Python: Print the number of prime numbers which are less than or equal to a given integer
Python Basic - 1: Exercise-38 with Solution
Write a Python program to print the number of prime numbers which are less than or equal to a given integer.
n (1 ≤ n ≤ 999,999)
Input the number(n): 35
Number of prime numbers which are less than or equal to n.: 11
primes =  * 500000 primes = 0 for i in range(3, 1000, 2): if primes[i // 2]: primes[(i * i) // 2::i] =  * len(primes[(i * i) // 2::i]) print("Input the number(n):") n=int(input()) if n < 4: print("Number of prime numbers which are less than or equal to n.:",n - 1) else: print("Number of prime numbers which are less than or equal to n.:",sum(primes[:(n + 1) // 2]) + 1)
Input the number(n): 35 Number of prime numbers which are less than or equal to n.: 11
Python Code Editor:
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Previous: Write a Python program which reads an integer n and find the number of combinations of a,b,c and d (0 ≤ a,b,c,d ≤ 9) where (a + b + c + d) will be equal to n.
Next: Write a program to compute the radius and the central coordinate (x, y) of a circle which is constructed by three given points on the plane surface.
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Python: Tips of the Day
How to make a flat list out of list of lists?
Given a list of lists l
flat_list = [item for sublist in l for item in sublist]
flat_list =  for sublist in l: for item in sublist: flat_list.append(item)
is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:
flatten = lambda l: [item for sublist in l for item in sublist]
As evidence, you can use the timeit module in the standard library:
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' '[item for sublist in l for item in sublist]' 10000 loops, best of 3: 143 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'sum(l, )' 1000 loops, best of 3: 969 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'reduce(lambda x,y: x+y,l)' 1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
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