直接判断数据过大
相对而言，由2，3，5，7组成的数更少，则先筛选出由2,3,5,7组成的数，再判断这些数中的质数个数即可

#include <iostream>
using namespace std;
int main()
{
  printf("1903");
  return 0;
}

#include <iostream>
#include<cmath>
using namespace std;
typedef long long ll;
const ll n = 20210605LL;
ll h[n + 10] = {0}, p[n + 10];
int main(){
  ll num = 0, count = 0, t, t1, sqr, flag;
  for(ll i = 2; i <= n; i++){
  	t1 = i;
  	while(t1){
  		t = t1 % 10;
  		if(t == 2 || t == 3 || t == 5 || t == 7){
  			t1 /= 10;
		  }
		else break;
	}
	if(t1 == 0) p[num++] = i;
  }
  for(ll i = 0; i < num; i++){
  	sqr = (ll)sqrt(p[i]*1.0);
  	flag = 0;
  	for(ll j = 2; j <= sqr; j++){
  		if(p[i] % j == 0) {
  			flag = 1;
  			break;
		  }
	  }
	  if(!flag) {
	  	count++;
	  	printf("%lld,", p[i]);
	  }
  }
  printf("%lld", count);
  return 0;
}