Count primes 题目连接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5901 Description Easy question! Calculate how many primes between [1...n]! Input Each line contain one integer n(1 <= n <= 1e11).Process to end of file. Output For each case, output…

题意:给求 1 - n 区间内的素数个数,n <= 1e11. 析:模板题. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstrin…

链接:传送门 题意:计算 [ 1 , n ] 之间素数的个数,(1 <= n <= 1e11) 思路:Meisell-Lehmer算法是计算超大范围内素数个数的一种算法,原理并不明白,由于英语太渣看不懂WIKI上的原理,附WIKI链接:Here /************************************************************************* > File Name: hdu5901.cpp > Author: WArobot &g…

Count primes Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2625    Accepted Submission(s): 1337 Problem Description Easy question! Calculate how many primes between [1...n]!   Input Each line…

原题地址:http://acm.hdu.edu.cn/showproblem.php?pid=5901 题意:输入n,输出n以内质数个数 模板题,模板我看不懂,只是存代码用. 官方题解链接:https://async.icpc-camp.org/d/560-2016 /************************************************************ 这个模板我一点都不会,代码是从codeforces上抄的,佚名 pi(i)表示i以内质数的个数 ******…

题目链接 题意:求[1,n]有多少个素数,1<=n<=10^11.时限为6000ms. 官方题解:一个模板题, 具体方法参考wiki或者Four Divisors. 题解:给出两种代码. 第一种方法Meisell-Lehmer算法只需265ms. 第二种方法不能运行但是能AC,只需35行. 第一种: //Meisell-Lehmer #include<cstdio> #include<cmath> using namespace std; #define LL long…

题意: 计数区间$[1, n](1 \leq n \leq 10^{11})$素数个数. 分析: 这里只介绍一种动态规划的做法. 首先要说一下[分层思想]在动态规划中非常重要,下面的做法也正是基于这一思想. 我们用$dp[i]$表示区间$[1, \frac{n}{i}]$中素数的个数,用$c[i]$表示区间$[1, i]$中素数个数. 那么我们要求的即是$dp[1]$.由于$n$最大是$10^{11}$,因此任何区间内合数的最小素因子不超过$\sqrt{10^{11}}$.为了筛选素数,只需从区…

转自:http://blog.csdn.net/chaiwenjun000/article/details/52589457 计从1到n的素数个数 两个模板 时间复杂度O(n^(3/4)) #include <bits/stdc++.h> #define ll long long using namespace std; ll f[],g[],n; void init(){ ll i,j,m; ;m*m<=n;++m)f[m]=n/m-; ;i<=m;++i)g[i]=i-; ;i…

#include<cstdio> #include<cmath> using namespace std; #define LL long long ; bool np[N]; int prime[N], pi[N]; int getprime() { ; np[] = np[] = true; pi[] = pi[] = ; ; i < N; ++i) { if(!np[i]) prime[++cnt] = i; pi[i] = cnt; ; j <= cnt &am…

题意:计算1~N间素数的个数(N<=1e11) 题解:题目要求很简单,作为论文题,模板有两种 \(O(n^\frac{3}{4} )\),另一种lehmer\(O(n^\frac{2}{3})\) link:https://zh.wikipedia.org/wiki/%E7%B4%A0%E6%95%B0%E8%AE%A1%E6%95%B0%E5%87%BD%E6%95%B0 /** @Date : 2016-11-18-13.59 * @Author : Lweleth (SoungEarlf@…

参考链接:https://blog.csdn.net/Dylan_Frank/article/details/54428481 #include <bits/stdc++.h> #define ll long long using namespace std; ll f[340000],g[340000],n; void init(){ ll i,j,m; for(m=1;m*m<=n;++m)f[m]=n/m-1; for(i=1;i<=m;++i)g[i]=i-1; for(i…

hdu5901题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5901 code vs 3223题目链接:http://codevs.cn/problem/3223/ 思路:主要是用了一个Meisell-Lehmer算法模板,复杂度O(n^(2/3)).讲道理,我不是很懂(瞎说什么大实话....),下面输出请自己改 #include<bits/stdc++.h> using namespace std; typedef long long LL;…

