PAT A1015 Reversible Primes （20 分）——进制转换，质数

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

 

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
#include <string>
using namespace std;
const int maxn = 1010;
int n, d;
int to_10(int i,int radix) {
    int res = 0,exp=0;
    while (i != 0) {
        res += i%10*pow(radix, exp);
        exp++;
        i /= 10;
    }
    return res;
}
bool is_prime(int i) {
    if (i == 0 || i == 1)return false;
    if (i == 2 || i == 3)return true;
    for (int j = 2; j*j <= i; j++) {
        if (i%j == 0)return false;
    }
    return true;
}
string to_s(int i,int radix) {
    string s = "";
    do {
        s += '0' + i % radix;
        i /= radix;
    } while (i != 0);
    reverse(s.begin(), s.end());
    return s;
}
int to_num(string s) {
    int res = 0;
    for (int i = s.length()-1; i >=0; i--) {
        res = res * 10 + s[i] - '0';
    }
    return res;
}
int main() {
    while (true) {
        scanf("%d", &n);
        if (n < 0)break;
        else {
            scanf("%d", &d);
            if (is_prime(n) && is_prime(to_10(to_num(to_s(n,d)), d))) printf("Yes
");
            else printf("No
");
        }
    }
    system("pause");
    return 0;
}

注意点：题目没看懂，导致结果一直错。题目的意思是一个10进制数如果他本身是素数，然后转换到给定进制下并将其反转，再转化到10进制，还是素数的话即为 Yes。