primes

joj2573: Product of two primes

  2573: Product of two primes Result TIME Limit MEMORY Limit Run Times AC Times JUDGE 3s
·2015-11-09 11:53

UVA 113 Power of Cryptography (数学)

(among other things) large prime numbers and computing powers of numbers modulo functions of these primes
·2015-11-08 16:09

位操作与空间压缩

#include <stdio.h> #include <memory.h> const int MAXN = 100; bool flag[MAXN]; int primes
·2015-11-08 15:58

Power of Cryptography

cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes
·2015-11-08 15:49

uva 10236 The Fibonacci Primes

数论题、 参考了  http://blog.csdn.net/clevermike/article/details/8177450 这道题的详细分析在黑书中（221页，第2章，数学方法与常见模型） 先看代码，代码中有注释 //费波那列素数定理 //费波那列素数：若某个费波那列数和比它小的所有费波那列数互质，则称它为费波那列素数 //若a是b的倍数，则fa是fb的倍数 //
·2015-11-08 14:08

USACO6.4-The Primes

The Primes IOI'94 In the square below, each row, each column and the two diagonals can be read as
·2015-11-08 14:22

hduoj 4715 Difference Between Primes 2013 ACM/ICPC Asia Regional Online —— Warmup

pid=4715   Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)   &
·2015-11-08 11:57

（Problem 47）Distinct primes factors

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 =
·2015-11-08 09:39

（Problem 37）Truncatable primes

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
·2015-11-08 09:38

（Problem 35）Circular primes

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71,
·2015-11-08 09:37

ACM/ICPC 2011 Asia-Amritapuri Site / E Distinct Primes（求数的素因子）

International Collegiate Programming Contest, Asia-Amritapuri Site, 2011   Problem E: Distinct Primes
·2015-11-07 15:27

Power of Cryptography

cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes
·2015-11-07 11:26

Count Primes

Count Primes Description: Count the number of prime numbers less than a non-negative number, n
·2015-11-07 11:47

(easy)LeetCode 204.Count Primes

Description: Count the number of prime numbers less than a non-negative number, n. Credits:Special thanks to @mithmatt for adding this problem and creating all test cases. 解析：大于1的自然数
·2015-11-06 07:28

Leetcode（204） Count Primes

题目Description:Countthenumberofprimenumberslessthananon-negativenumber,n.Credits:Specialthanksto@mithmattforaddingthisproblemandcreatingalltestcases.Hint:Let’sstartwithaisPrimefunction.Todetermineifanu
fly_yr·2015-11-05 15:00

hdu 5104 Primes Problem

pid=5104  Primes Problem Description Given a number n, please count how many tuple$(p_1, p_2, p
·2015-11-03 22:34

POJ2109——贪心——Power of Cryptography

cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes
·2015-11-03 22:12

欧拉计划 第10题

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.Find the sum of all the primes below two million
·2015-11-02 19:12

欧拉计划 第3题

  class Program { private static List<long> primes = new List&l
·2015-11-02 19:09

Count Primes

Count Primes 问题： Count the number of prime numbers less than a non-negative number, n 思路： 　　
·2015-11-02 14:43

Count Primes

https://leetcode.com/problems/count-primes/ Description: Count the number of prime numbers less than
·2015-11-02 12:56

UVa 1213 (01背包变形) Sum of Different Primes

题意： 选择K个质数使它们的和为N，求总的方案数。 分析： 虽然知道推出来了转移方程， 但还是没把代码敲出来，可能基本功还是不够吧。 d(i, j)表示i个素数的和为j的方案数，则 d(i, j) = sigma d(i-1, j-p[k]) ，其中p[k]表示第k个素数 注意递推的顺序是倒着推的，否则会计算重复的情况。 代码中第二重和第三重循环的顺序可互换。 1 #inc
·2015-11-02 11:50

[Project Euler]加入欧拉 Problem 10

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
·2015-11-02 09:39

求给定数目的前 n 个素数

bool is_prime (const vector<int>& primes, int num) { for (const auto& prime : primes) {
·2015-11-01 15:46

[Project Euler] Problem 10

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
·2015-11-01 13:29

[Project Euler] Problem 27

the remarkable quadratic formula: n² + n + 41 It turns out that the formula will produce 40 primes
·2015-11-01 10:15

LeetCode Count Primes 求素数个数（埃拉托色尼筛选法）

      题意：给一个数n，返回小于n的素数个数。 思路：设数字 k =from 2 to sqrt(n)，那么对于每个k，从k2开始，在[2,n)范围内只要是k的倍数的都删掉（也就是说[k,k2)是不用理的，若能被筛掉早就被筛了，保留下来的就是素数）。最后统计一下[2,n)内有多少个还存在的，都是素数。     &n
·2015-11-01 10:55

[Leetcode] Count Primes

Description: Count the number of prime numbers less than a non-negative number, n Hint: The number n could be in the order of 100,000 to 5,000,000. click to show more hints. Credits:Spec
·2015-11-01 08:05

