【模拟】【NOIP2011提高组Day1】铺地毯

【样例输入输出1】

carpet.in	carpet.out
3 1 0 2 3 0 2 3 3 2 1 3 3  2 2	3

 

【样例输入输出2】

carpet.in	carpet.out
3 1 0 2 3 0 2 3 3 2 1 3 3  4 5	-1

 

很明显的水题啦，读数O(n)，处理的时候循环一次O(n)就可得到答案，所以O(2n)就可以得到出解

判断一个点是否在矩形内如下图

所以这样代码就很好实现了

program carpet;

var
  n,g,k:longint;
  x,y,xx,yy:array[0..100000+10] of longint;
  ans:longint;

procedure init;
begin
  assign(input,'carpet.in');
  assign(output,'carpet.out');
  reset(input);
  rewrite(output);
end;

procedure outit;
begin
  close(input);
  close(output);
  halt;
end;

procedure readdata;
var
  i,a,b:longint;
begin
  read(n);
  for i:=1 to n do
  begin
    read(x[i],y[i],a,b);
    xx[i]:=x[i]+a;
    yy[i]:=y[i]+b;
  end;
  read(g,k);
end;

procedure main;
var
  i:longint;
begin
  ans:=-1;
  for i:=1 to n do
  begin
    if (g>=x[i])and(g<=xx[i])and(k>=y[i])and(k<=yy[i]) then
      ans:=i;
  end;
  writeln(ans);
end;


begin
  init;
  readdata;
  main;
  outit;
end.