简单，函数搞定！

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    int a=abs(x);
    int b=abs(y);
    if (a<=1 && b<=1) cout<<"yes";
    else cout<<"no";
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    int a=abs(x);
    int b=abs(y);
    if (a<=1 && b<=1)
    {
        cout<<"yes";
    }
    else
    {
        cout<<"no";
    }
    return 0;
} 

数学题就要从数学角度出发思考嘛！！

太简单了，随便写点题解：

在获取数据以后，只需要判断获取的x和y坐标是不是在1到-1之间就行

#include <iostream>
using namespace std;

int main()
{
    int x1 = -1,x2 = 1,y1 = -1,y2 = 1,x3,y3;
    cin >> x3 >> y3;
    if ((x1 <= x3) && (x3 <= x2) && (y1 <= y3) && (y3 <= y2))
        cout << "yes";
    else
        cout << "no";
    return 0;
}

#include <iostream>
 using namespace std;
int main() 
{
    int x, y;
 
    cin >> x >> y;
 
    if (x >= -1 && x <= 1 && y >= -1 && y<= 1) 
	{
	
	    cout << "yes";
    
    }
    
    else 
	{
	
	    cout << "no";
    
    }
    
    return 0;
}