PAT 1015. Reversible Primes (20)（d进制转化，质数判定（注意等于号））

题目

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

解题思路

• 1.是将这个数在d进制下反转，然后转化为十进制数，再判断它们俩是不是质数（题目有点烦），注意0，1不是质数。

AC代码

#include<iostream>#include<deque>#include<math.h>using namespace std;bool isPrim(int num){    if (num == 1) {        return false;    }    //！！！！注意有个等于号    for (int i = 2; i <= sqrt(num); ++i) {        if (num % i ==0) {            return false;        }    }    return true;}int main(int argc, char *argv[]){    int n;    int tem_n;    while (cin>>n) {        tem_n = n;        if (n<0) {            return 0;        }        int d;        cin >> d;        deque<int> dq;        while (n>0) {            dq.push_back(n % d);            n = n / d;        }        int reverse = 0,tp,i =0;        while (!dq.empty()) {            tp = dq.back();            reverse += pow(d,i) * tp;            i++;            dq.pop_back();        }       // cout << reverse << endl;        if (isPrim(reverse)&&isPrim(tem_n)) {            cout << "Yes" <<endl;        }else {            cout << "No" << endl;        }    }    return 0;}