USACO chapter 1 section 1..5 SuperPrime Rib

USACO chapter 1 section 1..5 SuperPrime Rib

USER: tianbing tianbing [tbbd4261]
TASK: sprime
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.011 secs, 2932 KB]
Test 2: TEST OK [0.000 secs, 2932 KB]
Test 3: TEST OK [0.000 secs, 2932 KB]
Test 4: TEST OK [0.011 secs, 2932 KB]
Test 5: TEST OK [0.011 secs, 2932 KB]
All tests OK.
Your program ('sprime') produced all correct answers!  This is your
submission #7 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
4
------- test 2 -------
5
------- test 3 -------
6
------- test 4 -------
7
------- test 5 -------
8

Keep up the good work!

Thanks for your submission!

代码：原来写的代码在怎么优化只能过四组数据。
             查了一下才知道素数有如下规律：
（来自NOCOW）            

从数学的角度：
1.首位只能是质数2 3 5 7

2.其余位只能是1，3，7，9

3.若n=1，直接输出2，3，5 ，7
这样很快的就过了

/*
ID:tbbd4261
PROG:sprime
LANG:C++
*/

#include < iostream >
#include < fstream >
#include < cmath >
#include < time.h >
using   namespace  std;
int  n;
int  isprime( int  n)
{    if (n < 2 ) return   0 ;
     if (n == 2 ) return   1 ;
     if (n % 2 == 0 ) return   0 ;
     for ( int  i = 3 ; i <= sqrt(n); i += 2 )
       if (n % i == 0 ) return   0 ;
     return   1 ;
}

void  dfs( int  s, int  k,ofstream  & cout)
{
  int  i;
  if ( ! isprime(s)) return ;
  if (k == n)cout << s << endl;
  else   for (i = 1 ; i <= 9 ; i += 2 )
       dfs(s * 10 + i,k + 1 ,cout);
}

int  main()
{
    ifstream cin( " sprime.in " );
    ofstream cout( " sprime.out " );
    int  t1 = 1 ,t2;
   cin >> n;
   dfs( 2 , 1 ,cout);dfs( 3 , 1 ,cout);dfs( 5 , 1 ,cout);dfs( 7 , 1 ,cout);
  
    // system("pause");
     return   0 ;
}