintAllWaysToMakeChange(15, 0, "");
// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"),
("dime", "dimes"),
("nickel", "nickels"),
("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList
// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
((List(amount) /: coinValues) {(list, coinValue) =>
val currentAmount = list.head
val numberOfCoins = currentAmount / coinValue
val remainingAmount = currentAmount % coinValue
remainingAmount :: numberOfCoins :: list.tail
}).tail.reverse.toArray
// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int],
quarters: Int,
dimes: Int,
nickels: Int,
pennies: Int): Array[Int] =
Array(base(quarter) + quarters,
base(dime) + dimes,
base(nickel) + nickels,
base(penny) + pennies)
// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
adjust(base, -1, +2, +1, 0)
// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
adjust(base, 0, -1, +2, 0)
// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
adjust(base, 0, 0, -1, +5)
// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
dime -> decreaseDime _,
nickel -> decreaseNickel _)
// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) =
(List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
decrease(whichCoin)(list.head) :: list
}
// Generate a pretty string
def coinsString(base: Array[Int]) = (
base
zip coinNames // join with names
map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
map (t => t._1 + " " + t._2)
mkString " "
)
// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
val base = leastCoins(amount)
val allQuarters = coinSpan(base, quarter)
val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
allNickels map coinsString mkString "\n"
}
Duh, I feel stupid right now. Below there is an overly complicated solution, which I'll preserve because it is a solution, after all. A simple solution would be this:
Here is the other solution. This solution is based on the observation that each coin is a multiple of the others, so they can be represented in terms of them.
// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"),
("dime", "dimes"),
("nickel", "nickels"),
("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList
// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
((List(amount) /: coinValues) {(list, coinValue) =>
val currentAmount = list.head
val numberOfCoins = currentAmount / coinValue
val remainingAmount = currentAmount % coinValue
remainingAmount :: numberOfCoins :: list.tail
}).tail.reverse.toArray
// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int],
quarters: Int,
dimes: Int,
nickels: Int,
pennies: Int): Array[Int] =
Array(base(quarter) + quarters,
base(dime) + dimes,
base(nickel) + nickels,
base(penny) + pennies)
// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
adjust(base, -1, +2, +1, 0)
// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
adjust(base, 0, -1, +2, 0)
// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
adjust(base, 0, 0, -1, +5)
// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
dime -> decreaseDime _,
nickel -> decreaseNickel _)
// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) =
(List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
decrease(whichCoin)(list.head) :: list
}
// Generate a pretty string
def coinsString(base: Array[Int]) = (
base
zip coinNames // join with names
map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
map (t => t._1 + " " + t._2)
mkString " "
)
// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
val base = leastCoins(amount)
val allQuarters = coinSpan(base, quarter)
val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
allNickels map coinsString mkString "\n"
}
This blog entry of mine solves this knapsack like problem for the figures from an XKCD comic. A simple change to the items dict and the exactcost value will yield all solutions for your problem too.
If the problem were to find the change that used the least cost, then a naive greedy algorithm that used as much of the highest value coin might well fail for some combinations of coins and target amount. For example if there are coins with values 1, 3, and 4; and the target amount is 6 then the greedy algorithm might suggest three coins of value 4, 1, and 1 when it is easy to see that you could use two coins each of value 3.
