USACO Prime Palindromes 构造回文数

这道题目一点也不卡素数的判断

就是朴素的sqrt(n) 也不卡

所以~放心的用吧。

 

构造回文的时候看了HINT

其中是这么写的：

Generate palindromes by combining digits properly. You might need more than one of the loops like below.



/* generate five digit palindrome: */

for (d1 = 1; d1 <= 9; d1+=2) {	/* only odd; evens aren't so prime */

    for (d2 = 0; d2 <= 9; d2++) {

        for (d3 = 0; d3 <= 9; d3++) {

	    palindrome = 10000*d1 + 1000*d2 +100*d3 + 10*d2 + d1;

	    ... deal with palindrome ...

	}

    }

}

　

Source Code:

/*

ID: wushuai2

PROG: pprime

LANG: C++

*/

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <list>

#include <queue>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int M = 660000         ;

const ll P = 10000000097ll   ;

const int INF = 0x3f3f3f3f   ;

const int MAX_N = 20         ;

const int MAXSIZE = 101000000;



bool primei(int n){

    int i, j;

    for(i = 2; i <= sqrt(n); ++i){

        if(n % i == 0)  return false;

    }

    return true;

}



ll b[10000]={5,7,11};

int p[8]={4,2,4,2,4,6,2,6};

bool prime(int n){

    int i = 7, j, q;

    if(n == 1)  return false;

    if(n == 2 || n == 5 || n == 3)  return true;

    if(n % 2 == 0 || n % 3 == 0 || n % 5 == 0)  return false;

    q = (int)sqrt((double)n);

    for(; i <= q; ){

        for(j = 0; j < 8; ++j){

            if(n % i == 0)  return false;

            i += p[j];

        }

        if(n % i == 0)  return false;

    }

    return true;

}

int creat(){

    int i, j, k, l, m, count = 3;

    for(i = 1; i <= 9; i += 2)

        for(j = 0; j <= 9; ++j)

            b[count++] = 100 * i + 10 * j + i;

    for(i = 1; i <= 9; i += 2)

        for(j = 0; j <= 9; ++j)

            for(k = 0; k <= 9; ++k)

              b[count++] = 10000 * i + 1000 * j + k * 100 + j * 10 + i;

        for(i = 1; i <= 9; i += 2)

         for(j = 0; j <= 9; ++j)

            for(k = 0; k <= 9; ++k)

                for(l = 0; l <= 9; ++l)

                  b[count++] = 1000000 * i + 100000 * j + k * 10000 + l * 1000 + k * 100 + j * 10 + i;

        return count - 1;

}



int main() {

    ofstream fout ("pprime.out");

    ifstream fin ("pprime.in");

    int i, j, k, t, n, s, c, w, q;

    int a;

    n = creat();

    fin >> a >> c;

    for(i = 0; i < n; ++i){

        if(b[i] >= a){

            if(b[i] > c) break;

            if(prime(b[i])){

                fout << b[i] << endl;

            }

        }

    }



    fin.close();

    fout.close();

    return 0;

}