primes

leetcode Count Primes

筛法求素数。没啥大不了的importjava.util.Arrays; publicclassSolution{ publicintcountPrimes(intn){ int[]array=newint[n]; Arrays.fill(array,1); intcount=0; for(inti=2;i
bleuesprit·2015-05-26 17:00

Reversible Primes (20)

题目如下：Areversibleprimeinanynumbersystemisaprimewhose"reverse"inthatnumbersystemisalsoaprime.Forexampleinthedecimalsystem73isareversibleprimebecauseitsreverse37isalsoaprime.Nowgivenanytwopositiveinteger
xyt8023y·2015-05-25 13:00

LeetCode204——Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,n.Credits:Specialthanksto@mithmattforaddingthisproblemandcreatingalltestcases.实现：intcountPrimes(intn){ bool*isprime=newbool[n]; for(
booirror·2015-05-24 22:00

欧拉工程第37题：Truncatable primes

题目链接：https://projecteuler.net/problem=37求出11个双剪质数的和例如：henumber3797hasaninterestingproperty.Beingprimeitself,itispossibletocontinuouslyremovedigitsfromlefttoright,andremainprimeateachstage:3797,797,97,
qunxingvip·2015-05-22 20:00

codeforces-230B T-primes

codeforces-230BT-primestimelimitpertest2secondsmemorylimitpertest256megabytes Weknowthatprimenumbersarepositiveintegersthathaveexactlytwodistinctpositivedivisors.Similarly,we’llcallapositiveintegertТ-
loy_184548·2015-05-22 16:00

欧拉工程第35题：Circular primes

题目链接：https://projecteuler.net/problem=35求100万以下循环质数的个数。注意：97，79是两个循环质数下面是两个有点不一样的方法：方法一：判断这个数是不是质数，如果是，在判断是不是循环质数。方法二：先把100万以下的质数存在布尔型数组中，是质数是true，再对每个数判断循环质数两个时间都是差不多的在600ms左右java代码：packageprojecteul
qunxingvip·2015-05-21 20:00

poj3132 Sum of Different Primes

65536KTotalSubmissions: 3293 Accepted: 2052DescriptionApositiveintegermaybeexpressedasasumofdifferentprimenumbers(primes
u010422038·2015-05-19 20:00

leetcode_Count Primes

描述：Countthenumberofprimenumberslessthananon-negativenumber, n思路：判断一个数是否为质数有两种方法，一种是判断它能否被２～(int)sqrt(n)+1之间的数整除，能被整除为合数，否则为质数但是，当n非常大的时候这种方法是非常费时间的。另外一种改进的方法是仅用n除以2~(int)sqrt(n)+1之间的所有质数，而2~(int)sqrt(
dfb198998·2015-05-17 14:00

【阔别许久的博】【我要开始攻数学和几何啦】【高精度取模+同余模定理，*】POJ 2365 The Embarrassed Cryptographer

100)与一整数L(2 <= L <= 106)，K为两个素数的乘积(The cryptographic keys are created from the product of two primes
·2015-05-13 20:00

Count Primes

References: How Many Primes Are There? Sieve of Eratosthenes 分析：运用Sieve of Eratos
·2015-05-09 12:00

LeetCode 204:Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,n分析：题目要求计算小于N的所有素数的个数。所有主程序很简单：判断是否是素数字，如是，count++（暂且忽略prim_vec,后面判断素数时会用到）：intcountPrimes(intn){ inti=0,count=0; for(i=1;iprim_vec;
sunao2002002·2015-05-08 18:00

Reversible Primes (20)

题目链接：http://www.patest.cn/contests/pat-a-practise/1015题目：时间限制400ms内存限制65536kB代码长度限制16000B判题程序Standard作者CHEN,YueA reversibleprime inanynumbersystemisaprimewhose"reverse"inthatnumbersystemisalsoaprime.F
Apie_CZX·2015-05-06 17:00

leetcode 204/187/205 Count Primes/Repeated DNA Sequences/Isomorphic Strings

一：leetcode204 CountPrimes题目：Description:Countthenumberofprimenumberslessthananon-negativenumber, n分析：此题的算法源码可以参看这里，http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes代码：classSolution{ public: intcountP
Lu597203933·2015-05-06 14:00

欧拉工程第27题：Quadratic primes

题目链接：https://projecteuler.net/problem=27n²+an+b,where|a|0){ n++; num=n*n+n*a+b; } if(n>length){ length=n; ab=a*b; //System.out.println(ab+","+a+","+b+","+length); } } } System.out.println(ab+","+lengt
qunxingvip·2015-05-05 19:00

HDU 5104 Primes Problem

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5104题面：PrimesProblemTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1776    AcceptedSubmission(s):813Proble
David_Jett·2015-05-04 23:00

Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber, nReferences:HowManyPrimesAreThere?SieveofEratosthenes题意很简单，求n以内的素数的个数注意：不包括n如果注意到reference的话，就能知道高效的求解方法如果没有注意到的话，就可能会依次判断。暴力破解就不说了
havedream_one·2015-05-03 08:00