public class Coins {
static int ac = 421;
static int bc = 311;
static int cc = 11;
static int target = 4000;
public static void main(String[] args) {
method2();
}
public static void method2(){
//running time n^2
int da = target/ac;
int db = target/bc;
for(int i=0;i<=da;i++){
for(int j=0;j<=db;j++){
int rem = target-(i*ac+j*bc);
if(rem < 0){
break;
}else{
if(rem%cc==0){
System.out.format("\n%d, %d, %d ---- %d + %d + %d = %d \n", i, j, rem/cc, i*ac, j*bc, (rem/cc)*cc, target);
}
}
}
}
}
}
public class Coins {
static int ac = 421;
static int bc = 311;
static int cc = 11;
static int target = 4000;
public static void main(String[] args) {
method2();
}
public static void method2(){
//running time n^2
int da = target/ac;
int db = target/bc;
for(int i=0;i<=da;i++){
for(int j=0;j<=db;j++){
int rem = target-(i*ac+j*bc);
if(rem < 0){
break;
}else{
if(rem%cc==0){
System.out.format("\n%d, %d, %d ---- %d + %d + %d = %d \n", i, j, rem/cc, i*ac, j*bc, (rem/cc)*cc, target);
}
}
}
}
}
}
我在 O'reily 的《Python For Data Analysis》一书中找到了这段简洁的代码。 它使用惰性实现和 int 比较，我认为可以使用小数将其修改为其他面额。 让我知道它对您有何作用！
def make_change(amount, coins=[1, 5, 10, 25], hand=None):
hand = [] if hand is None else hand
if amount == 0:
yield hand
for coin in coins:
# ensures we don't give too much change, and combinations are unique
if coin > amount or (len(hand) > 0 and hand[-1] < coin):
continue
for result in make_change(amount - coin, coins=coins,
hand=hand + [coin]):
yield result
I found this neat piece of code in the book "Python For Data Analysis" by O'reily. It uses lazy implementation and int comparison and i presume it can be modified for other denominations using decimals. Let me know how it works for you!
def make_change(amount, coins=[1, 5, 10, 25], hand=None):
hand = [] if hand is None else hand
if amount == 0:
yield hand
for coin in coins:
# ensures we don't give too much change, and combinations are unique
if coin > amount or (len(hand) > 0 and hand[-1] < coin):
continue
for result in make_change(amount - coin, coins=coins,
hand=hand + [coin]):
yield result
/*
* make a list of all distinct sets of coins of from the set of coins to
* sum up to the given target amount.
* Here the input set of coins is assumed yo be {1, 2, 4}, this set MUST
* have the coins sorted in ascending order.
* Outline of the algorithm:
*
* Keep track of what the current coin is, say ccn; current number of coins
* in the partial solution, say k; current sum, say sum, obtained by adding
* ccn; sum sofar, say accsum:
* 1) Use ccn as long as it can be added without exceeding the target
* a) if current sum equals target, add cc to solution coin set, increase
* coin coin in the solution by 1, and print it and return
* b) if current sum exceeds target, ccn can't be in the solution, so
* return
* c) if neither of the above, add current coin to partial solution,
* increase k by 1 (number of coins in partial solution), and recuse
* 2) When current denomination can no longer be used, start using the
* next higher denomination coins, just like in (1)
* 3) When all denominations have been used, we are done
*/
#include <iostream>
#include <cstdlib>
using namespace std;
// int num_calls = 0;
// int num_ways = 0;
void print(const int coins[], int n);
void combine_coins(
const int denoms[], // coins sorted in ascending order
int n, // number of denominations
int target, // target sum
int accsum, // accumulated sum
int coins[], // solution set, MUST equal
// target / lowest denom coin
int k // number of coins in coins[]
)
{
int ccn; // current coin
int sum; // current sum
// ++num_calls;
for (int i = 0; i < n; ++i) {
/*
* skip coins of lesser denomination: This is to be efficient
* and also avoid generating duplicate sequences. What we need
* is combinations and without this check we will generate
* permutations.
*/
if (k > 0 && denoms[i] < coins[k - 1])
continue; // skip coins of lesser denomination
ccn = denoms[i];
if ((sum = accsum + ccn) > target)
return; // no point trying higher denominations now
if (sum == target) {
// found yet another solution
coins[k] = ccn;
print(coins, k + 1);
// ++num_ways;
return;
}
coins[k] = ccn;
combine_coins(denoms, n, target, sum, coins, k + 1);
}
}
void print(const int coins[], int n)
{
int s = 0;
for (int i = 0; i < n; ++i) {
cout << coins[i] << " ";
s += coins[i];
}
cout << "\t = \t" << s << "\n";
}
int main(int argc, const char *argv[])
{
int denoms[] = {1, 2, 4};
int dsize = sizeof(denoms) / sizeof(denoms[0]);
int target;
if (argv[1])
target = atoi(argv[1]);
else
target = 8;
int *coins = new int[target];
combine_coins(denoms, dsize, target, 0, coins, 0);
// cout << "num calls = " << num_calls << ", num ways = " << num_ways << "\n";
return 0;
}
/*
* make a list of all distinct sets of coins of from the set of coins to
* sum up to the given target amount.
