用有限的硬币找零

Question

import sys

def minCoins(coins, m, how, V):
    # table[i] will be storing the minimum  
    # number of coins required for i value.  
    # So table[V] will have result 
    table = [0 for i in range(V + 1)] 
    index = []
    # Base case (If given value V is 0) 
    table[0] = 0

    # Initialize all table values as Infinite 
    for i in range(1, V + 1): 
        table[i] = sys.maxsize 

    # Compute minimum coins required  
    # for all values from 1 to V 
    for i in range(1, V + 1):
        # Go through all coins smaller than i 
        for j in range(m): 
            if (coins[j] <= i): 
                sub_res = table[i - coins[j]]
                if (sub_res != sys.maxsize and 
                    sub_res + 1 < table[i]):
                    if sub_res +1 <= how[j] :
                        table[i] = sub_res + 1

    return table[V]

coins = [200, 100, 50, 20, 10, 5, 2, 1]
howmanycoins = [0, 2, 2, 3, 0, 0, 7, 8]
m = len(coins)
V = 16
print("Minimum coins required is ", minCoins(coins, m, howmanycoins, V))   

我对此代码有疑问。 当表howmanycoins具有此值时，[0, 2, 2, 3, 0, 0, 7, 9]程序给出很好的答案7x“ 2” + 2x“ 1” = 9个硬币，但是当最后一个8位时，输出如下所示：

  

所需的最低硬币是9223372036854775807。