与小孩一起学编程11章
与孩子一起学编程11章
上一章的小游戏，玩的开心不，接下来，我们要讨论些其他方面的知识点
先说说循环，还是循环，形成所谓的嵌套循环
很简单的乘法表代码
multiplier = 5
for i in range (1, 11):
print i, "x", multiplier, "=", i * multiplier
IDLE运行结果>>> 1 x 5 = 5 2 x 5 = 10 3 x 5 = 15 4 x 5 = 20 5 x 5 = 25 6 x 5 = 30 7 x 5 = 35 8 x 5 = 40 9 x 5 = 45 10 x 5 = 50 >>>这是一个数的乘法表，如果我们要三个数的乘法表呢，这就是嵌套循环最擅长的
for multiplier in range (5, 8):
for i in range (1, 11):
print i, "x", multiplier, "=", i * multiplier
print
运用结果>>> 1 x 5 = 5 2 x 5 = 10 3 x 5 = 15 4 x 5 = 20 5 x 5 = 25 6 x 5 = 30 7 x 5 = 35 8 x 5 = 40 9 x 5 = 45 10 x 5 = 50 1 x 6 = 6 2 x 6 = 12 3 x 6 = 18 4 x 6 = 24 5 x 6 = 30 6 x 6 = 36 7 x 6 = 42 8 x 6 = 48 9 x 6 = 54 10 x 6 = 60 1 x 7 = 7 2 x 7 = 14 3 x 7 = 21 4 x 7 = 28 5 x 7 = 35 6 x 7 = 42 7 x 7 = 49 8 x 7 = 56 9 x 7 = 63 10 x 7 = 70 >>>上面写得在实际中都不实用的，为什么，可变循环可以由用户来控制
numStars = int(raw_input ("how many stars do you want? "))
for i in range (1, numStars):
print '*',运行结果>>> how many stars do you want? 6 * * * * * >>>发现没有，希望得到6个星星，结果输出了5个。为什么呢，想想。
numStars = int(raw_input ("how many stars do you want? "))
for i in range (1, numStars + 1):
print '*',这次结果>>> how many stars do you want? 6 * * * * * * >>>还有一种方法，for循环从0开始计数，也是可以的。
numStars = int(raw_input ("how many stars do you want? "))
for i in range (0, numStars):
print '*',那可变嵌套循环呢numLines = int(raw_input ('how many lines of stars do you want? '))
numStars = int(raw_input ("how many stars per line? "))
for line in range (0, numLines):
for star in range(0, numStars):
print '*',
print运行结果>>> how many lines of stars do you want? 6 how many stars per line? 5 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * >>>这样有点意思了吧，甚至，还有嵌套嵌套循环
numBlocks = int(raw_input ('how many blocks of stars do you want? '))
numLines = int(raw_input ('how many lines of stars do you want? '))
numStars = int(raw_input ("how many stars per line? "))
for block in range(0, numBlocks):
for line in range (0, numLines):
for star in range(0, numStars):
print '*',
print
print
运行结果如下>>> how many blocks of stars do you want? 3 how many lines of stars do you want? 5 how many stars per line? 4 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * >>>这也是双重嵌套循环，更多可变嵌套循环，还可以写个更复杂的可变嵌套循环
numBlocks = int(raw_input ('how many blocks of stars do you want? '))
for block in range(1, numBlocks + 1):
for line in range(1, block * 2 ):
for star in range(1, (block + line) * 2):
print '*',
print
print运行结果>>> how many blocks of stars do you want? 3 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * >>>还可以直接打印出循环变量的值
numBlocks = int(raw_input ('how many blocks of stars do you want? '))
for block in range(1, numBlocks + 1):
print 'block = ', block
for line in range(1, block * 2 ):
for star in range(1, (block + line) * 2):