* Here the input set of coins is assumed yo be {1, 2, 4}, this set MUST
* have the coins sorted in ascending order.
* Outline of the algorithm:
*
* Keep track of what the current coin is, say ccn; current number of coins
* in the partial solution, say k; current sum, say sum, obtained by adding
* ccn; sum sofar, say accsum:
* 1) Use ccn as long as it can be added without exceeding the target
* a) if current sum equals target, add cc to solution coin set, increase
* coin coin in the solution by 1, and print it and return
* b) if current sum exceeds target, ccn can't be in the solution, so
* return
* c) if neither of the above, add current coin to partial solution,
* increase k by 1 (number of coins in partial solution), and recuse
* 2) When current denomination can no longer be used, start using the
* next higher denomination coins, just like in (1)
* 3) When all denominations have been used, we are done
*/
#include <iostream>
#include <cstdlib>
using namespace std;
// int num_calls = 0;
// int num_ways = 0;
void print(const int coins[], int n);
void combine_coins(
const int denoms[], // coins sorted in ascending order
int n, // number of denominations
int target, // target sum
int accsum, // accumulated sum
int coins[], // solution set, MUST equal
// target / lowest denom coin
int k // number of coins in coins[]
)
{
int ccn; // current coin
int sum; // current sum
// ++num_calls;
for (int i = 0; i < n; ++i) {
/*
* skip coins of lesser denomination: This is to be efficient
* and also avoid generating duplicate sequences. What we need
* is combinations and without this check we will generate
* permutations.
*/
if (k > 0 && denoms[i] < coins[k - 1])
continue; // skip coins of lesser denomination
ccn = denoms[i];
if ((sum = accsum + ccn) > target)
return; // no point trying higher denominations now
if (sum == target) {
// found yet another solution
coins[k] = ccn;
print(coins, k + 1);
// ++num_ways;
return;
}
coins[k] = ccn;
combine_coins(denoms, n, target, sum, coins, k + 1);
}
}
void print(const int coins[], int n)
{
int s = 0;
for (int i = 0; i < n; ++i) {
cout << coins[i] << " ";
s += coins[i];
}
cout << "\t = \t" << s << "\n";
}
int main(int argc, const char *argv[])
{
int denoms[] = {1, 2, 4};
int dsize = sizeof(denoms) / sizeof(denoms[0]);
int target;
if (argv[1])
target = atoi(argv[1]);
else
target = 8;
int *coins = new int[target];
combine_coins(denoms, dsize, target, 0, coins, 0);
// cout << "num calls = " << num_calls << ", num ways = " << num_ways << "\n";
return 0;
}
def cntMoney(num):
mSz = len(money)
numbers = [[0]*(1+num) for _ in range(mSz)]
for mI in range(mSz): numbers[mI][0] = 1
for mI,m in enumerate(money):
for i in range(1,num+1):
numbers[mI][i] = numbers[mI][i-m] if i >= m else 0
if mI != 0: numbers[mI][i] += numbers[mI-1][i]
print('m,numbers',m,numbers[mI])
return numbers[mSz-1][num]
money = [1,5,10,25]
num = 12
print('money,combinations',num,cntMoney(num))
output:
('m,numbers', 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
('m,numbers', 5, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3])
('m,numbers', 10, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('m,numbers', 25, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('money,combinations', 12, 4)
Below is a python program to find all combinations of money. This is a dynamic programming solution with order(n) time. Money is 1,5,10,25
We traverse from row money 1 to row money 25 (4 rows). Row money 1 contains the count if we only consider money 1 in calculating the number of combinations. Row money 5 produces each column by taking the count in row money r for the same final money plus the previous 5 count in its own row (current position minus 5). Row money 10 uses row money 5, which contains counts for both 1,5 and adds in the previous 10 count (current position minus 10). Row money 25 uses row money 10, which contains counts for row money 1,5,10 plus the previous 25 count.
