[프로그래머스][Level2] 소수 찾기 c++

https://programmers.co.kr/learn/courses/30/lessons/42839 

프로그래머스

 

풀이 과정

흩어진 종이 조각을 붙여 소수를 만들어낼 수 있는 경우를 v 벡터에 담는다.  
이를 위해서 순열 알고리즘을 썼는데 이 알고리즘을 쓰기 전엔  
문자열이라도 항상 정렬을 해줘야 한다. 안 그럼 오류가 난다!

v벡터에 다 넣어줬으면 중복을 제거한다. 중복을 제거하기 전에도 항상 정렬을 해줘야 한다!  
그 다음 에라토스테네스의 체 알고리즘을 써 소수를 찾아준다.

 

풀이 1

#include <string>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

const int NUM = 10000001;
int prime[NUM];

void PrimeNum(){
    for(int i = 2; i <= NUM; i++){
        prime[i] = i;
    }
    for(int i = 2; i <= sqrt(NUM); i++){
        if(!prime[i]) continue;
        for(int j = i * i; j <= NUM; j += i){
            prime[j] = 0;
        }
    }
}

int solution(string numbers) {
    int answer = 0;

    vector<int> v;
    sort(numbers.begin(), numbers.end());
    do{
        for(int i = 0; i < numbers.size(); i++){
            string str = "";
            for(int j = i; j < numbers.size(); j++){
                str += numbers[j];
                v.push_back(stoi(str));
            }
        }
    }while(next_permutation(numbers.begin(), numbers.end()));

    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());

    PrimeNum();
    for(auto i : v){
        if(prime[i]) answer++;
    }
    return answer;
}

 

풀이 2

#include <string>
#include <vector>
#include <algorithm>

using namespace std;

const int NUM = 10000001;
int prime[NUM];
// sqrt 함수 구현
int sqrt(int n){
    int s = 0;
    int t = 0;
    s = n / 2;
    while(s != t){
        t = s;
        s = ((n / t) + t) / 2;
    }
    return s;
}

void PrimeNum(){
    for(int i = 2; i <= NUM; i++){
        prime[i] = i;
    }
    for(int i = 2; i <= sqrt(NUM); i++){
        if(!prime[i]) continue;
        for(int j = i * i; j <= NUM; j += i){
            prime[j] = 0;
        }
    }
}

int solution(string numbers) {
    int answer = 0;

    vector<int> v;
    sort(numbers.begin(), numbers.end());
    do{
        for(int i = 0; i < numbers.size(); i++){
            string str = "";
            for(int j = i; j < numbers.size(); j++){
                str += numbers[j];
                v.push_back(stoi(str));
            }
        }
    }while(next_permutation(numbers.begin(), numbers.end()));

    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());

    PrimeNum();
    for(auto i : v){
        if(prime[i]) answer++;
    }

    return answer;
}