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

最新推荐文章于 2022-06-03 16:26:36 发布

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最新推荐文章于 2022-06-03 16:26:36 发布

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原文链接：http://www.cnblogs.com/tccbj/p/10385391.html

本文介绍了一种算法，用于判断一个十进制数是否为可逆素数，即该数及其在给定进制下的反转数均为素数。通过转换进制、字符串反转和素数验证等步骤实现。

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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

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers

Output Specification:

For each test case, print in one line Yes if

Sample Input:

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\n");
            else printf("No\n");
        }
    }
    system("pause");
    return 0;
}