先递归求回文再判断素数。。。。。

第一章终于搞完了。。。。

The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 100,000,000); both a and b are considered to be within the range .

PROGRAM NAME: pprime

INPUT FORMAT

SAMPLE INPUT (file pprime.in)

5 500

OUTPUT FORMAT

SAMPLE OUTPUT (file pprime.out)

5
7
11
101
131
151
181
191
313
353
373
383

HINTS (use them carefully!)

/*
ID: qhn9992
PROG: pprime
LANG: C++11
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

bool isprime(int x)
{
    if(x==1) return false;
    int k=sqrt(x);
    for(int i=2;i<=k;i++)
    {
        if(x%i==0) return false;
    }
    return true;
}

int a[12],A,B,flag;

void dfs(int n,int x)
{
    if(!flag) return ;
    int ed=(n%2)? n/2+1:n/2;
    if(x==ed)
    {
        if(a[0]==0||a[n-1]==0) return ;
        int temp=0;
        for(int i=0;i<n;i++) temp=temp*10+a[i];
        if(temp<A) return ;
        if(temp>B)
        {
            flag=0;
            return ;
        }
        if(isprime(temp)) printf("%d\n",temp);
        return ;
    }
    for(int i=0;i<=9;i++)
    {
        a[x]=a[n-1-x]=i;
        if(x==0&&a[x]%2==0) continue;
        dfs(n,x+1);
    }
}

int main()
{
    freopen("pprime.in","r",stdin);
    freopen("pprime.out","w",stdout);
    scanf("%d%d",&A,&B);
    int w1=floor(log(A)/log(10))+1;
    int w2=floor(log(B)/log(10))+1;
    for(int i=w1;i<=w2;i++)
    {
        flag=1;
        dfs(i,0);
    }
    return 0;
}

USACO Prime Palindromes