for语句作业 - Khalil 吕龙

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一、冒泡排序和选择排序

# 冒泡排序
li = [34, 32, 67, 23, 12, 7, 4, 1]  # 原始排序
len1 = len(li)  # 获取列表长度
for i in range(0, len1):
    for j in range(1, len1 - i):
        if li[j - 1] > li[j]:  # 比较前后两元素哪个大
            temp = li[j - 1]  # 前一个大的放到临时暂存元素里
            li[j] = temp  # 把临时暂存元素里的大元素放到后面
print(li)

# 选择排序
li = [34, 32, 67, 23, 12, 7, 4, 1] #原始排序
for i in range(len(li)):           #遍历所有元素的索引
    max = i                        #定义已经完成排序的序列的最后一个数的索引
    for j in range(i + 1, len(li)): #遍历没有排序过的所有元素的索引
        if li[j] > li[max]:         #在未排序的序列里找最大的数
            max = j                 #保存未排序最大数的索引
    li[i], li[max] = li[max], li[i] #将找到的最大的数放到排好的序列末尾
print(li)                           #打印已排序好的序列

二、
1.九九乘法表

for i in range(1,10):
    for j in range(1,i):
        print(f"{j}*{i}={i*j}",end=' ')
    print()

2.完全数

num = int(input('please input a perfect number:'))
sum1 = 0
factor1 = []
for i in range(1, num):
    if num % i == 0:
        sum1 += i
        factor1.append(i)
factor1.append(num)
print(f"all factors of {num} are {factor1}")
if num == sum1:
    print(f'{num} is a perfect number')
else:
    print(f'{num} is not a perfect number')

3.三位数

n = 0
li = []
for i in range(1,5):
    for j in range(1,5):
        for k in range(1,5):
            if i != j and j !=k and i != k:
                li.append(100*i+10*j+k)
                n+=1
print(li,n)

#推导式 
li = [100 * i + 10 * j + k for i in range(1, 5) for j in range(1, 5) for k in range(1, 5) if i != j and j != k and i != k]
print(li, len(li))

# 疏影横斜水清浅
li = []
book = []
def dfs(step):
    if (step == 4):
        for i in range(3):
            print(li[i], end='')
        print()
        return
    for i in range(4):
        if book[i] == 0:     
            li[step] = i + 1
            book[i] = 1
            dfs(step + 1)
            book[i] = 0
    return 
if __name__ == '__main__':
    for i in range(4):
        li.append(0) 
        book.append(0)
    dfs(0)