For example, numbers[1][12] = numbers[0][12] + numbers[1][7] (7 = 12-5) which results in 3 = 1 + 2; numbers[3][12] = numbers[2][12] + numbers[3][9] (-13 = 12-25) which results in 4 = 0 + 4, since -13 is less than 0.
def cntMoney(num):
mSz = len(money)
numbers = [[0]*(1+num) for _ in range(mSz)]
for mI in range(mSz): numbers[mI][0] = 1
for mI,m in enumerate(money):
for i in range(1,num+1):
numbers[mI][i] = numbers[mI][i-m] if i >= m else 0
if mI != 0: numbers[mI][i] += numbers[mI-1][i]
print('m,numbers',m,numbers[mI])
return numbers[mSz-1][num]
money = [1,5,10,25]
num = 12
print('money,combinations',num,cntMoney(num))
output:
('m,numbers', 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
('m,numbers', 5, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3])
('m,numbers', 10, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('m,numbers', 25, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('money,combinations', 12, 4)
5 - 5(i) times 1 = 0
if(sum = 0)
print i times 1
5 - 4(i) times 1 = 1
5 - 3 times 1 = 2
2 - 1(j) times 2 = 0
if(sum = 0)
print i times 1 and j times 2
and so on......
如果每个循环中的剩余总和小于面额，即 如果剩余和1小于2，则中断循环
完整代码如下
如有错误请纠正
public class CoinCombinbationSimple {
public static void main(String[] args) {
int sum = 100000;
printCombination(sum);
}
static void printCombination(int sum) {
for (int i = sum; i >= 0; i--) {
int sumCopy1 = sum - i * 1;
if (sumCopy1 == 0) {
System.out.println(i + " 1 coins");
}
for (int j = sumCopy1 / 2; j >= 0; j--) {
int sumCopy2 = sumCopy1;
if (sumCopy2 < 2) {
break;
}
sumCopy2 = sumCopy1 - 2 * j;
if (sumCopy2 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins ");
}
for (int k = sumCopy2 / 5; k >= 0; k--) {
int sumCopy3 = sumCopy2;
if (sumCopy2 < 5) {
break;
}
sumCopy3 = sumCopy2 - 5 * k;
if (sumCopy3 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins "
+ k + " 5 coins");
}
}
}
}
}
}
The below java solution which will print the different combinations as well. Easy to understand. Idea is
for sum 5
The solution is
5 - 5(i) times 1 = 0
if(sum = 0)
print i times 1
5 - 4(i) times 1 = 1
5 - 3 times 1 = 2
2 - 1(j) times 2 = 0
if(sum = 0)
print i times 1 and j times 2
and so on......
If the remaining sum in each loop is lesser than the denomination ie if remaining sum 1 is lesser than 2, then just break the loop
The complete code below
Please correct me in case of any mistakes
public class CoinCombinbationSimple {
public static void main(String[] args) {
int sum = 100000;
printCombination(sum);
}
static void printCombination(int sum) {
for (int i = sum; i >= 0; i--) {
int sumCopy1 = sum - i * 1;
if (sumCopy1 == 0) {
System.out.println(i + " 1 coins");
}
for (int j = sumCopy1 / 2; j >= 0; j--) {
int sumCopy2 = sumCopy1;
if (sumCopy2 < 2) {
break;
}
sumCopy2 = sumCopy1 - 2 * j;
if (sumCopy2 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins ");
}
for (int k = sumCopy2 / 5; k >= 0; k--) {
int sumCopy3 = sumCopy2;
if (sumCopy2 < 5) {
break;
}
sumCopy3 = sumCopy2 - 5 * k;
if (sumCopy3 == 0) {
System.out.println(i + " 1 coins " + j + " 2 coins "
+ k + " 5 coins");
}
}
}
}
}
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评论（30）
我很久以前就研究过这个问题，你可以阅读我的关于它的小文章。 这是Mathematica 源代码。
通过使用生成函数，您可以获得该问题的封闭形式常量时间解。 Graham、Knuth 和 Patashnik 的具体数学就是针对此问题的书，并且包含对该问题的相当广泛的讨论。 本质上，您定义一个多项式，其中第 n 个系数是 n 美元找零的方式数量。
本文的第 4-5 页展示了如何使用 Mathematica（或任何其他方便的计算机代数系统）通过三行代码在几秒钟内计算出 10^10^6 美元的答案。
（这已经是很久以前的事了，在 75Mhz 奔腾上这只是几秒钟......）
I looked into this once a long time ago, and you can read my little write-up on it. Here’s the Mathematica source.