print '*',
print ' line =', line, 'star = ', star
print
运行结果>>> how many blocks of stars do you want? 3 block = 1 * * * line = 1 star = 3 block = 2 * * * * * line = 1 star = 5 * * * * * * * line = 2 star = 7 * * * * * * * * * line = 3 star = 9 block = 3 * * * * * * * line = 1 star = 7 * * * * * * * * * line = 2 star = 9 * * * * * * * * * * * line = 3 star = 11 * * * * * * * * * * * * * line = 4 star = 13 * * * * * * * * * * * * * * * line = 5 star = 15 >>>那使用嵌套循环能做些什么实际工作呢，他最擅长的其实就是得出一系列决定的所有可能的排列和组合。下面举个例子，热狗店里做一个广告海报，用数字显示如何订购热狗、小面包、番茄、芥末酱和洋葱的所有可能的组合。我们需要得出总共有多少种可能的组合。需要用到一种方法就是使用决策树(decision tree)，用代码实现
print "\tDog \tBun \tKetchup\tMustard\tOnions"
count = 1
for dog in [0, 1]:
for bun in [0, 1]:
for ketchup in [0, 1]:
for mustard in [0, 1]:
for onion in [0, 1]:
print "#", count, "\t",
print dog, "\t", bun, "\t", ketchup, "\t",
print mustard, "\t", onion
count = count + 1运行结果>>> Dog Bun Ketchup Mustard Onions # 1 0 0 0 0 0 # 2 0 0 0 0 1 # 3 0 0 0 1 0 # 4 0 0 0 1 1 # 5 0 0 1 0 0 # 6 0 0 1 0 1 # 7 0 0 1 1 0 # 8 0 0 1 1 1 # 9 0 1 0 0 0 # 10 0 1 0 0 1 # 11 0 1 0 1 0 # 12 0 1 0 1 1 # 13 0 1 1 0 0 # 14 0 1 1 0 1 # 15 0 1 1 1 0 # 16 0 1 1 1 1 # 17 1 0 0 0 0 # 18 1 0 0 0 1 # 19 1 0 0 1 0 # 20 1 0 0 1 1 # 21 1 0 1 0 0 # 22 1 0 1 0 1 # 23 1 0 1 1 0 # 24 1 0 1 1 1 # 25 1 1 0 0 0 # 26 1 1 0 0 1 # 27 1 1 0 1 0 # 28 1 1 0 1 1 # 29 1 1 1 0 0 # 30 1 1 1 0 1 # 31 1 1 1 1 0 # 32 1 1 1 1 1 >>>还可以把每样的卡路里值也跟在后面，
dog_cal = 140
bun_cal = 120
ket_cal = 80
mus_cal = 20
onion_cal = 40
print "\tDog \tBun \tKetchup\tMustard\tOnions\tCalories"
count = 1
for dog in [0, 1]:
for bun in [0, 1]:
for ketchup in [0, 1]:
for mustard in [0, 1]:
for onion in [0, 1]:
total_cal = (bun * bun_cal)+(dog * dog_cal) + \
(ketchup * ket_cal)+(mustard * mus_cal) + \
(onion * onion_cal)
print "#", count, "\t",
print dog, "\t", bun, "\t", ketchup, "\t",
print mustard, "\t", onion,
print "\t", total_cal
count = count + 1
运行结果>>> Dog Bun Ketchup Mustard Onions Calories # 1 0 0 0 0 0 0 # 2 0 0 0 0 1 40 # 3 0 0 0 1 0 20 # 4 0 0 0 1 1 60 # 5 0 0 1 0 0 80 # 6 0 0 1 0 1 120 # 7 0 0 1 1 0 100 # 8 0 0 1 1 1 140 # 9 0 1 0 0 0 120 # 10 0 1 0 0 1 160 # 11 0 1 0 1 0 140 # 12 0 1 0 1 1 180 # 13 0 1 1 0 0 200 # 14 0 1 1 0 1 240 # 15 0 1 1 1 0 220 # 16 0 1 1 1 1 260 # 17 1 0 0 0 0 140 # 18 1 0 0 0 1 180 # 19 1 0 0 1 0 160 # 20 1 0 0 1 1 200 # 21 1 0 1 0 0 220 # 22 1 0 1 0 1 260 # 23 1 0 1 1 0 240 # 24 1 0 1 1 1 280 # 25 1 1 0 0 0 260 # 26 1 1 0 0 1 300 # 27 1 1 0 1 0 280 # 28 1 1 0 1 1 320 # 29 1 1 1 0 0 340 # 30 1 1 1 0 1 380 # 31 1 1 1 1 0 360 # 32 1 1 1 1 1 400 >>>
里面用到了一些现在我们还没接触到的知识，没关系，接下来会一一来了解到的，很简单。我们用了制表符来对齐，也就是符号“\t”，还有个符号要注意，代码行末的反斜线“\”，表示这一行还没结束，下一行也是这一行的内容。
怎么样，这么枯燥的内容，编写一个程序来实现，是不是很有趣啊。