PAT 1015

1015. Reversible Primes (20)

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 10⁵) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line "Yes" if N is a reversible prime with radix D, or "No" if not.

Sample Input:

73 10 
23 2 
23 10 
-2 

Sample Output:

Yes 
Yes 
No 

采用素数筛选法，若$i$为素数，则$i*j$不是素数，其中$j=2,3,....$。这样可以不用判断哪个数为素数，因为非素数的都必定会被筛选出来。

代码

#include <stdio.h>
#include <math.h>

char primerTable[500000];

void calculatePrimerTable();
int num2array(int,int,int*);
int array2num(int*,int,int);
int main()
{
    calculatePrimerTable();
    int N,D,len;
    int data[32];
    while(scanf("%d",&N)){
        if(N < 0)
            break;
        scanf("%d",&D);
        if(primerTable[N] == 'N'){
            printf("No\n");
            continue;
        }
        len = num2array(N,D,data);
        if(primerTable[array2num(data,len,D)] == 'Y')
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

void calculatePrimerTable()
{
    int i;
    for(i=2;i<500000;i++){
        if(primerTable[i] != 'N'){
            primerTable[i] = 'Y';
            int j,n;
            for(j=2,n=2*i;n<500000;++j,n=j*i){
                primerTable[n] = 'N';
            }          
        }
    }
}

int num2array(int n,int base,int *s)
{
    if(n < 0)
        return 0;
    else if(n == 0){
        s[0] = 0;
        return 1;
    }
    int len = 0;
    while(n){
        s[len++] = n % base;
        n = n / base;
    }
    return len;
}

int array2num(int *s,int len,int base)
{
    int i,n = 0;
    for(i=0;i<len;++i){
        n = n * base + s[i];
    }
    return n;
}