By using generating functions, you can get a closed-form constant-time solution to the problem. Graham, Knuth, and Patashnik’s Concrete Mathematics is the book for this, and contains a fairly extensive discussion of the problem. Essentially you define a polynomial where the nth coefficient is the number of ways of making change for n dollars.
Pages 4-5 of the writeup show how you can use Mathematica (or any other convenient computer algebra system) to compute the answer for 10^10^6 dollars in a couple seconds in three lines of code.
(And this was long enough ago that that’s a couple of seconds on a 75Mhz Pentium...)
注意：这仅显示方式数量。
斯卡拉函数：
Note: This only shows the number of ways.
Scala function:
我倾向于递归解决方案。 您有一些面额列表，如果最小的面额可以均匀地划分任何剩余的货币金额，那么这应该可以正常工作。
基本上，您从最大面额转移到最小面额。
递归地，
如果只剩下 1 种面额，则只有一种方法可以填充总数。 您可以使用当前面额的 0 到 k 个副本，使得 k * 当前面额 <= 总数。
这是我针对你所提出问题的 python 版本，售价 200 美分。 我有1463种方法。 此版本打印所有组合和最终计数总数。
I would favor a recursive solution. You have some list of denominations, if the smallest one can evenly divide any remaining currency amount, this should work fine.
Basically, you move from largest to smallest denominations.
Recursively,
If there is only 1 denomination left, there is only one way to fill the total. You can use 0 to k copies of your current denomination such that k * cur denomination <= total.
Here's my python version of your stated problem, for 200 cents. I get 1463 ways. This version prints all the combinations and the final count total.
斯卡拉函数：
Scala function :
这里有一些绝对简单的 C++ 代码来解决要求显示所有组合的问题。
但我对计算组合数量的子问题非常感兴趣。 我怀疑它有一个封闭式方程。
Here's some absolutely straightforward C++ code to solve the problem which did ask for all the combinations to be shown.
But I'm quite intrigued about the sub problem of just calculating the number of combinations. I suspect there's a closed-form equation for it.
该子问题是一个典型的动态规划问题。
The sub problem is a typical Dynamic Programming problem.
该代码使用 Java 来解决这个问题，并且它也有效...由于循环太多，这种方法可能不是一个好主意，但它确实是一种直接的方法。
The code is using Java to solve this problem and it also works... This method may not be a good idea because of too many loops, but it's really a straight forward way.
这是一个非常老的问题，但我在 java 中提出了一个递归解决方案，它看起来比所有其他解决方案都要小，所以这里是 -
改进：
This is a really old question, but I came up with a recursive solution in java that seemed smaller than all the others, so here goes -
Improvements:
设 C(i,J) 是使用集合 J 中的值制作 i 美分的组合集合。
您可以将 C 定义为：
（first(J)以确定性方式获取集合中的一个元素）
事实证明，这是一个相当递归的函数......并且如果您使用记忆化，则效率相当高；）
Let C(i,J) the set of combinations of making i cents using the values in the set J.
