首页 > 其他分享 >POJ 3518 Prime Gap 素数打表

POJ 3518 Prime Gap 素数打表

时间:2023-02-07 12:36:46浏览次数:55  
标签:Prime prime 3518 int positive Gap length integer include


算法分析:

题意:

水题一枚:题意找到包含所给数的一个数列,该数列满足在两个相邻质数之间,数列包含最大的质数,举个例子,10,则7,8,9,10,11,则数列为8,9,10,11

分析

直接打表欧克

代码实现

Prime Gap

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 10711 Accepted: 6163

Description

The sequence of n − 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, ‹24, 25, 26, 27, 28› between 23 and 29 is a prime gap of length 6.

Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.

Input

The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.

Output

The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.

Sample Input

10
11
27
2
492170
0

Sample Output

4
0
6
0
114

Source

​Japan 2007​

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
const long N = 1399709;
int cnt[N];
long prime[N] = {0},num_prime = 0;
int isNotPrime[N] = {1, 1};
int Prime()
{
for(long i = 2 ; i < N ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
//关键处1
for(long j = 0 ; j < num_prime && i * prime[j] < N ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j] ) ) //关键处2
break;
}
}
return 0;
}
int main()
{
Prime();
int n;
while(scanf("%d",&n)!=EOF&&n!=0)
{
int ans=0;

for(int i=0;i<=1299709;i++)
{
if(prime[i]>n)
{
ans=prime[i]-prime[i-1];
break;
}
if(prime[i]==n)
{
ans=0;break;
}
}
printf("%d\n",ans);
}
return 0;
}

 

标签:Prime,prime,3518,int,positive,Gap,length,integer,include
From: https://blog.51cto.com/u_14932227/6041964

相关文章

  • POJ 3126 Prime Path 素数+BFS
    PrimePathTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 27754 Accepted: 15152DescriptionTheministersofthecabinetwerequiteupsetbythem......
  • C Primer Plus 第6版 电子书 pdf
    作者:普拉达(StephenPrata)出版社:人民邮电出版社出品方:异步图书副标题:第六版原作名:CPrimerPlus:6th译者:姜佑 关注公众号:红宸笑。回复:电子书即可 ......
  • C++ Primer 5th 阅读笔记:入门指南
    学习方法Thewaytolearnanewprogramminglanguageistowriteprograms.学习一门新编程语言的方式是编写程序。函数(Function)函数的四部分:返回类型;函数......
  • 题解 ARC155D Avoid Coprime Game
    题解ARC155DAvoidCoprimeGame题意给定一个可重集\(S\),保证\(\gcd_{x\inS}(x)=1\),维护一个初始为\(0\)的整数\(G\),双方轮流操作,每次每人选择\(S\)中一个数......
  • C++ Primer 5th 阅读笔记:前言
    机器效率和编程效率Itsfocus,andthatofitsprogrammingcommunity,haswidenedfromlookingmostlyatmachineefficiencytodevotingmoreattentiontoprogram......
  • 2018南京Gym - 101981J - Prime Game(计数)
    第一个元素的素因子2:它能贡献的区间有[1,1],[1,2],……,[1,10]10个区间第一个元素的素因子3:它能贡献的区间有[1,1],[1,2],……,[1,10]10个区间当前sum=10+10第二个元素......
  • C Primer Plus (7.12) 編程練習
    /*CPrimerPlus(7.11)3*/1#include<stdio.h>2intmain()3{4doubleweight,height;5printf("Pleaseenteryourweightandheight.\n");6......
  • A Primer on Boolean Satisfiability 布尔满足性入门介绍
    SAT问题:给定一个布尔表达式,看是否有一组赋值使得表达式为true.SAT属于典型的NP完全问题,决定SAT的最坏情况是2的n次方,尝试n个变量的所有可能的组合。涉及到的名词:语法:命......
  • 「解题报告」ARC136E Non-coprime DAG
    很妙啊这题。我们来分析\(x\)能到\(y\)的数有什么性质。既然是不互质,那么可以考虑看这个数的最小质因子是什么。记\(f(x)\)为\(x\)的最小质因子。我们将质因子......
  • 【SpringApplication】源码之【StackWalker】
     问题:SpringBoot是如何找到main方法的启动类的?  我们在SpringApplication275行看到有一个“探测Main”的方法,其中他使用了Java9的新特性:StackWalker。图1StackW......