# Problem

# For positive integers a and n, a modulo n (written amodn in shorthand) is the remainder when a is divided by n. For example, 29mod11=7 because 29=11×2+7.

# Modular arithmetic is the study of addition, subtraction, multiplication, and division with respect to the modulo operation. We say that a and b are congruent modulo n if amodn=bmodn; in this case, we use the notation a≡bmodn.

# Two useful facts in modular arithmetic are that if a≡bmodn and c≡dmodn, then a+c≡b+dmodn and a×c≡b×dmodn. To check your understanding of these rules, you may wish to verify these relationships for a=29, b=73, c=10, d=32, and n=11.

# As you will see in this exercise, some Rosalind problems will ask for a (very large) integer solution modulo a smaller number to avoid the computational pitfalls that arise with storing such large numbers.

#

# Given: A protein string of length at most 1000 aa.

# Return: The total number of different RNA strings from which the protein could have been translated, modulo 1,000,000. (Don't neglect the importance of the stop codon in protein translation.)

#

# Sample Dataset

# MA

# Sample Output

# 12

# 這道題目是關(guān)于模運算和蛋白質(zhì)翻譯的問題。給定一個不超過1000個氨基酸（字母“F”，“L”，“I”，“M”，“V”，“S”，“P”，“T”，“A”，“Y”，“H”，“Q”，“N”，“K”，“D”，“E”，“C”，“W”，“R”，“G”的字符串），求可以翻譯成該蛋白質(zhì)的不同RNA序列數(shù)目，答案對1,000,000取模。

# 簡單介紹一下蛋白質(zhì)翻譯。RNA是由四種堿基（腺嘌呤、尿嘧啶、胞嘧啶和鳥嘌呤）組成的。氨基酸則由RNA上的三個堿基編碼而來，其中有一個“起始密碼子”（AUG），還有三個“停止密碼子”（UAA，UAG和UGA），分別表示蛋白質(zhì)鏈的開端和結(jié)束。因此，一個蛋白質(zhì)可以用RAN序列編碼，一個RNA序列也可能被翻譯成不同的蛋白質(zhì)。

# 這里需要用到模運算。模運算是一種基本的整數(shù)運算，它給出了除法的余數(shù)。例如a mod n（記作amodn）表示a除以n的余數(shù)。當(dāng)比較大的整數(shù)進行計算時，可以使用模運算來簡化問題。如果兩個整數(shù)在模n時相等，則它們被稱為模n同余。

# 這道題目的思路如下：

# 準(zhǔn)備好DNA中20個氨基酸對應(yīng)的RNA序列表格；

# 對于給定的蛋白質(zhì)序列，根據(jù)RNA序列表格找到所有可能的RNA序列組合；

# 注意考慮到終止密碼子；

# 對于所有可能的RNA序列，計算它們的總數(shù)，并將結(jié)果對1,000,000取模。

# 計算RNA序列數(shù)

def count_rna_strings(protein_string):

? ? # 翻譯表

? ? codon_table = {

? ? ? ? "UUU": "F", "CUU": "L", "AUU": "I", "GUU": "V",

? ? ? ? "UUC": "F", "CUC": "L", "AUC": "I", "GUC": "V",

? ? ? ? "UUA": "L", "CUA": "L", "AUA": "I", "GUA": "V",

? ? ? ? "UUG": "L", "CUG": "L", "AUG": "M", "GUG": "V",

? ? ? ? "UCU": "S", "CCU": "P", "ACU": "T", "GCU": "A",

? ? ? ? "UCC": "S", "CCC": "P", "ACC": "T", "GCC": "A",

? ? ? ? "UCA": "S", "CCA": "P", "ACA": "T", "GCA": "A",

? ? ? ? "UCG": "S", "CCG": "P", "ACG": "T", "GCG": "A",

? ? ? ? "UAU": "Y", "CAU": "H", "AAU": "N", "GAU": "D",

? ? ? ? "UAC": "Y", "CAC": "H", "AAC": "N", "GAC": "D",

? ? ? ? "UAA": "*", "CAA": "Q", "AAA": "K", "GAA": "E",

? ? ? ? "UAG": "*", "CAG": "Q", "AAG": "K", "GAG": "E",

? ? ? ? "UGU": "C", "CGU": "R", "AGU": "S", "GGU": "G",

? ? ? ? "UGC": "C", "CGC": "R", "AGC": "S", "GGC": "G",

? ? ? ? "UGA": "*", "CGA": "R", "AGA": "R", "GGA": "G",

? ? ? ? "UGG": "W", "CGG": "R", "AGG": "R", "GGG": "G"

? ? }

? ? # 統(tǒng)計RNA序列數(shù)量

? ? rna_count = 1? # 起始為1，包含終止密碼子

? ? for aa in protein_string:

? ? ? ? codons_count = list(codon_table.values()).count(aa)

? ? ? ? rna_count = (rna_count * codons_count) % 1000000

? ? return rna_count * 3? # 乘上三個終止密碼子

# 給出蛋白質(zhì)序列

protein_string = "MA"

# 輸出結(jié)果

print(count_rna_strings(protein_string))