2570 absolute prime number, 2570 Prime Number

2570 absolute prime number, 2570 Prime Number
2570 absolute Prime Number

 

Time Limit: 1 s space limit: 128000 KB title level: Gold Title Description Description

A natural number is a prime number, and Its numeric position is still a prime number after any change, it is called an absolute prime number, such as 13. Find the number of absolute prime numbers of all x digits.

Input description Input Description

Input positive integer x

Output description Output Description

Number of absolute prime numbers of x digits

Sample Input Sample Input

1

Sample output Sample Output

4

Data range and prompt Data Size & Hint

X (x <= 6)

CATEGORY tag Tags click here to expand

Evaluate the prime number by DFS + sieve

Note the use of next_permutation and the control of the I boundary.

You only need to judge the number generated each time.

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 const int MAXN=100001; 8 int vis[MAXN]; 9 int bc[MAXN];10 int now=1;11 int tot=0;12 int f(int n)13 {14     if(n==0)return 1;15     else return n*f(n-1);16 }17 int main()18 {19     int n;20     scanf("%d",&n);21     double bg=pow(10,n-1);22     double ed=pow(10,n);23     vis[1]=1;24     for(int i=2;i<=sqrt(ed);i++)25     {26         if(vis[i]==0)27         {28             for(int j=i*i;j<=ed;j=j+i)29             {30                 vis[j]=1;31             }32         }33     }34     for(int i=(int)bg;i<=(int)ed-1;i++)35     {36         int flag=0;37         int p=i;38         if(vis[p]==1)continue;39         memset(bc,0,sizeof(bc));40         now=1;41         while(p!=0)42         {43             bc[now]=p%10;44             now++;45             p=p/10;46         }47         for(int i=1;i<=f(now-1);i++)48         {49             next_permutation(bc+1,bc+now);50             int ans=0;51             for(int j=1;j<=now-1;j++)52             {53                 ans=ans*10+bc[j];54             }55             if(vis[ans]==1)56             {57                 flag=1;58                 break;59             }60         }61         if(flag==0)62         tot++;63     }64     printf("%d",tot);65     return 0;66 }