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Problem solution · Go

Codeforces 618E — Robot Arm

Codeforces 618E — Robot Arm: a Go solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Sliding window or two pointers
Source
EndlessCheng Codeforces Go
Length
81 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Codeforces 618E — Robot Arm, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 81 lines of Go from the credited upstream file 618E.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from EndlessCheng Codeforces Go by Σndless (EndlessCheng) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCodeforces 618E — Robot Arm · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math"	"math/bits") // https://github.com/EndlessChengtype vec18 struct{ x, y float64 } func (a vec18) add(b vec18) vec18 { return vec18{a.x + b.x, a.y + b.y} }func (a vec18) rotateCCW(rad float64) vec18 {	sin, cos := math.Sincos(rad)	return vec18{a.x*cos - a.y*sin, a.x*sin + a.y*cos}} type seg18 []struct {	l, r int	v    vec18 // 在上一个向量的基础上，增加的偏移量	ang  int   // 下一个向量需要旋转的角度} func (t seg18) maintain(o int) {	a, b := t[o<<1], t[o<<1|1]	t[o].v = a.v.add(b.v.rotateCCW(float64(a.ang%360) / 180 * math.Pi))	t[o].ang = a.ang + b.ang} func (t seg18) build(o, l, r int) {	t[o].l, t[o].r = l, r	if l == r {		if l > 0 { // l=0 是原点，无偏移量			t[o].v.x = 1		}		return	}	m := (l + r) >> 1	t.build(o<<1, l, m)	t.build(o<<1|1, m+1, r)	t.maintain(o)} func (t seg18) update(o, i, incX, incAng int) {	if t[o].l == t[o].r {		t[o].v.x += float64(incX)		t[o].ang -= incAng		return	}	m := (t[o].l + t[o].r) >> 1	if i <= m {		t.update(o<<1, i, incX, incAng)	} else {		t.update(o<<1|1, i, incX, incAng)	}	t.maintain(o)} func cf618E(in io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	var n, m, op, i, inc int	Fscan(in, &n, &m)	t := make(seg18, 2<<bits.Len(uint(n)))	t.build(1, 0, n)	for range m {		Fscan(in, &op, &i, &inc)		if op == 1 {			t.update(1, i, inc, 0)		} else {			t.update(1, i-1, 0, inc)		}		p := t[1].v		Fprintf(out, "%.4f %.4f\n", p.x, p.y)	}} //func main() { cf618E(bufio.NewReader(os.Stdin), os.Stdout) }

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