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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
思路:此题有教训 ，一个是判断素数的时候要<=sqrt 另外一个需要注意的是在反转的时候也可能出现1所以判断素数的时候要注意<=1的情况，要将此情况写到函数中去。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std; 5 int str[200]; 6 bool Judge(int a) 7 { 8 if(a<=1) 9 return false; 10 int sqr=(int)sqrt(1.0*a); 11 for(int i=2;i<=sqr;i++) //此处有问题需要注意 12 { 13 if(a%i==0) 14 return false; 15 } 16 return true; 17 } 18 //进制转换 19 int Change(int a,int d) 20 { 21 int index=0; 22 while(a!=0) 23 { 24 int remain=a%d; 25 str[index++]=remain; 26 a=a/d; 27 } 28 //转换为十进制 29 int sum=0; 30 for(int i=0;i<index;i++) 31 { 32 sum=sum*d+str[i]; 33 } 34 return sum; 35 } 36 int main(int argc, char *argv[]) 37 { 38 int N,D; 39 while(scanf("%d",&N)!=EOF) 40 { 41 if(N<0) 42 break; 43 scanf("%d",&D); 44 if(!Judge(N)) 45 { 46 printf("No\n"); 47 continue; 48 } 49 int after=Change(N,D); 50 if(!Judge(after)) 51 { 52 printf("No\n"); 53 continue; 54 } 55 printf("Yes\n"); 56 } 57 return 0; 58 }
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原文地址：http://www.cnblogs.com/GoFly/p/4296839.html
