AtCode ABC149 C - Tour

• DFS、BFS

C - Next Prime

• https://atcoder.jp/contests/abc149/tasks/abc149_c?lang=en

Find the minimum prime number greater than or equal to X.

A prime number is an integer greater than 1 that cannot be evenly divided by any positive integer except 1 and itself.

For example, 2, 3, and 5 are prime numbers, while 4 and 6 are not.

• 2 ≤ le ≤ X ≤ le ≤ 10 5 ^5 5
• All values in input are integers.

Input is given from Standard Input in the following format:

X

Print the minimum prime number greater than or equal to X.

20

23

The minimum prime number greater than or equal to 2020 is 23.

2

2

X itself can be a prime number.

99992

100003

• 给一个整数N，求大于他的第一个质数

• 编写一个判断质数的函数
• 循环遍历大于N的数，是质数则退出循环

• Code Review

  • 函数名建议：is_prime

  • 1的判断最好加上（虽然题目 x > 2）

    if (x <= 1) return false;
    

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

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    for (int i = n; 1; i++) {
        if(prime_number(i)) {
            cout << i;
            break;
        }
    }
}

• 一般不要写死循环，容易出问题

• sqrt函数记不起来时，可以用i*i的形式代替

• 循环判断写的妙

#include
using namespace std;

bool is_prime(int x){
    if (x <= 1) return false;
    for (int i = 2; i * i <= x; i++) {
        if(x % i == 0) return false;
    }
    
    return true;
}

signed main(){
    int x;
    cin>>x;

    int p=x;
    while(!is_prime(p)) {
        p++;
    }

    cout< < p << endl;
    return 0;
}