Currency convertibility (dynamic programming) - Python realization

Currency exchange problem solving dynamic programming method #

There are n # currency coin denominations of v1, v2, v3 ... vn, where v1 = 1, the use of money with a total value of money exchange, how to find the minimum number of coins, i.e., x1, x2, x3 ... xn minimum sum

# Input: nominal value of various currencies v1, v2, v3 ... vn; To redeem the total value of money

# Output: get the least amount of currency exchange

 

1  # Denomination modified money system 
2 V = [1,2,5,10,50 ]
 . 3  # amount of money to be modified exchange Money 
. 4 Money = 253

1  # initial amount of money for each of 0 
2 X = [0] * len (V)
 . 3  # establishing a corresponding number of currency exchange table = Q [Money + 1] [len (V) + 1'd] 
. 4 Q = [([0 ] * (len (V) + 1'd)) for I in Range (0, + Money. 1 )]

. 1  # Q Table Initialization: The first row of the first column is set 0 
2  for I in Range (0, + Money. 1 ):
 . 3      Q [I] [0] = 0
 . 4  for I in Range (0, len (V) + 1'd ):
 . 5      Q [0] [I] = 0

1  # sheet process 
2  for I in Range (1, Money + 1 ):
 . 3      for J in Range (1, len (V) + 1'd ):
 . 4          IF V [J-1] == I:         # denomination = i 
. 5              Q [I] [J] =. 1
 . 6          elif V [J-. 1]> I:        # denomination> I 
. 7              Q [I] [J] = Q [I] [J-. 1 ]
 . 8          the else :                   # denomination <i 
. 9              Q [i] [J] = Q [IV [-J. 1]] [J] +. 1         # IV [J] is the nominal value of i

Function call:

. 1  Print (Q); Print ( ' \ n- ' )
 2  Print ( ' exchange to give a minimum number of currencies: ' , Q [Money] [len (V)])

operation result:

 A number obtained at least for currency conversion: 7