1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

Sample Output:

Yes
Yes

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 1000010
int a[MAXN];
long long y=0;
typedef long long ll;
bool is_prime[MAXN];
bool is_prime_small[MAXN];
void segment_sieve()//素数打表
{
    for(ll i=0; i*i<=MAXN; ++i)
        is_prime_small[i]=true;
    for(ll i=0; i<=MAXN; ++i)
        is_prime[i]=true;
    is_prime[0]=is_prime[1]=false;//0,1不是素数
    for(ll i=2; i*i<=MAXN; ++i)
    {
        if(is_prime_small[i])
        {
            for(ll j=2*i; j*j<=MAXN; j+=i)
                is_prime_small[j]=false;
            for(ll j=max(2LL,(i-1)/i)*i; j<=MAXN; j+=i)
                is_prime[j]=false;
        }
    }
}
void changeR(int x,int r)//逆置后转十进制判断是否素数
{
    y=0;
    int cnt=0;
    while(x>0)
    {
        a[cnt++]=x%r;
        x/=r;
    }
    int temp=1;
    for(int i=cnt-1; i>=0; --i)
    {
        y+=(a[i]*temp);
        temp*=r;
    }
    //cout<<y<<endl;
}
int main()
{
    segment_sieve();
    int x,r;
    while(cin>>x)
    {
        if(x<0) break;
        cin>>r;
        memset(a,0,sizeof(a));
        changeR(x,r);
        if(is_prime[x]&&is_prime[y]) cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}