ITP1_9の解答例(Python)

8に続いて今回も文字列の扱いについての問題です．

ITP1_9_A: Finding a Word

# ITP1_9_A
S = ""
W = input()
while True:
    s = input()
    if s == "END_OF_TEXT":
        break
    # 行の最後にスペースを入れないと次の行とつながる
    S += s + " "
    
# 小文字にしてから判定する
S = S.lower()
S_list = S.split(" ")
# 完全一致を探したいのでSのままcountはできない
print(S_list.count(W))

ITP1_9_B: Shuffle

# ITP1_9_B
while True:
    s = input()
    # 終了条件
    if s == "-":
        break
    n = int(input())
    for _ in range(n):
        h = int(input())
        s = s[h:] + s[:h]
    print(s)

# ITP1_9_B
while True:
    s = input()
    # 終了条件
    if s == "-":
        break
    n = int(input())
    H = 0
    for _ in range(n):
        H += int(input())
        
    h = H%(len(s))
    print(s[h:]+s[:h])
        

ITP1_9_C: Card Game

# ITP1_9_C
n = int(input())
taro = 0
hanako = 0
for _ in range(n):
    s1,s2 = input().split(" ")
    s1 = s1.lower()
    s2 = s2.lower()
    done = False
    for i in range(min(len(s1),len(s2))):
        if ord(s1[i]) < ord(s2[i]):
            hanako += 3
            done = True
            break
        elif ord(s1[i]) > ord(s2[i]):
            taro += 3
            done = True
            break
    
    # 既に決着がついている場合はこれ以降の判定はしない
    if done:
        continue
            
    # 最後の文字まで同じだった場合は文字数
    if len(s1) < len(s2):
        hanako += 3
    elif len(s1) > len(s2):
        taro += 3
    else:
        taro += 1
        hanako += 1
        
print(taro,hanako)

ITP1_9_D: Transformation

# ITP1_9_D
S = input()
n = int(input())
for _ in range(n):
    command = input().split(" ")
    opp = command[0]
    a = int(command[1])
    b = int(command[2])+1
    if opp == "print":
        print(S[a:b])
    elif opp == "reverse":
        S = S[:a] + "".join(list(reversed(S[a:b]))) + S[b:]
    elif opp == "replace":
        p = command[3]
        S = S[:a] + p + S[b:]

def str_reverse(S):
    new_S = ""
    for i in range(len(S)):
        new_S += S[len(S)-i-1]
        
    return new_S