求全部的代码

980404453

Description
写一个判断素数的函数，在主函数输入一个整数，输出是否是素数的消息。

Input
一个数(<=1000)

Output
如果是素数输出prime 如果不是输出not prime




Description
求π的近似值，计算公式为π/4=1-1/3+1/5-1/7+...，直到当前项的绝对值恰好小于（10^-n）为止（该项不计入总和）。

Input
输入n。（1<=n<=8）

多组测试数据。

Output
输出π的值。

杨扬阳羊洋

1. def prime(n):
   
2.     flag = 1
   
3.     for i in range(2,int(n/2)):
   
4.         if n % i == 0 :
   
5.             flag = 0
   
6.         return flag
   

复制代码

1. import prime_judge
   
2. n = input ("input an integer less than 1000:")
   
3. n = int(n)
   
4. if n <= 1000:
   
5.     if prime_judge.prime(n):
   
6.         print ("%d is a prime"%(n))
   
7.     else :
   
8.         print ("%d is not a prime"%(n))
   
9. else:
   
10.     print("%d is bigger than 1000"%(n))
    

复制代码

杨扬阳羊洋

求π（pi）的近似值看成求n的近似值，搞了好久都不对啊，很难受。改成求pi的近似值很快就算出来了。以下是代码

1. n = input ("input an number n that 1 <= n <= 8 :")
   
2. n = int(n)
   
3. pi = 1
   
4. i = 1
   
5. j = 1
   
6. while abs(1/i) >= 10**(-n):
   
7.     i += 2
   
8.     j += 1
   
9.     if j % 2 == 0 :   
   
10.         pi -= 1/i
    
11.     else :
    
12.         pi += 1/i
    
13. pi = pi * 4
    
14. print(pi)
    
15. print(j)

复制代码

顺便把测试结果也发一下好了
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :1
2.9760461760461765
6
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :2
3.1611986129870506
51
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :3
3.143588659585789
501
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :4
3.1417926135957908
5001
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :5
3.141612653189785
50001
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :6
3.1415946535856922
500001
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :7
3.1415928535897395
5000001
>>>
=========== RESTART: F:/python3/file cord/Approximate summation.py ===========
input an number n that 1 <= n <= 8 :8
3.1415926735902504
50000001
>>>

980404453

杨扬阳羊洋 发表于 2019-3-31 12:19
求π（pi）的近似值看成求n的近似值，搞了好久都不对啊，很难受。改成求pi的近似值很快就算出来了。以下是 ...

感谢

980404453

杨扬阳羊洋 发表于 2019-3-31 10:38

感谢 大佬