传送门
https://www.patest.cn/contests/pat-a-practise/1015
题目
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
分析
题意：
给出一个10进制数N和进制D，先化成给定进制D的数，然后翻转，再化成10进制数N'进行验证其是否是质数（素数），若翻转前的数N和翻转后的数N'均为质数，则输出Yes，否则，输出No。
我单独写了一个翻转的函数，先计算出10进制N的D进制数，然后再计算翻转后D进制数的10进制数N'，最后用开始建立好的质数表来判断其是否为质数，最后按要求输出即可。
源代码
//C/C++实现
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
bool isNotPrime[100000] = {true, true};
int reverse(int n, int d){
int tmp = n;
vector<int> v;
while(tmp){
v.push_back(tmp % d);
tmp /= d;
}
int count = 0, sum = 0, size = v.size();
for(int i = 0; i < size; ++i){
sum += v[i] * pow(d, size - i - 1);
}
return sum;
}
int main(){
//create prime table
for(int i = 2; i < 50000; ++i){
for(int j = 2; i * j < 100000; ++j){
isNotPrime[i * j] = true;
}
}
int n, d;
scanf("%d", &n);
while(n > 0){
scanf("%d", &d);
if(!isNotPrime[n] && !isNotPrime[reverse(n, d)]){
printf("Yes\n");
}
else{
printf("No\n");
}
scanf("%d", &n);
}
return 0;
}
