这道题只需要从min到max循环中判断i能否被factor整除，如果i能被factor整除，那么输出

上代码！(AC过了哦)

#include <iostream>
using namespace std;
int main()
{
    int min, max, factor;
    cin >> min >> max >> factor; //输入
    for (int i = min; i <= max; i += 1) //从min到max循环
    {
        if (i % factor == 0) //如果i能被factor整除
        {
            cout << i << " "; //输出
        }
    }
    return 0;
}

点个赞吧！求求啦！

方法一：

#include <iostream>
using namespace std;
int main()
{
    int min , max , factor;
    cin >> min >> max >> factor;
    for (int i = min ; i <= max ; i += 1)
        if (i % factor == 0)
            cout << i << " ";
    return 0;
}

方法二：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int min , max , factor;
    cin >> min >> max >> factor;
    for (int i = ceil((double)min / factor) * factor ; i <= max / factor * factor; i += factor)
        cout << i << " ";
    return 0;
}

#include<iostream>
using namespace std;
int main()
{
    int min,max,factor;
    cin >> min >> max >> factor;
    for(int i=min;i<=max;i++)
    {
        if(i%factor==0)
        {
            cout << i << " ";
        }
    }
    return 0;
}
求赞，感谢❤️

#include<iostream>
using namespace std;
int minn,maxx,factor;
int main(){
    cin>>minn>>maxx>>factor;
    for(int i=minn;i<=maxx;i++){
        if(i%factor==0)
            cout<<i<<" ";
    }
    return 0;
}//编者：Royal

a = [int(i)for i in input().split()]
min = a[0]
max = a[1]
factor = a[2]
ans = []
for i in range(min,max + 1):
    if i % factor == 0:
        ans.append(i)
print(*ans,sep = ' ')