Count Primes —— Leetcode

Description:Countthenumberofprimenumberslessthananon-negativenumber, nclicktoshowmorehints.References:HowManyPrimesAreThere?SieveofEratosthenes参考References第二个链接，wikipedia上的动图，筛选算法，最后得出素数个数，算法如下：注意int*
BlitzSkies·2015-05-02 22:00

leetcode--Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,npublicclassSolution{ publicintcountPrimes(intn){ if(n<2)return0; boolean[]flag=newboolean[n+1]; intcount=0; for(inti=2;i
kangaroo835127729·2015-05-02 15:00

Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber, nSolution1:createadicttostorethenumberofprimelessthann.TimeComplexity:O(n*n)classSolution: #@param{integer}n #@return{integer} defc
myself9711·2015-05-02 08:00

LeetCode 204 - Count Primes

: Count the number of prime numbers less than a non-negative number, n References: How Many Primes
yuanhsh·2015-05-01 02:00

LeetCode 204 - Count Primes

: Count the number of prime numbers less than a non-negative number, n References: How Many Primes
yuanhsh·2015-05-01 02:00

HDU 4715 Difference Between Primes

ProblemDescriptionAllyouknowGoldbachconjecture.Thatistosay,Everyevenintegergreaterthan2canbeexpressedasthesumoftwoprimes.Today,skywindpresentanewconjecture:everyevenintegercanbeexpressedasthedifferenc
u012970471·2015-04-29 20:00

leetcode 204: Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,n[思路]素数不能被比它小的整数整除,建一个boolean数组,从2开始,把其倍数小于N的都删掉.注意innerloop从i开始,比i小的会在以前就被check过.[CODE]publicclassSolution{ publicintcountPrimes(in
xudli·2015-04-29 08:00

Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,nclassSolution{ public: intcountPrimes(intn){ if(n<2) { return0; } boolprime[n]; memset(prime,true,n*sizeof(bool)); prime[0]=false;
brucehb·2015-04-28 22:00

leetcode 204题求素数个数

Countthenumberofprimenumberslessthananon-negativenumber,n 提示晒数法：http://en.wikipedia.org/wiki/Sieve_of_Eratostheneshttps://primes.utm.edu
wangyaninglm·2015-04-28 22:00

[LeetCode]Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber, nclicktoshowmorehints.References:HowManyPrimesAreThere?SieveofEratosthenesCredits:Specialthanksto @mithmatt foraddingthisproblemand
u014691362·2015-04-28 21:00

LeetCode #Count Primes#

一般的我的第一感觉还是搓了...时间复杂度还是比较大会写成这样:""" Programmer:EOF Date:2015.04.28 File:cp_unac.py E-mail:[email protected] """ importmath classSolution: defcountPrimes(self,n): ifn<=2: return0 elifn==3: retur
u011368821·2015-04-28 11:00

Leetcode＃204Count Primes

publicclassSolution{  publicintcountPrimes(intn){    intc=0;    if(n<=1)      returnc;    else{      for(intj=2;j<=n;j++)      {        intv=0;        for(inti=2;i<=Math.sqrt(n);i++)        {         
谧晦·2015-04-28 11:09

Leetcode＃204Count Primes

publicclassSolution{publicintcountPrimes(intn){intc=0;if(n<=1)returnc;else{for(intj=2;j<=n;j++){intv=0;for(inti=2;i<=Math.sqrt(n);i++){if(j%i==0){v=1;break;}}if(v==0)c++;}returnc;}}}提交算法后显示Status:Time
谧晦·2015-04-28 11:09

[LeetCode] Count Primes

Description:Countthenumberofprimenumberslessthananon-negativenumber,n解题思路采用Eratosthenes筛选法，依次分别去掉2的倍数，3的倍数，5的倍数，……，最后剩下的即为素数。实现代码//Rumtime:83ms classSolution{ public: intcountPrimes(intn){ intcount=0;
u011331383·2015-04-28 10:00

LeetCode 204 - Count Primes

一、问题描述Description:Countthenumberofprimenumberslessthananon-negativenumber,nHint:Thenumberncouldbeintheorderof100,000to5,000,000.二、解题报告解法一：首先，我试图在遍历的过程中保存之前所有的质数，然后对于对于一个自然数N：只需用小于N的素数去除就可以了。而且，不必用从2到N
lisong694767315·2015-04-27 15:00

Primes Problem HDU　５１０４ 打表大法好啊~~

PrimesProblemTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1439    AcceptedSubmission(s):610ProblemDescriptionGivenanumbern,pleasecounthowmanytuple(p1
Litter_Limbo·2015-04-15 20:00

Primes HDU 2161

PrimesTimeLimit:1000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):7791    AcceptedSubmission(s):3195ProblemDescriptionWriteaprogramtoreadinalistofintegersanddetermi
Litter_Limbo·2015-04-07 21:00

