题解 | #牛牛的线段#

#include<iostream>
using namespace std;
int main()
{
    int x1,x2,y1,y2;
    cin>>x1>>y1>>x2>>y2;
    cout<<(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    return 0;
}