C :

#include <stdio.h>
int iszs(int n){
	if(n<2){
		return 0;
	}
	int i ;
	for(i=2;i<=n-1;i++){
		if(n%i==0){
			return 0;
		}
	}
	return 1;
}
int main()
{
	int k,i,j,h,s,a,d,c;
	for(i=1001;i<3000;i++){
		k=i/1000;
		j=i/100%10;
		h=i/10%10;
		a=i%10;
		s=k*1000+j*100+h*10+a;
		c=j*100+h*10+a;
		d=h*10+a;
		if(iszs(s)==1&&iszs(c)==1&&iszs(d)==1&&iszs(a)==1){
			printf("%d\n",i);
		}
	}
	return 0;
}

C++ :

#include <iostream>
#include <cmath>
using namespace std;
//判断素数
bool sushu(int n){
	bool r=true;
	int i;
	for(i=2;i<=sqrt(n);i++){
		if(n%i==0){
			r=false;
			break;
		}

	}
	if(n<=1){
		r=false;
	}
	return r;
}

//输出1000~3000的纯粹素数
void chuncui(){
	int a,b,c,i;
	for(i=1000;i<=3000;i++){
		a=i%1000;
		b=i%100;
		c=i%10;
		if(sushu(i)==true&&sushu(a)==true&&sushu(b)==true&&sushu(c)==true){
			cout<<i<<endl;
		}
	}
} 

int main(){
	chuncui();
}
 

Python :

import math;
def suShu(n):
   f = True;
   for i in range(2,int(math.sqrt(n))+1):
          if n % i == 0:
               f = False;
               break;
   return f and n != 1;

def ccsuShu(n):
   a = n % 1000;
   b = n % 100;
   c = n % 10;
   return suShu(n) and suShu(a) and suShu(b) and suShu(c);

for i in range(1000,3000):
    if ccsuShu(i):
       print(i);