PAT-甲级-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 1023 223 10-2

Sample Output:

YesYesNo

【解析】题目的意思就是叫你判断输入的n和n经过m进制转换后得到的数是不是素数如果都是素数就输出yes否则就是no所以我们就要注意一下进制转换的概念了，排列的话是越先算出来的越在后面。
#include<cstdio>#include<iostream>using namespace std;int isprime(int n){    int i;    if(n<2)        return 0;    for(i=2;i*i<=n;i++)//判断素数    {        if(n%i==0)            return 0;    }    return 1;}int change(int n,int d){    int i,j,a[110]={0};    int cnt=0,sum=0,temp;    while(n)    {        a[cnt++]=n%d;        n=n/d;    }    for(i=cnt-1;i>=0;i--)    {        temp=1;        for(j=0;j<cnt-1-i;j++)//进制转换，数组当中越在前面的d的指数越大        {            temp=temp*d;        }    sum=sum+a[i]*temp;    }    return sum;}int main(){    int i,j,n,d,m;    while(scanf("%d",&n))    {        if(n<0)            break;        scanf("%d",&d);        m=change(n,d);        if(isprime(n)==1&&isprime(m)==1)            printf("Yes\n");        else            printf("No\n");    }    return 0;}