HDU 1431 素数回文(打表)

素数回文 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16056    Accepted Submission(s): 3564     Problem Description xiaoou33对既是素数又是回文的数特别感兴趣。比如说151既是素数又是个回文。现在xiaoou333想要你帮助他找出某个范围内的素数回文数，请你写个程序找出 a 跟b 之间满足条件的数。(5 <= a < b <= 100,000,000);      Input 这里有许多组数据，每组包括两组数据a跟b。     Output 对每一组数据,按从小到大输出a，b之间所有满足条件的素数回文数（包括a跟b）每组数据之后空一行。     Sample Input 5 500     Sample Output 5 7 11 101 131 151 181 191 313 353 373 383     Author xiaoou333     Source zjut

 

 

 1 #include <stdio.h>  
 2 #include <stdio.h>  
 3 using namespace std;  
 4 bool a[9989900];  
 5 int prime[1005];  
 6 void set()  
 7 {  
 8     int i,j;  
 9     for(i = 2; i<=9989899; i+=2)  
10         a[i] = true;  
11     for(i = 3; i<=3161; i++){  
12         if(a[i])  
13             continue;  
14         for(j = i+i; j<=9989899; j+=i)  
15             a[j] = true;  
16     }  
17 }  
18 int huiwen(int n)  
19 {  
20     int a = 0,b = n;
21     int r;  
22     while(b){  
23         r = b%10;  
24         a = a*10+r;  
25         b/=10;  
26     }  
27     if(a == n)  
28         return 1;  
29     return 0;  
30 }   
31 int main()  
32 {  
33     int n,m,i,k = 2;  
34     set();  
35     prime[0] = 5;  
36     prime[1] = 7;  
37     for(i = 11; i<=9989899; i+=2){  
38         if(!a[i] && huiwen(i))  
39             prime[k++] = i;  
40     }  
41     while(~scanf("%d%d",&n,&m)){  
42         for(i = 0; i<k; i++){  
43             if(prime[i]<n)  
44                 continue;  
45             else if(prime[i]>m)  
46                 break;  
47             else  
48                 printf("%d\n",prime[i]);  
49         }  
50         printf("\n");  
51     }  
52     return 0;  
53 }