LeetCode Count Primes

References: How Many Primes Are There? Sieve of Eratosthenes   上面的这两个参考内容可以
·2015-10-31 18:44

Count Primes

好久不练手了，注意boolean array default 是false public class Solution { public int countPrimes(int n) { if(n<=2) return 0; boolean[] a = new boolean[n]; for
·2015-10-31 17:28

Count Primes

Description: Count the number of prime numbers less than a non-negative number, n Hint: The number n could be in the order of 100,000 to 5,000,000. C++实现代码： #include<new> #includ
·2015-10-31 14:09

POJ 3132 Sum of Different Primes ( 满背包问题)

Sum of Different Primes Time Limit: 5000MS   Memory Limit: 65536K Total Submissions
·2015-10-31 14:19

UVA 10236 The Fibonacci Primes

UVA_10236     这个题目AC得太辛苦了，也再一次加深了对浮点数误差的理解，库函数能不调还是最好别调……其实后来感觉就像UVA论坛上一个人说的，AC这个题有两个准则，第一个是要用long double，第二个是不要相信fmod、sqrt、pow等这些库函数，它们都会带来精度上的问题。     下面回到题目，这个题目涉及的知识还是满多的，又让我学
·2015-10-31 11:45

C++学习之【使用位操作符求素数分析】

gt; using namespace std; void getPrime_1() { const int MAXN = 100; bool flag[MAXN]; int primes
·2015-10-31 11:05

Count Primes - LeetCode

Description: Count the number of prime numbers less than a non-negative number, n References: How Many Primes
·2015-10-31 11:00

（Problem 10）Summation of primes

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
·2015-10-31 10:59

（Problem 37）Truncatable primes

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
·2015-10-31 10:57

（Problem 35）Circular primes

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71,
·2015-10-31 10:57

Primes on Interval

AC代码： #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn =&nbs
·2015-10-31 10:58

Reversible Primes (20)

the problem is from PAT,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1015 this problem is relatively easy.if there must be something which need to be noticed, i think “1 is not a
·2015-10-31 10:23

python例程之质数

/usr/bin/python#find primes#usage: .
·2015-10-31 10:02

BestCoder Round #18

1001 Primes Problem  打个10^4的素数表，枚举前两个搞一下就好了。
·2015-10-31 09:17

筛素数法小结

根据这样很容易写出代码，下面代码就是是筛素数法得到100以内的素数并保存到primes[]数组中。
·2015-10-31 09:53

Distinct Primes

Arithmancy is Draco Malfoy's favorite subject, but what spoils it for him is that Hermione Granger is in his class, and she is better than him at it.  Prime numbers are of mystical importance in
·2015-10-31 08:00

UVA-10168 Summation of Four Primes 哥德巴赫猜想

这题如果是要输出所有的解的情况的话，用两个有序表查找可以优化到O(n^3)，幸好这题只是要求出一种方案，那么我们就有以下结论： 当 N < 8 的时候是无解的 当 N > 8 并且 N是一个奇数的话，那么就可以拆成 2 + 5 + 一个偶数，根据哥德巴赫猜想，一个合数一定能够分解成两个素数之和，所以只要遍历一遍素数表即可 当 N > 8 并且 N是一个偶数的话，那么就可以拆
·2015-10-30 14:38

* SPOJ PGCD Primes in GCD Table （需要自己推线性筛函数，好题）

题目大意： 给定n，m，求有多少组(a,b) 0<a<=n , 0<b<=m ， 使得gcd（a，b）= p ， p是一个素数   这里本来利用枚举一个个素数，然后利用莫比乌斯反演可以很方便得到答案，但是数据量过大，完全水不过去 题目分析过程（从别人地方抄来的） ans = sigma(p, sigma(d, μ(d) * (n/pd) * (m/pd)
·2015-10-30 14:06

快速切题 sgu113 Nearly prime numbers 难度:0

KB   Nearly prime number is an integer positive number for which it is possible to find such primes
·2015-10-30 13:14

[LeetCode] Count Primes 质数的个数

References: How Many Primes Are There? Sieve of Eratosthenes Credits:Spe
·2015-10-28 08:14

[LeetCode] Count Primes

Description: Count the number of prime numbers less than a non-negative number, n. 思路：一个一个判断， 时间复杂度：O(n^2) 代码：TLE public class Solution { public int countPrimes(int n) { int c
·2015-10-27 16:57

【leetcode】Count Primes（easy）

Count the number of prime numbers less than a non-negative number, n   思路：数质数的个数 开始写了个蛮力的，存储已有质数，判断新数字是否可以整除已有质数。然后妥妥的超时了(⊙v⊙)。 看提示，发现有个Eratosthenes算法找质数的，说白了就是给所有的数字一个标记，把质数的整数倍标为false，
·2015-10-27 15:13
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