usaco 1.5 pprime...

hint1就是解决问题的关键：Generate the palindromes and see if they are prime.

有目的的生成要比无目的逐一验证更有效率，当然前提是得存在有效的生成方法。

/*
    ID:chenjiong
    PROG:pprime
    LANG:C++
*/

#include <stdio.h>
#include <string.h>
#include <math.h>

const int MAXN = 20000;

int palin[MAXN];
int cnt;
int a,b;

void p1()
{
    palin[cnt++] = 5;
    palin[cnt++] = 7;
}

void p2()
{
    palin[cnt++] = 11;
}

void p3()
{
    int i,j;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            palin[cnt++] = i * 100 + j * 10 + i;
}

void p4()
{
    int i,j;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            palin[cnt++] = i * 1000 + j * 100 + j * 10 + i;
}

void p5()
{
    int i,j,k;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            for ( k = 0; k <= 9; k++)
                palin[cnt++] = i * 10000 + j * 1000 + k * 100 + j * 10 + i;
}

void p6()
{
    int i,j,k;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            for ( k = 0; k <= 9; k++)
                palin[cnt++] = i * 100000 + j * 10000 + k * 1000 + k * 100 + j * 10 + i;
}

void p7()
{
    int i,j,k,w;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            for ( k = 0; k <= 9; k++)
                for ( w = 0; w <= 9; w++)
                    palin[cnt++] = i * 1000000 + j * 100000 + k * 10000 + w * 1000 + k * 100 + j * 10 + i;
}

void p8()
{
    int i,j,k,w;
    for ( i = 1; i <= 9; i += 2)
        for ( j = 0; j <= 9; j++)
            for ( k = 0; k <= 9; k++)
                for ( w = 0; w <= 9; w++)
                    palin[cnt++] = i * 10000000 + j * 1000000 + k * 100000 + w * 10000 + w * 1000 + k * 100 + j * 10 + i;
}

void palindrome(int x)
{
    switch ( x )
    {
        case 1 : p1(); break;
        case 2 : p2(); break;
        case 3 : p3(); break;
        case 4 : p4(); break;
        case 5 : p5(); break;
        case 6 : p6(); break;
        case 7 : p7(); break;
        case 8 : p8(); break;
    }
}

bool is_prime(int a)
{
    if ( a == 1 )
        return false;
    if ( a == 2 || a == 3 )
        return true;
    if ( a % 2 == 0 )
        return false;
    int i;
    for ( i = 3; i <= sqrt(a); i += 2)
        if ( a % i == 0 )
            return false;
    return true;
}

int digit_num(int a)
{
    int ans = 0;
    while ( a > 0 )
    {
        a /= 10;
        ans++;
    }
    return ans;
}

int main()
{
    freopen("pprime.in","r",stdin);
    freopen("pprime.out","w",stdout);

    scanf("%d%d",&a,&b);
    int s = digit_num(a);
    int d = digit_num(b);
    
    int i,j;    
    cnt = 0;
    for ( i = s; i <= d; i++)
        palindrome(i);

    for ( j = 0; j < cnt; j++)
    {
        if ( a <= palin[j] && palin[j] <= b )
        {
            if ( is_prime(palin[j]) )
                printf("%d\n",palin[j]);
        }
    }

    return 0;
}