HDU-2161，判断是否为素数，素数筛法

最新推荐文章于 2019-08-11 23:53:46 发布

luer9

最新推荐文章于 2019-08-11 23:53:46 发布

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分类专栏：素数

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本文介绍了一种读取整数列表并判断每个数是否为素数的程序设计方法。通过使用素数筛算法，可以高效地找出指定范围内的所有素数。文中提供了多种实现素数筛的代码示例，并详细解释了每种方法的工作原理。

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Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers 1 and 2 are not considered primes.

InputEach input line contains a single integer. The list of integers is terminated with a number<= 0. You may assume that the input contains at most 250 numbers and each number is less than or equal to 16000.
OutputThe output should consists of one line for every number, where each line first lists the problem number, followed by a colon and space, followed by "yes" or "no".
Sample Input

Sample Output

1: no
2: no
3: yes
4: no
5: yes
6: yes

代码如下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
bool p[18000];
void isprime()//素数筛
{
   memset(p, true, sizeof(p));
   p[1] = false;
   for(int i=2;i<=16000;i++)
   {
      if(p[i]==true)
      {
         for(int j=i+i;j<=16000;j+=i)
         {
           p[j] = false;
         }
      }
   }
}
int main()
{
  int n;
  int t = 1;
  isprime();
  while(1)
  {
     scanf("%d", &n);
     if(n<=0)//结束条件
     break;
     if(p[n]==true&&n!=2)//2不是素数
     printf("%d: yes\n", t);
     else
     printf("%d: no\n", t);
     t++;
  }
   return 0;
}

大佬们的素数筛选法

①

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define ll long long
#define N 100
bool isprime[N];／／1
int pri[N];／／2
void prime()
{
    int cnt=1;
    memset(pri,0,sizeof(pri));
    isprime[1]=1;
    for(int i=2;i<=sqrt(N);i++)／／3
    {
        if(!isprime[i])
        {
            pri[cnt++]=i;／／4
            for(int k=i*i;k<=N;k+=i)／／5
            {
                isprime[k]=1;／／表示k是一个合数
            }
        }
    }
    return;
}

②

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define ll long long
#define N 100
bool isprime[N];
int pri[N];
void prime2()
{
    int cnt=1;
    memset(pri,0,sizeof(pri));
    isprime[1]=1;
    for(int i=2;i<=N;i++)
    {
        if(!isprime[i])
        {
            pri[cnt++]=i;
        }

        for(int j=1;j<cnt&&i*pri[j]<=N;j++)／／1
        {
            isprime[i*pri[j]]=1;
            if(i%pri[j]==0)／／2
                break;
        }
    }
    return;
}

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath> 
#include <cstring>
#include <string>
#define ll long long
#define maxx 100000000
using namespace std;
int pri[maxx];
bool isprime[maxx];
void prime()
{
	int cnt=1;
	memset(pri,0,sizeof(pri));
	isprime[1]=1;
	for(int i=2;i<=sqrt(maxx);i++)
	{
		if(!isprime[i])
		{
			pri[cnt++]=i;
			for(int k=i*i;k<=maxx;k+=i)
			{
				isprime[k]=1;
			}
		}
	}
}
int main()
{
	prime();
	int n,m;
	cin>>n>>m;
	for(int i=n;i<=m;i++)
	{
		cout<<pri[i]<<" ";
	}
	return 0;
}
/*1000000 168
  10000000 446
  100000000 1229
*/