-1 
b=int(1)

if b == 1:
     b=2
     c = "on my thumb"
elif b== 2:
    b=3
    c = "on my shoe"
elif b== 3:
    b=4
    c = "on my knee"
elif b== 4:
    b+1
    c = "on my door"
elif b== 5:
    b+1
    c = "on my hive"
elif b== 6:
    b+1
    c = "on my sticks"
elif b== 7:
    b+1
    c = "up in heaven"
elif b== 8:
    b+1
    c = "on my gate"
elif b== 9:
    b+1
    c = "on my spine"
else:
    c = "once again"

for r in range(10):
    print("This old man, he played one He played knick-knack " + c +" Knick-knack paddywhack, give your dog a bone This old man came rolling home")
    b+1

コーディングは比較的新しいので、自分が何をしているのかよくわかりませんが、rが印刷されるたびにcを変更して、古い童謡を完成させようとしています。私はこれをPythonでコーディングしています...

4

3 に答える 3

4 
parts = ["on my thumb", "on my shoe", ...]
numerators = ["one", "two", "three", ...]

for num, part in zip(numerators, parts):
    print "This old man, he played " + num
    print "He played knick-knack " + part
    print "Knick-knack paddywhack, give your dog a bone"
    print "This old man came rolling home"
于 2012-10-09T15:14:54.543 に答える
0

多分このようなもの：

string b[10] = {"on my thumb","on my shoe","on my knee","on my door","on my hive","on my sticks","up in heaven","on my gate","on my spine","once again"};

for (i=0; i<b.length; i++) {
   print("This old man, he played one He played knick-knack " + b[i] +" Knick-knack paddywhack, give your dog a bone This old man came rolling home.\n");
}

これらの値は変更されないため、値を保持する静的配列を作成してから、結果を表示しながら内容を循環させることができます。

于 2012-10-09T15:17:57.050 に答える
0

文字列を配列またはリストに保存するだけです

lines = ["on my thumb", "on my shoe", "...."....]

次に、C# っぽい構文の for ループを使用して、その配列 (または list ) を反復処理します。

foreach(line in lines)
{
  print("bla" + line + "blubb");
}
于 2012-10-09T15:14:49.037 に答える