title:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

翻译:

10下面的质数的和为2 + 3 + 5 + 7 = 17。

请求出200,0000下面全部质数的和。

import math,time
def isOk(a):
for i in range(2,int(math.sqrt(a))+1):
if a%i==0:
return False
return True
s=0
begin=time.time()
for i in range(2,2000000):
if isOk(i):
s += i
print s
end = time.time()
print end-begin

projecteuler---->problem=10----Summation of primes的更多相关文章

  1. Problem 10: Summation of primes

    def primeslist(max): ''' 求max值以内的质数序列 ''' a = [True]*(max+1) a[0],a[1]=False,False for index in rang ...

  2. Project Euler Problem 10

    Summation of primes Problem 10 The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of ...

  3. Problem 10

    Problem 10 # Problem_10.py """ The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. ...

  4. (Problem 47)Distinct primes factors

    The first two consecutive numbers to have two distinct prime factors are: 14 = 2  7 15 = 3  5 The fi ...

  5. (Problem 37)Truncatable primes

    The number 3797 has an interesting property. Being prime itself, it is possible to continuously remo ...

  6. (Problem 35)Circular primes

    The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, ...

  7. uoj problem 10

    uoj problem 10 题目大意: 给定任务若干,每个任务在\(t_i\)收到,需要\(s_i\)秒去完成,优先级为\(p_i\) 你采用如下策略: 每一秒开始时,先收到所有在该秒出现的任务,然 ...

  8. (Problem 10)Summation of primes

    The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two milli ...

  9. projecteuler Summation of primes

    The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two milli ...

  10. Summation of primes

    是我算法不对,还是笔记本CPU太差? 我优化了两次,还是花了三四个小时来得到结果. 在输出上加1就是最终结果. The sum of the primes below 10 is 2 + 3 + 5 ...

随机推荐

  1. LeetCode 344. Reverse String(反转字符串)

    题目描述 LeetCode 344. 反转字符串 请编写一个函数,其功能是将输入的字符串反转过来. 示例 输入: s = "hello" 返回: "olleh" ...

  2. python之sqlite3使用详解

    Python SQLITE数据库是一款非常小巧的嵌入式开源数据库软件,也就是说没有独立的维护进程,所有的维护都来自于程序本身.它使用一个文件存储整个数据库,操 作十分方便.它的最大优点是使用方便,功能 ...

  3. onethink 重写URL后,apache提示No input file specified

    <IfModule mod_rewrite.c> RewriteEngine on RewriteCond %{REQUEST_FILENAME} !-d RewriteCond %{RE ...

  4. 876. Middle of the Linked List【Easy】【单链表中点】

    Given a non-empty, singly linked list with head node head, return a middle node of linked list. If t ...

  5. poj 2940

    Wine Trading in Gergovia Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3187   Accepte ...

  6. WQS二分题集

    WQS二分,一种优化一类特殊DP的方法. 很多最优化问题都是形如“一堆物品,取与不取之间有限制.现在规定只取k个,最大/小化总收益”. 这类问题最自然的想法是:设f[i][j]表示前i个取j个的最大收 ...

  7. ARC-100 C - Linear Approximation

    题面在这里! 可以看成点集{a[i]-i}和b之间距离的和,于是找到中位数就可以直接算了2333. #include<bits/stdc++.h> #define ll long long ...

  8. BZOJ 2296【POJ Challenge】随机种子(构造)

    [题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=2296 [题目大意] 给出一个数x,求一个10的16次以内的数使得其被x整除并且数字包含 ...

  9. [TCO2009]NumberGraph

    题意:给你一些带权的节点和一个正整数集合$S$,$S$中每一个数的二进制后缀$0$个数相同,节点$x$的权值为$v_x$,如果对于$x,y$存在$t\in S$使得$|v_x-v_y|=t$,那么连边 ...

  10. 【贪心】hdu5969 最大的位或

    对于右端点r和左端点l,考虑他们的二进制位从高到低,直到第一位不同的为止. 更高的都取成相同的,更低的都取成1. 比如 101011110001 101011101001 101011111111 # ...