4190. Prime Palindromes 一亿以内的质数回文数

Description

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 .

Input

There are multiple test cases.

Each case contains two integers, a and b.

a=b=0 indicates the end of input.

Output

For each test case, output the list of palindromic primes in numerical order, one per line.

 

Sample Input

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5 500

0 0

Sample Output

5

7

11

101

131

151

181

191

313

353

373

383



题目链接：4190. Prime Palindromes
题意分析：求给定区间内的质数回文数
题目分析：
1.多组测试数据，所以先打表。
2.先求质数再判断回文，效率低下；所以先构造回文数，再判断质数。
3.偶数位的回文数都能被11整除，自己证明去。所以，偶数位的回文数除了11都是合数。
4.一个k位数，可以构造出一个奇数位的回文数。比如13，可以构造131；189可以构造18981.所以100000000内的只要从1构造到9999即可。
5.若范围为1000000000，那么明显超出int范围，要用long long。当然此题无此陷阱。

//http://soj.me/show_problem.php?pid=4190

#include <iostream>

#include <cstdio>

#include <vector>

#include <cmath>

#include <algorithm>

using namespace std;



vector<int> v;



bool prime(int x)

{

    for (int i = 2; i < sqrt(x+0.5); i++)

        if (x%i == 0)

            return 0;

        

    return 1;

}



void get()

{

    // 11是唯一的偶数位的质数回文数。

    v.push_back(11);

    

    //构造奇数位的回文数

    for (int i = 2; i <= 10000; ++i)

    {

        int tmp = i/10, sum;

        

        for (sum = i; tmp != 0; tmp /= 10)

        {

            sum = sum*10 + tmp%10;

        }

        

        if (prime(sum))

            v.push_back(sum);

    }

}

int main()

{

    get();

    sort(v.begin(), v.end());    //因为是不按顺序push，所以要sort

    int a, b;

    while(cin >> a >> b, a||b)

    {

        for (int i = 0; i < v.size(); ++i)

        {

            if (v[i] >= a)

            {

                if (v[i] > b)

                    break;

                

                printf("%d\n", v[i]);

            }

        }

    }

}

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