1015 Reversible Primes (20 分)可逆质数

1015 Reversible Primes (20 分)

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 (<10​5​​) 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

思路：

1.题目中有多个输入但没有明确输入的个数，并且有负数来表示输入的结束，可以用while(scanf("%d",&n)!=EOF)来控制输入的结束。

2.输入的n为负数，直接跳出while循环，因为输入已到达尾部。

判断输入的n是否为质数，不是输出No，是则进行下一步。

3.将n转换为radix进制存入d[]数组，再将d[]数组从0到len-1   进行 n=n*radix+d[i]转换为逆的radix进制n。

4.判断n是否为质数。

#include <cstdio> #include <cmath> bool isprime(int c){ if(c<=1)return false; int sqr=(int)sqrt(c); for(int i=2;i<=sqr;i++){ if(c%i==0)return false; } return true; } int main(){ int n,radix; int d[100]; while(scanf("%d",&n)!=EOF){ if(n<0)break; scanf("%d",&radix); if(isprime(n)==false){printf("No\n");} else { int len=0; while(n!=0){ d[len++]=n%radix; n/=radix; } for(int i=0;i<len;i++){ n=n*radix+d[i]; } if(isprime(n)==false)printf("No\n"); else printf("Yes\n"); } } return 0; }