You can define C as that:
(first(J) takes in a deterministic way an element of a set)
It turns out a pretty recursive function... and reasonably efficient if you use memoization ;)
半黑客解决独特的组合问题 - 强制降序：
$denoms = [1,5,10,25] def all_combs(sum,last) return 1 if sum == 0 return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom| total+all_combs(sum-denom,denom)} end这会运行得很慢，因为它不会被记住，但你明白了。
semi-hack to get around the unique combination problem - force descending order:
$denoms = [1,5,10,25] def all_combs(sum,last) return 1 if sum == 0 return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom| total+all_combs(sum-denom,denom)} endThis will run slow since it won't be memoized, but you get the idea.
这是我用 Python 给出的答案。 它不使用递归：
示例输出
This is my answer in Python. It does not use recursion:
Example output
两者：从高到低迭代所有面额，取其中一种面额，从所需总数中减去，然后对余数进行递归（将可用面额限制为等于或低于当前迭代值。）
Both: iterate through all denominations from high to low, take one of denomination, subtract from requried total, then recurse on remainder (constraining avilable denominations to be equal or lower to current iteration value.)
如果货币系统允许，可以采用简单的贪婪算法，尽可能多地获取每种硬币，从最高价值的货币开始。
否则，需要动态规划来快速找到最优解，因为这个问题本质上是背包问题 。
例如，如果货币系统有硬币：
{13, 8, 1}，则贪心解决方案会将 24 找零为{13, 8, 1, 1, 1}< /code>，但真正的最佳解决方案是{8, 8, 8}编辑：我认为我们正在以最佳方式进行更改，而不是列出以一美元进行更改的所有方法。 我最近的采访询问了如何做出改变，所以我在读完问题之前就跳到了前面。
If the currency system allows it, a simple greedy algorithm that takes as many of each coin as possible, starting with the highest value currency.
Otherwise, dynamic programming is required to find an optimal solution quickly since this problem is essentially the knapsack problem.
For example, if a currency system has the coins:
{13, 8, 1}, the greedy solution would make change for 24 as{13, 8, 1, 1, 1}, but the true optimal solution is{8, 8, 8}Edit: I thought we were making change optimally, not listing all the ways to make change for a dollar. My recent interview asked how to make change so I jumped ahead before finishing to read the question.
我知道这是一个非常古老的问题。 我正在寻找正确的答案，但找不到任何简单且令人满意的答案。 我花了一些时间，但还是记下了一些东西。
这是一个 javascript 解决方案并使用递归。
I know this is a very old question. I was searching through the proper answer and couldn't find anything that is simple and satisfactory. Took me some time but was able to jot down something.
This is a javascript solution and uses recursion.
在 Scala 编程语言中我会这样做：
In Scala Programming language i would do it like this:
这是一个简单的递归算法，先取出一张钞票，然后递归取出一张较小的钞票，直到达到总和，然后取出另一张相同面额的钞票，再次递归。 请参阅下面的示例输出以进行说明。
打印以下内容：
This is a simple recursive algorithm that takes a bill, then takes a smaller bill recursively until it reaches the sum, it then takes another bill of same denomination, and recurses again. See sample output below for illustration.
Prints the following:
呃，我现在感觉自己很蠢。 下面有一个过于复杂的解决方案，我将保留它，因为它毕竟是一个解决方案。 一个简单的解决方案是这样的：
这是另一个解决方案。 该解决方案基于以下观察：每个硬币都是其他硬币的倍数，因此可以用它们来表示。
因此，以 37 个硬币为例：
Duh, I feel stupid right now. Below there is an overly complicated solution, which I'll preserve because it is a solution, after all. A simple solution would be this:
Here is the other solution. This solution is based on the observation that each coin is a multiple of the others, so they can be represented in terms of them.
So, for 37 coins, for example:
我的这篇博客文章解决了这个类似背包的问题来自XKCD 漫画的人物。 对
items字典和exactcost值进行简单更改也将为您的问题提供所有解决方案。如果问题是找到使用最少成本的找零，那么使用尽可能多的最高价值硬币的天真贪婪算法很可能会因硬币和目标金额的某些组合而失败。 例如，如果有值为 1、3 和 4 的硬币； 如果目标金额为 6，则贪心算法可能会建议使用价值 4、1 和 1 的三枚硬币，因为很容易看出您可以使用价值 3.
