这道题只需要使用三次 if 语句，依次判断能否整除这 $3$ 个数，如果能就输出即可。

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,sum=0;
	cin>>n;
    if(n%3==0){
        cout<<"3"<<" ";
        sum++;
    }
    if(n%5==0){
        cout<<"5"<<" ";
        sum++;
    }
    if(n%7==0){
        cout<<"7"<<" ";
        sum++;
    }
    if(sum==0) cout<<"n"<<endl;
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {
	int n;
	cin >> n;
	if (n % 3 == 0 && n % 5 == 0 && n % 7 == 0) {
		cout << "3 5 7";
	} else if (n % 3 == 0 and n % 5 == 0) {
		cout << "3 5";
	} else if (n % 3 == 0 and n % 7 == 0) {
		cout << "3 7";
	} else if (n % 5 == 0 and n % 7 == 0) {
		cout << "5 7";
	} else if (n % 3 == 0) {
		cout << "3";
	} else if (n % 5 == 0) {
		cout << "5";
	} else if (n % 7 == 0) {
		cout << "7";
	} else {
		cout << 'n';
	}
	
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n%3==0) cout<<3<<" ";
    if(n%5==0) cout<<5<<" ";
    if(n%7==0) cout<<7<<" ";
    if(n%3!=0&&n%5!=0&&n%7!=0) cout<<"n";
    return 0;
}

#include<iostream> 
#include<cstdio> 
#include<cmath> 
#include<algorithm>
using namespace std;
int main(){
	int a;
	cin >> a;
	if (a%3 == 0 && a%5 == 0 && a%7 == 0){
		cout << "3 5 7";
	}else if (a%3 == 0 && a%5 == 0){
		cout << "3 5";
	}else if (a%3 == 0 && a%7 == 0){
		cout << "3 7";
	}else if (a%5 == 0 && a%7 == 0){
		cout << "5 7";
	}else if (a%3 == 0){
		cout << "3";
	}else if (a%5 == 0){
		cout << "5";
	}else if (a%7 == 0){
		cout << "7";
	}else{
		cout << "n";
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
long long q,w,e,r,t,y,u,i,o,p,d,f,g,h,j,k,l,z,x,c,v,b,n,m;
string s;
long long a[100000],as[100000],asd[100000];
char cc;
double dd;
int main()
{
	cin>>x;
	if(x%3==0)
	cout<<"3 ";
	if(x%5==0)
	cout<<"5 ";
	if(x%7==0)
	cout<<"7 ";
	if(x%3!=0&&x%5!=0&&x%7!=0)
	cout<<"n";
    return 0;
}