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500-yen Saving

"500-yen Saving" is one of Japanese famous methods to save money. The method is quite simple.
whenever you receive a 500-yen coin in your change of shopping, put the coin to your 500-yen saving box. Typically, you will find more than one million yen in your saving box in ten years.

Some Japanese people are addicted to the 500-yen saving. They try their best to collect 500-yen coins efficiently by using 1000-yen bills and some coins effectively in their purchasing. For example, you will give 1320 yen (one 1000-yen bill, three 100-yen coins and two 10-yen coins) to pay 817 yen, to receive one 500-yen coin (and three 1-yen coins) in the change.

A friend of yours is one of these 500-yen saving addicts. He is planning a sightseeing trip and wants to visit a number of souvenir shops along his way. He will visit souvenir shops one by one according to the trip plan. Every souvenir shop sells only one kind of souvenir goods, and he has the complete list of their prices. He wants to collect as many 500-yen coins as possible through buying at most one souvenir from a shop. On his departure, he will start with sufficiently many 1000-yen bills and no coins at all. The order of shops to visit cannot be changed. As far as he can collect the same number of 500-yen coins, he wants to cut his expenses as much as possible.

Let's say that he is visiting shops with their souvenir prices of 800 yen, 700 yen, 1600 yen, and 600 yen, in this order. He can collect at most two 500-yen coins spending 2900 yen, the least expenses to collect two 500-yen coins, in this case. After skipping the first shop, the way of spending 700-yen at the second shop is by handing over a 1000-yen bill and receiving three 100-yen coins. In the next shop, handing over one of these 100-yen coins and two 1000-yen bills for buying a 1600-yen souvenir will make him receive one 500-yen coin. In almost the same way, he can obtain another 500-yen coin at the last shop. He can also collect two 500-yen coins buying at the first shop, but his total...
Using python3 can you solve the prior task?
 ... What is a Python code to find more than one million yen in your saving box in ten years?
assistant
#Here is Python code to find more than one million yen in your saving box in ten years.

import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+9
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)

def main():
    rr = []

    def f(n):
        a = [I() for _ in range(n)]
        t = [(0,0,0)]
        for c in a:
            nt = collections.defaultdict(lambda: inf)
            c1 = c % 1000
            c5 = c % 500
            mc5 = 500 - c5
            for g,k,p in t:
                nt[(g,k)] = p
            if c1 == 0:
                for g,k,p in t:
                    if k >= 500 and nt[(g+1,k-500)] > p + c:
                        nt[(g+1,k-500)] = p + c
            elif c1 == 500:
                for g,k,p in t:
                    if nt[(g+1, k)] > p + c:
                        nt[(g+1, k)] = p + c
            elif c1 < 500:
                for g,k,p in t:
                    if nt[(g+1, k+mc5)] > p + c:
                        nt[(g+1, k+mc5)] = p + c
            else:
                for g,k,p in t:
                    if k + mc5 >= 500 and nt[(g+1,k+mc5-500)] > p + c:
                        nt[(g+1,k+mc5-500)] = p + c
                    if nt[(g, k+mc5)] > p + c:
                        nt[(g, k+mc5)] = p + c
            t = []
            cg = -1
            mk = -1
            mp = inf
            # print('nt',nt)
            for g,k in sorted(nt.keys(), reverse=True):
                p = nt[(g,k)]
                if p == inf:
                    continue
                if cg != g:
                    mp = inf
                  ...