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

aaarticlea/jpeg;base64,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" alt="" /> Copy sample input to clipboard

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 = ; i < sqrt(x+0.5); i++)
        if (x%i == )
            return ;

    return ;
}

void get()
{
    // 11是唯一的偶数位的质数回文数。
    v.push_back();

    //构造奇数位的回文数
    for (int i = ; i <= ; ++i)
    {
        int tmp = i/, sum;

        for (sum = i; tmp != ; tmp /= )
        {
            sum = sum* + tmp%;
        }

        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 = ; i < v.size(); ++i)
        {
            if (v[i] >= a)
            {
                if (v[i] > b)
                    break;

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