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：
我的方法比较笨，没有建立新的函数，只是把验证是否为素数和倒置数值的算法全部整合到主函数里了，而且还重复了一次，就供小白们参考吧~
#include<stdio.h>
#include<iostream>
#include<math.h>using namespace std;int main()
{int n,m;scanf("%d",&n);while(n>=0){scanf("%d",&m);bool flag=false;if(n==1||n==0)flag=true;for(int k=2;k*k<=n;k++){if(n%k==0){flag=true;break;}} //注意先判断这个数本身是不是素数 int consult[100],remainder[100];//设置短除法中的商和余数记录数组 int i=0;consult[i]=n;for(;consult[i]>=m;i++){consult[i+1]=consult[i]/m;remainder[i]=consult[i]%m;}remainder[i]=consult[i]%m;int sum=0;int j=0; for(j=0;j<=i;j++){sum=sum+remainder[i-j]*pow(m,j);//求十进制下倒置的值为多少 }if(sum==1||sum==0)flag=true;for(int k=2;k*k<=sum;k++){if(sum%k==0){flag=true;break;}} if(flag)printf("No\n");elseprintf("Yes\n");scanf("%d",&n); //更新m和n的值 }return 0;}
结束。