判断质数

  1. num = int(input('请输入数字:'))
  2. startnum = 2
  3. endnum = num - 1
  4. for i in range(startnum, endnum):
  5.     if num%i == 0:
  6.         print('非质数')
  7.         break
  8.     else:
  9.         print('质数')

九九乘法表

  1. count = 1
  2. for i in range(1,10):
  3.     for j in range(1,i+1):
  4.         print(str(j)+"x"+str(i)+"="+str(i*j),end='')
  5.         if i==j:
  6.             print('\n')
  9. # 列表解析式方式
  10. print('\n'.join(["".join(["{}*{}={:<4}".format(j, i, i * j) for i in range(1, j + 1)]) for j in range(1, 10)]))
  12. [print('{}*{}={:<4}{}'.format(j, i, i * j, '\n' if i == j else ''), end='') for i in range(1, 10) for j in range(1, i + 1)]

斐波那契数列

  1. # for循环迭代
  2. i = 0
  3. pre = 0
  4. nxt = 1
  5. print(pre)
  6. print(nxt)
  7. while i < 10:
  8.     count = pre + nxt
  9.     pre, nxt = nxt, pre + nxt
  10.     i += 1
  11.     print(nxt)
  13. # 递归方法
  14. i = 0
  15. pre = 0
  16. nxt = 1
  17. print(pre)
  18. print(nxt)
  19. def fib(n, pre=0, nxt=1):
  20.     pre, nxt = nxt, pre + nxt
  21.     print(nxt)
  22.     if n == 2:
  23.         return
  24.     fib(n - 1, pre, nxt)
  25. fib(10)

求阶乘

  1. # 循环方式
  2. # 获取用户输入的数字
  3. num = int(input("请输入一个数字: "))
  4. factorial = 1
  6. # 查看数字是负数，0 或 正数
  7. if num < 0:
  8.    print("负数没有阶乘")
  9. elif num == 0:
  10.    print("0 的阶乘为 1")
  11. else:
  12.     for i in range(1,num + 1):
  13.        factorial *= i
  14.     print("%d 的阶乘为 %d" %(num,factorial))
  17. # 递归方式  
  18. def factorial(num):
  19.     if num == 1:
  20.         return 1
  21.     else:
  22.         return num * factorial(num-1)
  24. print(factorial(10))

杨辉三角

  1. # -*- coding: utf-8 -*-
  2. # @Time    : 2019/7/12 16:11
  3. # @File    : testPython.py
  4. __author__ = 'liangliang.yu@onebank.com.cn'
  6. all = []
  7. n = 100
  8. for i in range(n):
  9.     row = [1]*(i+1)  # 开辟内存空间，全部补1，后续判断修改列表中内容
  10.     all.append(row)
  12.     # n=3才会进入下面循环
  13.     for j in range(1, i//2+1): # 根据杨辉三角规律缩小取值范围为一半，后面根据奇偶判断补充中点右侧一半
  14.         val = all[i-1][j-1] + all[i-1][j]  # 根据上一行内容计算当前行的值
  15.         row[j] = val
  17.         if j != i-j:  # 判断奇偶，奇数个数的中点跳过
  18.             row[-j-1] = val
  19.     print(row)

使用列表求矩阵倒置

  1. # 方法一：
  2. m = [[1, 2], [3, 4], [5, 6]]
  3. matrix = [[] for _ in m[0]]
  4. for i in m:
  5.     for j in range(len(i)):
  6.         matrix[j].append(i[j])
  7. print(matrix)
  9. # 方法二
  10. matrix = [[1, 2], [3, 4], [5, 6]]
  11. new_matrix = []
  12. for i in range(len(matrix[0])):
  13.     col = []
  14.     for j in range(len(matrix)):
  15.         col.append(matrix[j][i])
  16.     new_matrix.append(col)
  17. print(new_matrix)