PAT甲级 1015

Reversible Primes

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

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1. 考点

   • 判断数是否为质数

2. 题解

   • 先将十进制的n是否为质数
   • 将十进制的n化为D进制的数，然后反转，再转为十进制的数
   • 判断该数是否为质数

#include <stdio.h>#include <vector>using namespace std;vector<bool> rst;vector<int> r;bool isPrime(int n){  if(n==1)return false;  int i;  for(i=2;i*i<=n;i++){    if(n%i==0)return false;  }  return true;}bool isReversible(int n,int d){  if(!isPrime(n))return false;  while(n>0){    r.push_back(n%d);    n/=d;  }  int i=0;  for(auto it=r.begin();it!=r.end();it++){    n=n*d+(*it);  }  r.clear();  if(!isPrime(n))return false;  return true;}int main(){  //freopen("./in","r",stdin);  int n,d;  scanf("%d",&n);  while(n>0){    scanf("%d",&d);    if(isReversible(n,d)){      rst.push_back(true);    }else{      rst.push_back(false);    }    scanf("%d",&n);  }  for(auto it=rst.begin();it!=rst.end();it++){    if(*it)printf("Yes\n");    else printf("No\n");  }  return 0;}