PAT-A 1015. Reversible Primes (20)

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;
}

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