AtCode ABC149 C - Tour

标签

• DFS、BFS

题目地址

C - Next Prime

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

问题描述

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

Notes

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.

Constraints

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

Input

Input is given from Standard Input in the following format:

X

Output

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

Sample Input 1

20

Sample Output 1

23

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

Sample Input 2

2

Sample Output 2

2

X itself can be a prime number.

Sample Input 3

99992

Sample Output 3

100003

题意

• 给一个整数N，求大于他的第一个质数

思路

• 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

题解

小码匠

Code Review

• 函数名建议：is_prime

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;
        }
    }
}

官方题解

• 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<iostream>
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;
}