PAT 甲级 1015. Reversible Primes (20)
2017-09-20 18:07
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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:
Sample Output:
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
#include<iostream> #include<string> #include<cstdio> #include<vector> #include<algorithm> #include<queue> #include <cmath> using namespace std; bool isprime(int n) { if (n <= 1) return false; int sqr = int(sqrt(n*1.0)); for (int i = 2; i <= sqr; i++) { if (n%i == 0) return false; } return true; } int main() { int n, d; while (scanf("%d", &n) != EOF) { if (n < 0) break; scanf("%d", &d); if (isprime(n) == false) { printf("No\n"); continue; } int len = 0; int arr[100]; do { arr[len++] = n%d; n = n / d; } while (n != 0); for (int i = 0; i < len; i++) { n = n*d + arr[i]; } if (isprime(n) == false) { printf("No\n"); } else { printf("Yes\n"); } } return 0; }
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