题目没什么好说的，我的方法中只求解了[0,3400)的素数，算得上是一个优化吧。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 10000000; 7 const int M = 8; 8 const int P = 3400; 9 bool mark[N]; 10 bool used[M]; 11 int divide[M]; 12 int digit[M]; 13 char str[M]; 14 int len, ans; 15 bool flag[P]; 16 int prime[P]; 17 int cnt; 18 19 void get() 20 { 21 memset( flag, 1, sizeof(flag) ); 22 flag[0] = flag[1] = false; 23 cnt = 0; 24 for ( int i = 2; i < P; i++ ) 25 { 26 if ( flag[i] ) 27 { 28 prime[cnt++] = i; 29 for ( int j = i + i; j < P; j += i ) 30 { 31 flag[j] = false; 32 } 33 } 34 } 35 } 36 37 bool isPrime( int a ) 38 { 39 if ( a == 0 || a == 1 ) return false; 40 for ( int i = 0; i < cnt; i++ ) 41 { 42 if ( prime[i] * prime[i] > a ) break; 43 if ( a % prime[i] == 0 ) return false; 44 } 45 return true; 46 } 47 48 void judge( int depth ) 49 { 50 int num = 0; 51 for ( int i = 0; i <= depth; i++ ) 52 { 53 num = num * 10 + digit[i]; 54 } 55 if ( isPrime(num) && !mark[num] ) 56 { 57 ans++; 58 mark[num] = 1; 59 } 60 } 61 62 void dfs( int depth ) 63 { 64 if ( depth >= len ) return ; 65 for ( int i = 0; i < len; i++ ) 66 { 67 if ( !used[i] ) 68 { 69 digit[depth] = divide[i]; 70 judge(depth); 71 used[i] = 1; 72 dfs( depth + 1 ); 73 used[i] = 0; 74 } 75 } 76 } 77 78 int main () 79 { 80 get(); 81 int t; 82 scanf("%d", &t); 83 while ( t-- ) 84 { 85 scanf("%s", str); 86 len = strlen(str); 87 for ( int i = 0; i < len; i++ ) 88 { 89 divide[i] = str[i] - '0'; 90 } 91 ans = 0; 92 memset( used, 0, sizeof(used) ); 93 memset( mark, 0, sizeof(mark) ); 94 dfs(0); 95 printf("%d ", ans); 96 } 97 return 0; 98 }
据说把1000W以内的素数表打出来都能过...