2428: Super Prime 求超级素数

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本文介绍了一个算法问题——寻找指定序号的超级素数。超级素数是指从最低位开始的所有子串均为素数的整数。文章提供了一段C++代码实现，包括素数判断、超级素数生成等关键步骤。

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2428: Super Prime

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	313	88	Standard

In these days, skywind had thought a series of new problem. But he must keep these problem for the coming important contests such as "JiLin Provincial Collegiate Programming Contest" and "The 2007 ACM Asia Programming Contest Changchun Site Network Preliminary Contest". This time, he brings out a easy problem.

Superprime is such integer that all part of it from lowest digit is prime. For example, 9547 is a superprime because 7, 47, 547 and 9547 is all prime. Given a index i, your task is to find the i-th superprime. Noting that in this problem, 1 is considered as a prime, but 2 is NOT.

Input and Output

Each line of input is a integer i, that you should print the i-th superprime in each line. The maximal superprime in output is less than 10^9;

Sample Input

1
2
212

Sample Output

1
3
9547

Problem Source: skywind

#include<iostream>
#include<cmath>
using namespace std;
int prime[40000],is[40000],flag[40000];
int pl;
int isprime(int n)
{
    double k=sqrt(n);
    for(int i=0;is[i]<=k;i++)
    {
        if(n%is[i]==0) return 0;
    }
    return 1;
}
int main()
{
    for(int i=2;i<40000;i++)
    {
        if(flag[i]==0) is[pl++]=i;
        for(int j=0;j<pl&&i*is[j]<40000;j++)
        {
            flag[i*is[j]]=1;
            if(i%is[j]==0) break;
        }
    }
    int base[10];
    base[1]=1;
    for(int i=2;i<10;i++) base[i]=base[i-1]*10;
    prime[0]=1;
    for(int i=1;i<4;i++) prime[i]=is[i];
    int temp_begin=0,temp_end=4,cnt=4;
    for(int t=2;t<10;t++)
    {
        for(int s=1;s<10;s++)
        {
            int p=s*base[t];
            for(int i=temp_begin;i<temp_end;i++)
            {
                int m=p+prime[i];
                if(isprime(m)) prime[cnt++]=m;
            }
        }
        temp_begin=temp_end;temp_end=cnt;
    }
    int n;
    while(cin>>n)
    {
        cout<<prime[n-1]<<endl;
    }
    return 0;
}