- def hanno(n,A,B,C):
- if n >0:
- hanno(n-1,A,C,B)#先把n-1个盘子,从柱子A移动到柱子B
- print(f"盘子{n},从{A}移动到{C}")
- hanno(n-1,B,A,C) #柱子A上面有n-1个盘子,再将盘子从A,借助A,移动到C
-
- hanno(3,'柱子A','柱子B','柱子c')
- def fun_01(n):
- if n <= 2:
- return 1
- else:
- return (fun_01(n - 1) + fun_01(n - 2))
-
-
- num = int(input("输出几项:"))
- a = fun_01(num)
- for i in range(1, num+1):
- print(fun_01(i))
- #已知鸡和兔子共几只、鸡和兔子的脚共几只
-
- sum = int(input("鸡和兔子共有:")) #35
- num = int(input("鸡和兔子脚共有:")) #94
- for i in range(sum+1):
- y = sum - i
- if 2*x + 4*y == num:
- print(f"鸡仔有{i}只,兔子有{y}只")
-
- 运行结果
- 鸡和兔子共有:35
- 鸡和兔子脚共有:94
- 鸡仔有23只,兔子有12只
- def iszhi(a):
- if a == 1:
- print("1既不是质数也不是合数")
- for i in range(2,a):
- if a % i == 0:
- return False
-
- else:
- return True
-
- a = iszhi(5)
- print(a)