题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2161 题意:判断n是不是素数,输入到0停止.题目规定1 2 都不是素数. 题解:筛素数.老题目.不过这次是普通筛23333..之前做的题了. #include<iostream> #include<cmath> #include<cstdio> using namespace std; #define N 16001 bool prime[N]; void init(){…

http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description All you know Goldbach conjecture.That is to say, Every even integer great…

Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits:Special thanks to @mithmatt for adding this problem and creating all test…

http://acm.hdu.edu.cn/showproblem.php?pid=1455 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=243 uva开头描述: 307 - Sticks Time limit: 3.000 seconds hduoj 开头描述: E - Sticks Time Limit:3000MS     Memo…

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[]…

题目大家都非常熟悉,求小于n的所有素数的个数. 自己写的python 代码老是通不过时间门槛,无奈去看了看大家写的code.下面是我看到的投票最高的code: class Solution: # @param {integer} n # @return {integer} def countPrimes(self, n): if n < 3: return 0 primes = [True] * n primes[0] = primes[1] = False for i in range(2,…

examination questions Description: Count the number of prime numbers less than a non-negative number, n References: How Many Primes Are There? Sieve of Eratosthenes Please use the following function to solve the problem: int countPrimes(int n){ } 解题代…

Count Primes Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits: Special thanks to @mithmatt for adding this problem and cre…

Count the number of prime numbers less than a non-negative number, n public int countPrimes(int n) { ArrayList<Integer> primes = new ArrayList<Integer>(); ) , n - ); primes.add(); primes.add(); ; i <= n; i++) { boolean isPrime = true; for (…

The Android University ACM Team Selection Contest Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述  Now it's 20000 A.D., and the androids also participate in the ACM Inter-national Collegiate Programming Contest (ACM/ICPC). In order to sele…