This blog entry of mine solves this knapsack like problem for the figures from an XKCD comic. A simple change to the
itemsdict and theexactcostvalue will yield all solutions for your problem too.If the problem were to find the change that used the least cost, then a naive greedy algorithm that used as much of the highest value coin might well fail for some combinations of coins and target amount. For example if there are coins with values 1, 3, and 4; and the target amount is 6 then the greedy algorithm might suggest three coins of value 4, 1, and 1 when it is easy to see that you could use two coins each of value 3.
我在 O'reily 的《Python For Data Analysis》一书中找到了这段简洁的代码。 它使用惰性实现和 int 比较，我认为可以使用小数将其修改为其他面额。 让我知道它对您有何作用！
I found this neat piece of code in the book "Python For Data Analysis" by O'reily. It uses lazy implementation and int comparison and i presume it can be modified for other denominations using decimals. Let me know how it works for you!
这是子涵答案的改进。 当面额仅为 1 美分时，就会出现大量不必要的循环。
它是直观且非递归的。
This is the improvement of Zihan's answer. The great deal of unnecessary loops comes when the denomination is just 1 cent.
It's intuitive and non-recursive.
简单的java解决方案：
Straightforward java solution:
这里有很多变体，但找不到任何地方的组合数量的 PHP 解决方案，所以我将添加一个。
Lots of variations here but couldn't find a PHP solution for the number of combinations anywhere so I'll add one in.
这是一个 C# 函数：
像这样使用它：
它打印：
Here's a C# function:
Use it like this:
It prints:
下面是一个查找所有货币组合的 python 程序。 这是一个具有 order(n) 时间的动态规划解决方案。
Money 是 1,5,10,25
我们从 Money 1 行遍历到 Money 25 行（4 行）。 如果我们只考虑其中的钱 1，则行钱 1 包含计数
计算组合的数量。 行钱 5 通过获取行钱 r 中的计数来生成每一列
相同的最终金额加上其所在行中的前 5 个计数（当前位置减 5）。 排钱10使用排钱5，
其中包含 1,5 的计数并添加前 10 个计数（当前位置减 10）。 排款25用排
Money 10，其中包含第 1、5、10 行 Money 的计数以及前 25 个计数。
例如，数字[1][12] = 数字[0][12] + 数字[1][7] (7 = 12-5)，结果为 3 = 1 + 2； 数字[3][12] =
数字[2][12] + 数字[3][9] (-13 = 12-25) 结果为 4 = 0 + 4，因为 -13 小于 0。
Below is a python program to find all combinations of money. This is a dynamic programming solution with order(n) time.
Money is 1,5,10,25
We traverse from row money 1 to row money 25 (4 rows). Row money 1 contains the count if we only consider money 1 in
calculating the number of combinations. Row money 5 produces each column by taking the count in row money r for the
same final money plus the previous 5 count in its own row (current position minus 5). Row money 10 uses row money 5,
which contains counts for both 1,5 and adds in the previous 10 count (current position minus 10). Row money 25 uses row
money 10, which contains counts for row money 1,5,10 plus the previous 25 count.
For example, numbers[1][12] = numbers[0][12] + numbers[1][7] (7 = 12-5) which results in 3 = 1 + 2; numbers[3][12] =
numbers[2][12] + numbers[3][9] (-13 = 12-25) which results in 4 = 0 + 4, since -13 is less than 0.
Java解决方案
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Java solution
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下面的 java 解决方案也将打印不同的组合。 容易明白。 想法是
求和 5
解决方案是
如果每个循环中的剩余总和小于面额，即
如果剩余和1小于2，则中断循环
完整代码如下
如有错误请纠正
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The below java solution which will print the different combinations as well. Easy to understand. Idea is
for sum 5
The solution is
If the remaining sum in each loop is lesser than the denomination ie
if remaining sum 1 is lesser than 2, then just break the loop
The complete code below
Please correct me in case of any mistakes
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