![[java] 백준 1978번 소수 찾기](https://img1.daumcdn.net/thumb/R750x0/?scode=mtistory2&fname=https%3A%2F%2Fblog.kakaocdn.net%2Fdna%2Fmvc7h%2FbtrackPD0yb%2FAAAAAAAAAAAAAAAAAAAAAMHNxMtfoDC11KmniaPEhQl3hZi8TkcMyKaTWAKs3rIl%2Fimg.jpg%3Fcredential%3DyqXZFxpELC7KVnFOS48ylbz2pIh7yKj8%26expires%3D1761922799%26allow_ip%3D%26allow_referer%3D%26signature%3D3XgjKD20eNk%252FT%252FE2TKLOjmx5eZM%253D)
[java] 백준 1978번 소수 찾기알고리즘/백준2021. 7. 22. 21:09
      Table of Contents
        
import java.util.Scanner;
class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int sosu=0;
        for(int i=0; i<N; i++){
            int a = sc.nextInt();
            int count = 0;
            for(int j=1; j<=a; j++){
                if(a % j == 0){
                    count++;
                }
            }
            if(count == 2) {
                sosu+=1;
            }
        }
        System.out.println(sosu);
    }
}N : 입력 받을 개수
a : 입력 받을 수
for(int j=1; j<=a; j++){
    if(a % j == 0){
        count++;
    }
}
if(count == 2) {
    sosu+=1;
}예를 들어 a에 3이 입력되어 있다면 j=1부터 3까지 반복문을 돌게된다.
조건문에서 1과 자신으로 밖에 나누어지지 않는다면 count는 2가 될 것이다.
그래서 다음 조건문에서 count == 2 이다면 sosu(소수)의 개수를 1개 추가 시키라고 해놓았다.
마지막에 소수의 개수를 출력
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@펄찌 :: Pearl's Story
            많은 생각들을 가지고 있고 사람들과 대화를 하는 것이 즐거운 펄의 스페이스입니다.
즐거운 하루 되셨으면 좋겠습니다😊
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