Prime Palindromes（暴力 + 剪枝）

Prime Palindromes

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

Line 1:

Two integers, a and b

SAMPLE INPUT (file pprime.in)

5 500

OUTPUT FORMAT

The list of palindromic primes in numerical order, one per line.

SAMPLE OUTPUT (file pprime.out)

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

HINTS (use them carefully!)

 

   题意：

   给出a和b，表示a到b这个范围（5<=a<b<=100000000），输出在这个范围内的所有回文串素数。

 

   思路：

   暴搜，剪枝。

   1.回文串数字的长度可能是1,2,3,4,5,6,7,8。但是长度为偶数的时候，一定能被11整除，故长度只可能是1,3,5,7，还有长度为2时的11；

   2.回文串的首尾数字必须相同，那么首尾和末尾只能是1,3,7,9，还有当只有一位数的时候的5；

   3.素数筛选，小于10000时直接判断，大于10000时用10000内的素数是否能整除这个数来判断。

 

   AC：

 

/*   
TASK:pprime   
LANG:C++   
ID:sum-g1   
*/ 
#include<stdio.h>
#include<math.h>
#include<string.h>
int a,b;
int pri[15000],prinum[15000],ans=0;

int test(int num)  //10000内的素数判断
{
    if(num == 1) return 0;
    for(int i = 2;i <= num/2;i += 1)
        if(!(num % i)) return 0;
    return 1;
}

int test2(int num)  //10000外的素数判断
{
    for(int i = 0;i < ans;i++)
    {
        if(prinum[i] > num/2) break;
        if(!(num % prinum[i])) return 0;
    }
    return 1;
}

void pal()
{
    for(int i = 1;i <= 9;i += 2)   //1位
        if(pri[i] && i <= b && i >= a) printf("%d\n",i);

    if(11 <= b && 11 >= a) printf("%d\n",11);   //2位

    for(int i = 1 ;i <= 9;i += 2)   //3位
    {
        if(i == 5) continue;
        for(int j = 0;j <= 9;j++)
        {
            int k = i * pow(10,2) + 
                    j * pow(10,1) + 
                    i * pow(10,0);
            if(k > b || k < a) continue;
            if(pri[k]) printf("%d\n",k);
        }
    }

    for(int i = 1 ;i <= 9;i += 2)  //5位
    {
        if(i == 5) continue;
        for(int j = 0;j <= 9;j++)
            for(int p = 0;p <= 9;p++)
            {
                int k = i * pow(10,4) + 
                        j * pow(10,3) + 
                        p * pow(10,2) + 
                        j * pow(10,1) + 
                        i * pow(10,0);
                if(k > b || k < a) continue;
                if(test2(k)) printf("%d\n",k);
            }
    }

    for(int i = 1 ;i <= 9;i += 2)   //7位
    {
        if(i == 5) continue;
        for(int j = 0;j <= 9;j++)
            for(int p = 0;p <= 9;p++)
                for(int q = 0;q <= 9;q++)
                {

                    int k = i * pow(10,6) + 
                            j * pow(10,5) + 
                            p * pow(10,4) + 
                            q * pow(10,3) + 
                            p * pow(10,2) + 
                            j * pow(10,1) + 
                            i * pow(10,0);
                    if(k > b || k < a) continue;
                    if(test2(k)) printf("%d\n",k);
                }
    }
}

int main()
{
    freopen("pprime.in","r",stdin);      
    freopen("pprime.out","w",stdout);
    for(int i = 1;i <= 10000;i++)
        if(test(i))
        {
            pri[i]=1;
            prinum[ans++]=i;
        }
    scanf("%d%d",&a,&b);
    pal();
    return 0;
}