Count Primes Count the number of prime numbers less than a non-negative number, *n*. Example: Input: 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7. 解法1 埃氏筛法 class Solution { public: int countPrimes(int n) { if(…

263. Ugly Number 注意:1.小于等于0都不属于丑数 2.while循环的判断不是num >= 0, 而是能被2 .3.5整除,即能被整除才去除这些数 class Solution { public: bool isUgly(int num) { ) return false; == ) num /= ; == ) num /= ; == ) num /= ; ? true : false; } }; 264. Ugly Number II 用一个数组去存第n个前面的所有整数,然后…

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…

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.…

204. Count Primes Easy Count the number of prime numbers less than a non-negative number, n. Example: Input: 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7. package leetcode.easy; public class CountPrimes { publ…

title: woj1012 Thingk and Count DP好题 date: 2020-03-12 categories: acm tags: [acm,dp,woj] 难题,dp好题,几何题(?数学题) 想明白了其实思路很简单,但是因为数据量很大优化需要考虑. 滚动数组. description You are given a chessboard made up of N squares by M squares. Some of the squares are colored bl…

Queston 204. Count Primes Solution 题目大意:给一个数,求小于这个数的素数的个数 思路:初始化一个boolean数组,初始设置为true,先遍历将2的倍数设置为false,再遍历3并将3的倍数置为false... Java实现: 此法超时 public int countPrimes(int n) { // 2, 3, 5, 7 boolean[] candidate = new boolean[n]; Arrays.fill(candidate, Boolea…

这是悦乐书的第190次更新,第193篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第49题(顺位题号是204).计算小于非负数n的素数的数量.例如: 输入:10 输出:4 说明:有4个素数小于10,它们是2,3,5,7. 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法 判断一个数n是否为素数,就需要判断1到n-1之间的数能否被n整除,能够被整除说明不是素数,否则就是素数.…

解题报告:题目大意,给你n个球,要将这n个球从下到上按层次排列,要求同一个层次的的每一个分支的数量都必须相等,问有多少种排列的方法. 此题的一个DP题,假设现在有n个球,要将这n个球排列好,我们就必须将n个球的问题转化成小于n个球的子问题 ,我们可以很明显地观察到,对于每一种排列 的方法,它的最上面的那一层总是只有一个球,所以我们就可以这样出发,将这n个球先减掉一个球,还剩下n-1个球,然后就是下一层怎么排列 的问题了.由于现在还剩下n-1个球,那么我们现在要想的就是把这n-1个球应该怎么排列,…

Count The Pairs Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 277    Accepted Submission(s): 150 Problem Description   With the 60th anniversary celebration of Nanjing University of Science…

1.HDU 1251 统计难题  Trie树模板题,或者map 2.总结:用C++过了,G++就爆内存.. 题意:查找给定前缀的单词数量. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b #define F(i,a,b…

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 the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8845    Accepted Submission(s): 4104 Problem Description It is well known that AekdyCoin is good at string problems as well as n…

题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6470 Count Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 301    Accepted Submission(s): 127 Problem Description Farmer John有n头奶牛.某天奶牛想要数一数有多少头奶牛…

Jesus Is Here Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 512    Accepted Submission(s): 368 Problem Description I've sent Fang Fang around 201314 text messages in almost 5 years. Why can't…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6016 题意:给定男羊和女羊的朋友关系,即给定一个图,问从任意一只羊开始连续数四只不相同的羊的方法数. 这题挺简单的就是需要动一下脑子,我们只要找到关系A-B-C-D即可,且关系址建立在男女羊之间. 就是说我们只要找到 男羊-女羊-男羊-女羊,或者 女羊-男羊-女羊-男羊 即可. 于是那只要找到任意一种方式的总数然后乘2即可. 拿后一种关系为例. ans[i]表示i号女羊有几个男羊朋友vc[i]存储与…

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…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=2222 AC自动机模板题 我现在对AC自动机的理解还一般,就贴一下我参考学习的两篇博客的链接: http://blog.csdn.net/niushuai666/article/details/7002823 http://www.cppblog.com/menjitianya/archive/2014/07/10/207604.html #include<stdio.h> #include<s…