Sdut 3305 Prime number in prime (prime sieve)

Prime numbers in prime numbers Time limit:1000ms Memory limit:65536k have questions? Dot here ^_^ Title Description If a prime number, it is also prime in prime numbers, which is called prime numbers in prime numbers. For example 3 is a 2nd prime number, so 3 is prime number, although 7 is a prime number, but 7 in 4th place, so 7 is not prime numbers in the prime to give you a count of N, to find out the >=n of a minimum prime number of prime numbers in the input input one n (0<=n <= 10^6)

Prime number in the smallest prime number of the output >=n

Sample input

234

Sample output

335

Hint Source

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Cmath>using namespace Std;int b[1500001];int n;int vis[5000010];int a[1500001];int pan ()                     //prime sieve {    int m=sqrt ( 2000000+0.5);    memset (vis,0,sizeof (Vis));    for (int i=2;i<=m;i++)        if (!vis[i]) for    (int j=i*i;j<=2000000;j+=i)             vis[j]=1;} int P () {    int num=85714;<span id= "Transmark" ></span> for    (int i=1100003;i>=2;i--)    {        if (vis[i]==0)        {            b[num]=i;            if (vis[num]==0)            {              a[i]=num;            }            else                a[i]=a[i+1];            num--;        }        else            {a[i]=a[i+1];}}}    int main () {    pan ();    P ();   while (~SCANF ("%d", &n))   {       if (n==0| | n==1)        printf ("2\n");      cout<<a[n];      else      printf ("%d\n", B[a[n]]);   }    

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Sdut 3305 Prime number in prime (prime sieve)