PAT

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 1023 223 10-2

Sample Output:

YesYesNo

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给定条件：
1.一个十进制数n
2.一个整数d ，1<d<=10

要求：
1.判断n是否为素数
2.判断n转换成d进制后，将数字变为“倒序”（如十进制1234倒过来为4321二进制1101倒过来为1011）然后再转换为10进制，是否为素数
3.如果两个都为素数，输出Yes否则输出No

求解：
1.按照题目要求进行编码即可

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#include <cstdio>#include <vector>#include <cmath>#include <algorithm>using namespace std;int n, d, revNum;vector<int> v;bool isPrimeNum(int num) {if(num <= 1) return false;for(int i = 2; i <= sqrt(num); i++) {if(num % i == 0) {return false;}}return true;}int main() {while(scanf("%d", &n) != EOF) {if(n < 0) return 0;v.clear();scanf("%d", &d);if(isPrimeNum(n) == false) {printf("No\n");continue;}int temp = n;while(temp != 0) {v.push_back(temp%d);temp /= d;}revNum = 0;for(int i = 0; i < v.size(); i++) {revNum = revNum * d + v[i];}if(isPrimeNum(revNum)) {printf("Yes\n");} else {printf("No\n");}}return 0;}

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