Prime Gap

最新推荐文章于 2024-02-04 09:43:07 发布

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最新推荐文章于 2024-02-04 09:43:07 发布

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Prime Gap

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 40 Accepted Submission(s) : 31

Problem 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 + nis 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

<p>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.</p>

Output

<p>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.</p>

Sample Input

Sample Output

4
0
6
0
114
题意：给一个n  是素数直接输出0
不是素数则输出比他大的最小的素数与比他小的最大的素数的差 
思路：筛素数
之前有简单看过欧拉筛法和埃拉托斯特拉筛法
埃拉托斯特拉筛法就是找到第一个新素数然后将其倍数全部筛掉，做很多无用功，但是足够了
欧拉筛法的时间复杂度更小 具体问度娘
int Eratosthenes (int n){  
        int i, j, k;  
        phi[1] = 1;  
        for (i = 2; i < n; ++i){  
                if (!vis[i]) p[cnt++] = i;  
                for (j = i; j < n; j += i) {  
                        if (!phi[j]) phi[j] = j;  
                        phi[j] = phi[j] / i * (i - 1);  
                        vis[j] = true;  
                }  
        }  
        return cnt;  
}  
Source Code#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int p[1299709+10];
void find()
{
    int i,j;
    for(i=2;i<1299709+10;i++)
    {
        if(!p[i])
        {
            for(j=i+i;j<1299709+10;j+=i)
            {
                p[j]=1;
                }
            }
    }
    }
int judge(int x)
{
    for(int i=2;i<x;i++)
    {
        if(x%i==0)
        return 0;
        }
    return 1;
    }
int main()
{
    find();
    int i,j;
    //cout<<judge(11)<<endl;
    //for(i=1;i<20;i++)
    //cout<<i<<" "<<p[i]<<endl;
    int n;
    int num1,num2;
    while(cin>>n)
    {
        if(n==0)break;
        if(p[n]==0)
        cout<<0<<endl;
        else
        {
            for(i=n;i<1299709+10;i++)
            {
                if(p[i]==0)
                {
                    num1=i;
                    break;
                    }
                }
            for(i=n;i>=1;i--)
            {
                if(p[i]==0)
                {
                    num2=i;
                    break;
                    }
                }
            cout<<num1-num2<<endl;
            }
        }
    return 0;
    }