USACO 1.5 Prime Palindromes

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!)

題解：判斷迴文素數

第一遍 用素數篩 篩出素數，然後判斷是否迴文。 超時.........

第二遍 找出迴文數 然後判斷是否爲素數 超時....

之後才發現不用挨個找回文數 直接創造迴文數 然後判斷其是否爲素數。超級快

/*
ID: cxq_xia1
PROG: pprime
LANG: C++
*/

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int a,b;
int lena,lenb;
char creatN[10];
bool isprime(int n)                     //最樸素的方法判斷素數
{
    if(n<a||n>b)
        return false;
    for(int i=2;i<=sqrt(n+0.0);i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

int getNumLen(int n)                    //計算一個數的位數
{
    int len=0;

    while(n!=0)
    {
        n/=10; len++;
    }
    return len;
}
int getStrValue(int nowLen)
{
    int value=0;
    for(int i=1;i<=nowLen;i++)
    {
     //   value+=creatN[i]*pow(10.0,i-1.0);           不是很清楚pow函數，101他算出來是100。無奈用了下面的方法
        int tmp=1;
        for(int j=1;j<i;j++)
            tmp*=10;
        value+=creatN[i]*tmp;

    }
    return value;
}
void creatPD2(int nowLen,int harfNowlen,int nowdigit)
{
    for(int i=0;i<10;i++)
    {
        if(nowdigit==1&&i==0)
            continue;
        creatN[nowdigit]=i;
        creatN[nowLen+1-nowdigit]=i;
        if(nowdigit<harfNowlen)
        {
            nowdigit++;
            creatPD2(nowLen,harfNowlen,nowdigit);
            nowdigit--;
        }
        else
        {
            int tmp=getStrValue(nowLen);
            if(isprime(tmp))
                cout << tmp << endl;
        }
    }

}

void creatPD1(int nowLen)
{
    int harfNowLen;
    memset(creatN,0,sizeof(creatN));

    harfNowLen=(nowLen+1)/2;

    creatPD2(nowLen,harfNowLen,1);

    nowLen++;
    if(nowLen>lenb)
        return;
    else
        creatPD1(nowLen);

}

int main()
{

    freopen("pprime.in","r",stdin);
    freopen("pprime.out","w",stdout);
    cin >> a>> b;
    lena=getNumLen(a); lenb=getNumLen(b);

    creatPD1(lena);             //創造迴文數
    return 0;
}