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 range(2, len(a)): if a[index]==True: for i in range(index+index,len(a),index): a[i]=False return [i for i, j in enumerate(a) if j==True] temp = primeslist…

Summation of primes Problem 10 The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. The code resemble : import math limit = 2000000 crosslimit = int(math.sqrt(limit)) #sieve = [False] * limit sieve =…

Problem 10 # Problem_10.py """ The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. 找出两百万一下的质数之和 """ import math primes = [] for i in range(2, 2000001): flag = True for x in…

The first two consecutive numbers to have two distinct prime factors are: 14 = 2  7 15 = 3  5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2²  7  23 645 = 3  5  43 646 = 2  17  19. Find the first four consecutiv…

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3. Fi…

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. How many circular primes are there…

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,然后取出当前优先级最高的任务,一直工作这个任务到下一秒,该任务的需要的时间-1s,如此循环进行,直到任务全部完成. 现在有一任务的优先级未知,但知道其被完成的时间点.确定一个合法的优先级.并计算出所有任务完成的时间 题解: 其实vfk的题解很详细. 那我就写一写我的理解吧. 首先拿到这道题我们就会去想…

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. #include<stdio.h> #include<math.h> #include<stdbool.h> #define N 2000000 bool prim(int n) { int i; ; i*i<=n; i++) { ) return false…

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 + 7 = 17. Find the sum of all the primes below two million. def isprime(n): boolisprime = True for i in xrange(3,n): if n % i == 0: boolisprime = Fals…

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. 译文: 10以下的素数之和为17,求出2000000以下的素数之和. ======================= 第一次code: import java.util.Scanner; public class Main { public static void main(String[]…