coin-change

问题 The classic coin change problem is well described here: http://www.algorithmist.com/index.php/Coin_Change Here I want to not only know how many combinations there are, but also print out all of them. I'm using the same DP algorithm in that link in my implementation but instead of recording how many combinations in the DP table for DP[i][j] = count , I store the combinations in the table. So I'm using a 3D vector for this DP table. I tried to improve my implementation noticing that when

问题 The problem requires to count number of coin changes for a particular cost. For example, if I have coin values of 50, 20, 10, 5, 1 , I can form costs of: 5 => (5), (11111), which are 2 ways. 10 => (10), (5, 5), (5, 11111), (11111, 11111), which are 4 ways. Here is my function. It is returning wrong results begging from cost of 10 (returns 9 ways while the actual number of ways is only 4) int dp[10000]; int coins[] = { 50, 20, 10, 5, 1 }; int rec(int n) { if (n == 0) return 1; if (dp[n] != -1)

问题 Given a set of denominations and the sum required I have to find the minimum number of coins that make that sum and also the number of coins for each denominations Please help!! 回答1: The pseudo-code to determine minimum number of coins needed to make that sum is: Procedure coinChange(coins, total): n := coins.length dp[n][total + 1] for i from 0 to n dp[i][0] := 0 end for for i from 1 to (total + 1) dp[0][i] := i end for for i from 1 to n for j from 1 to (total + 1) if coins[i] > j //if the

问题 I have a small problem understanding the coin change problem in dynamic programming. Simply put, I have to change a sum using a minimum number of coins. I have n denominations of coins of values 1 = v1 < v2 < ... < vn, and we note M(j) the minimum number of coins required to make change for amount j. In the above formula I don't understand what M(j-vi) means. vi has to be the maximum value of the coins used in j-1? 回答1: You're making piles of coins for different values j, named M(j). The

问题 The code I have written solves the basic coin change problem using dynamic programming and gives the minimum number of coins required to make the change. But I want to store the count of each coin playing part in the minimum number. What I am trying to do is initializing an array count[] and just like hashing it increments the number of coin[j] whenever min is found, i.e count[coin[j]]++ . But this is not working the way I wanted because it adds the coin every time it finds min corresponding

问题 I am interested in solving a variant of the coin exchange problem. Recall the formal definition of the coin exchange problem: Given a value N , if we want to make change for N cents, and we have infinite supply of each of S = {S1, S2, .. , Sm} integral-valued coins, how many ways can we make the change? The order of coins doesn’t matter. For example, for N = 4 and S = {1,2,3} , there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6} ,

问题 I am exploring how a Dynamic Programming design approach relates to the underlying combinatorial properties of problems. For this, I am looking at the canonical instance of the coin change problem : Let S = [d_1, d_2, ..., d_m] and n > 0 be a requested amount. In how many ways can we add up to n using nothing but the elements in S ? If we follow a Dynamic Programming approach to design an algorithm for this problem that would allow for a solution with polynomial complexity, we would start by

问题 Given a target amount and a list of coin denominations, my code is supposed to find the fewest coins needed to reach the target amount. Examples: C(78, [1, 5, 10, 25, 50]) = 6 we can make 78 from 3x 25 + 3x 1 , so 6 coins are required C(48, [1, 7, 24, 42]) = 2 48 = 2x 24 , so 2 coins are sufficient C(35, [1, 3, 16, 30, 50]) = 3 we can make 35 from 2x 16 + 1x 3 , so 3 coins suffice I made the code with for loops, but how do I make it recursive? def C(i, coins, cdict = None): if cdict == None:

问题 I'm looking for the most efficient way to figure out a change amount (Quarters, dimes, nickels, and pennies) from a purchase amount. The purchase amount must be less than $1, and the change is from one dollar. I need to know how many quarters, dimes, nickels, and pennies someone would get back. Would it be best to set up a dictionary? 回答1: Gee, you mean this isn't problem 2b in every programming course any more? Eh, probably not, they don't seem to teach people how to make change any more

问题 My apologies in advance. I'm aware that this question has been asked before with answers that have not produced the results I want/need. I am making an attempt to write a function that does the following in Python3: I need a recursive function that returns all the number of ways (coin combinations) that produce a specified amount. This function can only contain two arguments, amount and coins. I've had a difficult time wrapping my mind around recursion, so explanations would also be greatly