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

 

 

#include <stdio.h>  
#include <stdio.h>  
using namespace std;  
bool a[9989900];  
int prime[1005];  
void set()  
{  
    int i,j;  
    for(i = 2; i<=9989899; i+=2)  
        a[i] = true;  
    for(i = 3; i<=3161; i++){  
        if(a[i])  
            continue;  
        for(j = i+i; j<=9989899; j+=i)  
            a[j] = true;  
    }  
}  
int huiwen(int n)  
{  
    int a = 0,b = n;
    int r;  
    while(b){  
        r = b%10;  
        a = a*10+r;  
        b/=10;  
    }  
    if(a == n)  
        return 1;  
    return 0;  
}   
int main()  
{  
    int n,m,i,k = 2;  
    set();  
    prime[0] = 5;  
    prime[1] = 7;  
    for(i = 11; i<=9989899; i+=2){  
        if(!a[i] && huiwen(i))  
            prime[k++] = i;  
    }  
    while(~scanf("%d%d",&n,&m)){  
        for(i = 0; i<k; i++){  
            if(prime[i]<n)  
                continue;  
            else if(prime[i]>m)  
                break;  
            else  
                printf("%d\n",prime[i]);  
        }  
        printf("\n");  
    }  
    return 0;  
}