1015. Reversible Primes (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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 10
23 2
23 10
-2
Sample Output:
Yes
Yes
No

2、思路

1）思路很清晰，但运行的时候发现不少问题；
2）第一次运行的时候数组越界，找了半天，发现#define N 数字太大，尽量使用vector
3）超时：两个地方，判断质数的时候使用了sqrt开方操作；在进制转换的过程中，可以直接转换，而不是使用数组作为中间状态。

3、代码

#include<iostream>
using namespace std;

//判断一个正整数是不是质数
bool is_prime(int n) {
    int j;
    if (n < 2) return false;
    if (n == 2 || n == 3) return true;
    for (j = 2; j*j<= n  ; ++j) {
        if (n%j == 0) {
            return false;
        }
    }
    return true;
}

//将一个10进制的数字转为d进制、倒序、再转为10进制
int change(int n,int d) {
    int m=0;
    int i, j;
    while (n != 0) {
        m *= d;
        m += n%d;
        n = n / d;
    }
    return m;
}

int main()
{
    int n, d,m;
    int i, j;
    while (scanf("%d", &n)!=EOF) {
        if (n < 0)
            return 0;
        scanf("%d", &d);
        //判断n是否为质数
        if (!is_prime(n)) {
            printf("No\n");
            continue;
        }
        
        //进制转换
        m = change(n, d);

        //判断
        if (is_prime(m))
            printf("Yes\n");
        else
            printf("No\n");
    }
    system("pause");
    return 0;
}