Algorithm Notes CodeUp Problem A: Prime Numbers

topic

Description

Input an integer n (2<=n<=10000), it is required to output all prime numbers from 1 to this integer (excluding 1 and this integer) with one digit 1, if not, output -1.

Input

There are multiple sets of data entered.
One line per group, enter n.

Output

Output all prime numbers from 1 to this integer (not including 1 and this integer) with one digit 1 (the prime numbers are separated by spaces, and there is no space after the last prime number). If not, output -1.

Ideas

Since n<=10000, you can consider typing the prime number table before outputting.
Note: If there is no prime number (that is, n<=11), output -1

Code

#include <iostream>
#include <stdio.h>
#include <vector>
#include <math.h>
using namespace std;

vector<int> primes;

bool isPrime(int n){
    
    
    int sqr=sqrt(1.0*n);
    for(int i=2;i<=sqr;i++)
        if(n%i==0)
            return false;
    return true;
}

void FindPrime(){
    
    
    for(int i=11;i<10000;i+=10)//注意只需要个位为1的素数
        if(isPrime(i))
            primes.push_back(i);
}

int main(){
    
    
    int n;
    FindPrime();
    while(scanf("%d",&n)!=EOF){
    
    
        if(n<=11){
    
    
            printf("%d\n",-1);
            continue;
        }
        for(int i=0;primes[i]<n&&i<=305;i++){
    
    
            if(primes[i+1]>=n||i==305)//注意最后一个素数后面没有空格
                printf("%d",primes[i]);
            else
                printf("%d ",primes[i]);
        }
        printf("\n");
    }
    return 0;
}