題目

題目連結:https://pintia.cn/problem-sets/994805342720868352/problems/994805495863296000

思路:

這道題的思路很簡單, 檢查輸入數是否是質數, 如果是的話, 把它按進位制轉換, 再翻轉, 轉換回10, 判斷是否為質數.

坑點:

1. 別忘了1不是質數
2. 檢查輸入的N是否是質數
3. 本題中不涉及的可能存在的坑, 如果D是一個大於10的數, 不能需要將每一位單獨儲存, 然後翻轉, 轉制回10

程式碼

#include<iostream>
#include<string>
 

#include<cmath>
#include<vector>
using namespace std;
vector<int> reverse_trans(int dec, int base_to);
bool verify_prime(int);
void solution();
void Test_verify_prime();
void Test_reverse_trans();
int vec_to_10(vector<int>, int);
int main()
{
    //Test_verify_prime();
    //Test_reverse_trans();
 

    solution();
    getchar();
    getchar();
    return 0;
}


string reverse_trans1(int dec, int base_to)
{
    string s;
    if (dec == 0) {
        return "0";
    };
    int exp = log(dec) / log(base_to); // find the max log(N)/ log(base)
    int base = pow(base_to, exp);
    while (base) {
        int
 
 digit = dec / base;
        dec -= digit * base;
        base /= base_to;
        s = to_string(digit) + s;
    }
    return s;
}

bool verify_prime(int item)
{
    if (item <2) return false;
    if (item == 2) return true;
    for (int i = 2; i <= sqrt(item); i++)
    {
        if (item % i == 0) return false;
    }
    return true;
}

void solution() {
    int N, D;
    while (1)
    {
        cin >> N;
        if (N < 0) break;
        cin >> D;
        if (!verify_prime(N)) {
          cout << "No\n";
          continue;
        }
        // vector<int> s_trans = reverse_trans(N, D); // trans a N to the number based on D and reverse it.
        // int reverse_int = vec_to_10(s_trans, D);
        string s_trans = reverse_trans1(N, D);
        int reverse_int = stoi(s_trans, nullptr, D);
        if (verify_prime(reverse_int)) cout << "Yes\n";
        else cout << "No\n";
    }
}

void Test_verify_prime() {
    cout << "test verify_prime: " << endl;
    int test[15] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 27};
    for(int i = 0; i < 15; i++) {
        cout << verify_prime(test[i]) << " ";
    }
    cout << "\nN N Y Y N Y N Y N N Y N Y N N" << endl;
}

void Test_reverse_trans() {
    cout << "test reverse_trans: " << endl;
}

vector<int> reverse_trans(int dec, int base_to) // abandon. Not efficient.
{
    vector<int> s;
    if (dec == 0) {
        s.push_back(0);
        return s;
    };
    int base = 1;
    while (dec / base) base *= base_to;
    base = base / base_to;
    while (base) {
        int digit = dec / base;
        dec -= digit * base;
        base /= base_to;
        s.push_back(digit);
    }
    return s;
}

int vec_to_10(vector<int> s, int base)
{
    int sum = 0;
    
    for(auto f = s.rbegin(); f != s.rend(); f++)
    {
        sum = sum * base + *f;
    }
    return sum;
}