Poj Kadj Squares (Geometry)

• Kadj Squares

• The meaning of the question is
  to give the side lengths of N squares, rotate 45 degrees counterclockwise along the lower left vertex, and one vertex is on the x-axis. One vertex of the first square is placed on the y-axis, from left to right at a time. The sides are allowed to overlap but the areas are not allowed to intersect. Ask those squares to be seen when looking down from the top.

• Idea
  At the beginning, I first obtained the coordinates of b, and then knew the coordinates of the left and right endpoints, and then judged whether the rectangle was covered. I don't know if it was written wrong or the accuracy was stuck. Later found that we can expand the side length by $\sqrt{2}$At this time, the left and right endpoints are all integers. Then judge whether it is covered according to the left and right endpoints.
  

• Code

#pragma GCC optimize(2)
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
//#include<cmath> 

using namespace std;

typedef long long ll;
typedef unsigned long ul;
typedef unsigned long long ull;
#define pi acos(-1.0)
#define e exp(1.0)
#define pb push_back
#define mk make_pair
#define fir first
#define sec second
#define scf scanf
#define prf printf
#define sqt sqrt(2.0)
typedef pair<ll,ll> pa;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int MAX_N=55;
int N;
int a[MAX_N],l[MAX_N],r[MAX_N];
int abs(int n){
    
    
	return n>0?n:-n;
} 
int main()
{
    
    
//  freopen(".../.txt","w",stdout);
//  freopen(".../.txt","r",stdin);
	ios::sync_with_stdio(false);
	while(cin>>N&&N){
    
    
		int i,j,k,L,R;
		for(i=1;i<=N;i++)
		cin>>a[i];
		for(i=1;i<=N;i++){
    
    
			L=0;
			for(j=1;j<i;j++){
    
    
				L=max(L,r[j]-abs(a[i]-a[j]));
			}
			l[i]=L;
			r[i]=L+2*a[i];
		} 
		for(i=1;i<=N;i++){
    
    
			L=l[i];
			R=r[i];			
			for(j=1;j<=N;j++){
    
    
				if(i==j)
				continue;
				if(j<i)//左边的正方形能覆盖的最大坐标 
				L=max(L,r[j]);
				else//右边的正方形能覆盖的最小坐标 
				R=min(R,l[j]);
			}
			if(L<R)
			cout<<i<<' ';
		}
		cout<<endl;
	}
	return 0;
}