Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. Credits:Special thanks to @mithmatt for adding this problem and creating all test cases. 求n以内的所有素数,以前看过的一道题目,通过将所有非素数标记出来,再找出素数,代码如下: public i…

题目大意 https://leetcode.com/problems/count-primes/description/ 204. Count Primes Count the number of prime numbers less than a non-negative number, n. Example: Input: 10Output: 4Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.…

Sieve of Eratosthenes (素数筛选算法) Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number. For example, if n is 10, the output should be “2, 3, 5, 7″. If n is 20, the output should be “2, 3, 5, 7, 11, 13,…

Problem Description Give you a lot of positive integers, just to find out how many prime numbers there are. Input There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won't exceed 32-…

How many prime numbers Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12955    Accepted Submission(s): 4490 Problem Description   Give you a lot of positive integers, just to find out how many…

Description: Count the number of prime numbers less than a non-negative number, n. 解题思路: 空间换时间,开一个空间为n的数组,因为非素数至少可以分解为一个素数,因此遇到素数的时候,将其有限倍置为非素数,这样动态遍历+构造下来,没有被设置的就是素数. public int countPrimes(int n) { if (n <= 2) return 0; boolean[] notPrime = new boo…

1- 问题描述 Count the number of prime numbers less than a non-negative number, n 2- 算法思想 给出要筛数值的范围 $n$,找出 $\sqrt{n}$ 以内的素数 $p_{1}, p_{2}, \cdots, p_{k}$.先用2去筛,即把2留下,把2的倍数剔除掉:再用下一个素数,也就是3筛,把3留下,把3的倍数剔除掉:接下去用下一个素数5筛,把5留下,把5的倍数剔除掉:不断重复下去....... 3- Python实现…

How many prime numbers Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14684    Accepted Submission(s): 5091 Problem Description   Give you a lot of positive integers, just to find out how many…

第一眼看这道题目的时候觉得可能会很难也看不太懂,但是看了给出的Hint之后思路就十分清晰了 Consider the first sample. Overall, the first sample has 3 queries. The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9. The second query comes…

385C - Bear and Prime Numbers 思路:记录数组中1-1e7中每个数出现的次数,然后用素数筛看哪些能被素数整除,并加到记录该素数的数组中,然后1-1e7求一遍前缀和. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset((a),(b),sizeof(a)) const int INF=0x3f…