1015 Reversible Primes (20)(20 分)提问
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
radix是进制的意思,判断一个数是否是素数,如果不是则输出no,如果是则在D进制下进行Reverse,反转之后如果还是素数,那么就输出yes.
//如何求d进制的逆,参考:https://blog.csdn.net/sunbaigui/article/details/8657051
#include<iostream> #include<stdio.h> #include<math.h> using namespace std; bool isPrime(int n){ int m=sqrt(n); if(n==0||n==1)return false; for(int i=2;i<=m;i++){ if(n%i==0)return false; } return true; } int rever(int n,int d){ // string s="";求d进制,逆的方法 // int m; // while(n!=0){ // m=n%d; // n/=d; // s+=(m+'0'); // } // cout<<s<<" "; // n=0; // m=s.size(); // for(int i=0;i<m;i++){ // n=n*d+(s[i]-'0');//这样求也是对的,但是明显比较复杂。 // } int sum=0; do{ sum=sum*d+n%d;//求逆! n/=d; }while(n!=0); return sum; } int main() { int n,d; while(scanf("%d%d",&n,&d)!=1){ //判断n是否是素数 if(isPrime(n)){ n=rever(n,d); if(isPrime(n)) printf("Yes\n"); else printf("No\n"); }else printf("No\n"); } return 0; }
1.求逆一个do-while循环太厉害了,注释掉的部分是我写的,不太对。
2.要注意特殊条件的判断,n==0||n==1直接返回false;不是素数,要不然,样例只能通过两个,另两个通不过!special case.