0x00 题目
H4x0r曾在他的博客上说道，“时间最终赶上了每个人,除了H4x0r。如果你能挑战我!”
来吧: 时光穿梭门
0x01 解题打开传送门之后，题目要求在3秒内提交前N(比如：30个)素数的和。
第一种方法的算法思想就是简单的素数判断方法
import re import httplib import urllib, urllib2, cookielib def find_sum_of_first_N_prime_numbers(N): sum = 0 n = 0 num = 1 while n < N: # prime numbers are greater than 1 if num > 1: for i in range(2,num): if (num % i) == 0: break else: sum += num n = n+1 num = num+1 return sum conn = httplib.HTTPConnection("hack.bckdr.in") response = conn.request("GET", 'http://hack.bckdr.in/2013-MISC-75/misc75.php') response = conn.getresponse() text = response.read() num_in_page = [int(s) for s in text.split() if s.isdigit()] N = num_in_page[1] sum = find_sum_of_first_N_prime_numbers(N) headers = response.getheaders() cookie = headers[2][2] headers = {"Content-type": "application/x-www-form-urlencoded" , "Cookie": cookie , "Host": "hack.bckdr.in" , "Connection" : "keep-alive" , "Cache-Control" : "max-age=0" , "Origin": "http://hack.bckdr.in" , "Referer": "http://hack.bckdr.in/2013-MISC-75/misc75.php" , "User-Agent": "Mozilla/5.0 (windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2272.118 Safari/537.36"} params = urllib.urlencode({'answer': sum, 'submit': 'Submit'}) response = conn.request("POST", 'http://hack.bckdr.in/2013-MISC-75/misc75.php', params, headers) response = conn.getresponse() print(response.read())但我感觉这个算法效率太低，如果题目给的数字很大，验证时间再调小，这个算法就不再适用了，根据 素数判断算法(高效率） 一文编写了另一个参考示例：
首先生成质数列表：
import math import time start=time.clock() n=10000000 result = list() result.append(2) result.append(3) for i in xrange(5,n+1,2): for j in xrange(3,int(math.sqrt(i))+1): if i%j == 0: break else: result.append(i) end=time.clock() print end-start f=open(r'prime_number.txt','w') for x in result: f.write(str(x)+'\n') f.close主要用到算法思想:
判断一个数是否为素数只需除以这个数的平方根+1之间的所有的数
从2开始往后依次增1的数列里所有偶数一定不是素数(这能减少一半的时间)
素数快速筛选法
测试了一下(CPU:Intel(R) Core(TM) i5-4200H CPU @ 2.80GHz System:Windows 10 Pronfessional x64)，生成前10000以内的素数平均0.02秒，生成前100000以内的素数平均0.3秒，生成前1000000以内的素数平均7秒，生成前10000000以内的素数平均140秒，速度还是可以的，但是这肯定不是最优的，应该还可以进行优化。
然后是读取素数表计算前N个素数的和并提交结果：
import requests import re import time def main(): starttime=time.clock() url = "http://hack.bckdr.in/2013-MISC-75/misc75.php" req = requests.get(url) tmpcookies = req.cookies match = re.findall(r'[0-9]+', req.text); num = int(match[1]) try: src = open("prime_number.txt", "r") data = src.read() finally: src.close() primelist = data.split() sum = 0 for i in range(0, num): sum += int(primelist[i]) print("Sum is {}.".format(sum)) data = {"answer": sum} req2 = requests.post(url, data=data, cookies=tmpcookies) print(req2.text) endtime=time.clock() print endtime - starttime if __name__ == "__main__": main()也可以将两部分写在一起
import requests import re import math def prime_generate(n): result = list() result.append(2) result.append(3) for i in xrange(5,n+1,2): for j in xrange(3,int(math.sqrt(i))+1): if i%j == 0: break else: result.append(i) return result def calc_sum(num): finsum=0 for x in prime_generate(10000)[:num]: finsum=finsum+x return finsum def main(): url = "http://hack.bckdr.in/2013-MISC-75/misc75.php" req = requests.get(url) tmpcookies = req.cookies # Set rundom N. match = re.findall(r'[0-9]+', req.text); num = int(match[1]) sum=calc_sum(num) data = {"answer": sum} req2 = requests.post(url, data=data, cookies=tmpcookies) print(req2.text) if __name__ == "__main__": main()小伙伴有什么优化的算法还请告诉我，一起交流学习:P