【斐波拉契+数论+同余】【ZOJ3707】Calculate Prime S

题目大意：S[n]表示集合{1,2,3,4,5.......n}不存在连续元素的子集个数PrimeS表示S[n]与之前的所有S[i]互质;问找到大于第K个PrimeS能整除X的第一个S[n]并且输出(S
zy691357966·2015-04-03 17:00

当我真正理解素数线性筛法

参考自：点击链接主要代码：constintMAXN=10000010; boolcom[MAXN]; intprimes,prime[MAXN/10]; voidsolve(intn) { primes
u014427196·2015-03-19 22:00

NYOJ 187 快速查找素数

输入给出一个正整数数N(N #include #definemax2000020 inta[max]={1,1}; voidis_primes()//素数打表，牢记！！！
zwj1452267376·2015-03-15 18:00

hdu 2161 Primes 筛法求素数 大水题

PrimesTimeLimit:1000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):7659AcceptedSubmission(s):3130ProblemDescriptionWriteaprogramtoreadinalistofintegersanddeterminewhethero
Lionel_D·2015-03-13 13:12

hdu 2161 Primes 筛法求素数 大水题

PrimesTimeLimit:1000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):7659    AcceptedSubmission(s):3130ProblemDescriptionWriteaprogramtoreadinalistofintegersanddetermine
Lionel_D·2015-03-13 13:00

Facebook interview - Prime Number Combination Product

  public List<Integer> combinePrimeProduct(int[] primes) { Arrays
yuanhsh·2015-03-13 00:00

Facebook interview - Prime Number Combination Product

  public List<Integer> combinePrimeProduct(int[] primes) { Arrays
yuanhsh·2015-03-13 00:00

Reversible Primes (20)

http://www.patest.cn/contests/pat-a-practise/1015A reversibleprime inanynumbersystemisaprimewhose"reverse"inthatnumbersystemisalsoaprime.Forexampleinthedecimalsystem73isareversibleprimebecauseitsrever
a_big_pig·2015-03-08 22:00

Reversible Primes (20)

Areversibleprimeinanynumbersystemisaprimewhose"reverse"inthatnumbersystemisalsoaprime.Forexampleinthedecimalsystem73isareversibleprimebecauseitsreverse37isalsoaprime.NowgivenanytwopositiveintegersN( #
oFengWuYu1·2015-02-25 20:00

HDU 5104 Primes Problem（数学）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5104ProblemDescriptionGivenanumbern,pleasecounthowmanytuple(p1,p2,p3)satisfiedthatp1 #include #include intp[10017]; voidinit() { memset(p,0,sizeof(p)); f
u012860063·2015-01-25 10:00

poj3292 Semi-prime H-numbers

H-primes是只可以被1和本身整除的H-numbers(在4n+1的数域内)。H-semi-primes是有且仅有两个H-numbers因子(除了1和本身)。
ocgcn2010·2015-01-21 14:00

Facebook Hacker Cup 2015 Round 1 Homework（附带测试数据）

                                  Yourfirst-grademathteacher,Mr.Book,hasjustintroducedyoutoanamazingnewconcept—primes
lin375691011·2015-01-21 08:00

hdu 5104 Primes Problem

hdu5104 PrimesProblem#include intprime(intm) { inti; for(i=2;i*i<=m;i++) { if(m%i==0)return0; } return1; } intmain() { inta[10005],i,n,count,j,k; memset(a,0,sizeof(a)); for(i=2;i<10005;i++) { if(prime
xinag578·2014-12-06 12:00

hash实现--开放寻址方式

下面是我的简单的实现：staticconstint__stl_num_primes=28; staticconstunsignedlong__stl_prime_list[__stl_num_primes
hustyangju·2014-11-25 10:00

HDU-5104-Primes Problem （BestCoder Round #18！！）

PrimesProblemTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):817    AcceptedSubmission(s):382ProblemDescriptionGivenanumbern,pleasecounthowmanytuple(p1,p2
u014355480·2014-11-22 23:00

HDU5104-Primes Problem

题意：输入一个数字n，找出三个数字p1,p2,p3，满足p1 #include #defineMax12000 intispri[Max]; voidInitPrime() { inti,j; memset(ispri,0,sizeof(ispri)); ispri[1]=1; for(i=2;i
KJBU2·2014-11-17 20:00

HDU5104 Primes Problem（素数筛选法）

  PrimesProblem题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5104解题思路：BestCoder官方题解：首先将10000内的素数筛出来(约1000个），(p1,p2,p3)枚举三元组前两个p1,p2，可知若存在p3满足条件，必有p3=n−p1−p2，故令t=n−p1−p2。若t为不小于p2的素数，则t满足p3的条件。则答案加一。要用
piaocoder·2014-11-16 16:00
上一页 8 9 10 11 12 13 14 15 下一页
按字母分类： ABCDEFGHIJKLMNOPQRSTUVWXYZ其他