都怪这个小星上课不好好听讲，让我们来做这道题，还好这题不难，我一个刚学到Level4的都会~只要先判断这个数列是等差数列还是等比数列，再根据情况求出后三个数，最后打印出来就行了O(∩_∩)O

已AC，请放心食用

#include <iostream>
using namespace std;
int main()
{
    double a, b, c, d, e, f;
    cin >> a >> b >> c;
    if (b / a == c / b)
    {
        d = c * (c / b);
        e = d * (c / b);
        f = e * (c / b);
    }
    else if (b - a == c - b)
    {
        d = c + (c - b);
        e = d + (c - b);
        f = e + (c - b);
    }
    cout << d << " " << e << " " << f;
    return 0;
}

养成好习惯，看后点个赞( •̀ ω •́ )✧！

#include <iostream>
using namespace std;
int main()
{
	double a, b, c;//设置为double是为了方便除法，如果是int的话除出来都是整数
	cin >> a >> b >> c;
	int num;
	if (c - b == b - a && c / b != b / a)//等差数列
	{
		num = c - b;
		cout << c + num << " " << c + 2 * num << " " << c + 3 * num;
	}
	if (c - b != b - a && c / b == b / a)//等比数列
	{
		num = c / b;
		cout << c * num << " " << c * num * num << " " << c * num * num * num;
	}
	if (a == b && b == c)//常数列，既是等差数列又是等比数列
	{
		cout << a << " " << a << " " << a;
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int a[10];
int main()
{
    for(int i=1;i<=3;i++)
    {
        cin>>a[i];
    }
    if(a[2]-a[1]==a[3]-a[2])
    {
        for(int i=1;i<=3;i++)
        {
            a[3]+=a[2]-a[1];
            cout<<a[3]<<" ";
        }
    }
    else
    {
        
        for(int i=1;i<=3;i++)
        {
            a[3]*=a[2]/a[1];
            cout<<a[3]<<" ";
        }
    }
    return 0;
}

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int a, b, c;
    cin >> a >> b >> c;
    if (b - a == c - b)
    {
        int d = c + (b - a);
        for (int i = 1; i <= 3; i++)
        {
            cout << d << " ";
            d += b - a;
        }
    }
    if (b / a == c / b)
    {
        int d = c * (b / a);
        for (int i = 1; i <= 3; i++)
        {
            cout << d << " ";
            d *= b / a;
        }
    }
    return 0;
}

因为只用输出后三个数可以直接算

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    if (b-a==c-b)
        cout<<c+(b-a)<<" "<<c+2*(b-a)<<" "<<c+3*(b-a);
    else
        cout<<c*(b/a)<<" "<<c*(b/a)*(b/a)<<" "<<c*(b/a)*(b/a)*(b/a);
    return 0;
}

号没了

#include <iostream>
using namespace std;
int main()
{
    double a1,a2,a3;
    cin >> a1 >> a2 >> a3;
    if(a2/a1==a3/a2)
    {
        cout << a3*(a3/a2) << " " << a3*(a3/a2)*(a3/a2) << " " << a3*(a3/a2)*(a3/a2)*(a3/a2);
    }
    else
    {
        cout << a3+(a3-a2) << " " << a3+(a3-a2)+(a3-a2) << " " << a3+(a3-a2)+(a3-a2)+(a3-a2); 
    }
}