CPMagical CP Library

Problem solution · Go

Codeforces 1486C2 — Guessing the Greatest (hard version)

Codeforces 1486C2 — Guessing the Greatest (hard version): a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
Source
EndlessCheng Codeforces Go
Length
100 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

Direct simulation

For Codeforces 1486C2 — Guessing the Greatest (hard version), the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 100 lines of Go from the credited upstream file 1486C2.go.
  • The implementation keeps its working state in language-native values and containers.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 1486C2 — Guessing the Greatest (hard version) · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"os") type (	input86 struct{ n int }	req86   struct{ l, r int }	resp86  struct{ v int }	guess86 struct{ ans int }) // github.com/EndlessCheng/codeforces-gofunc CF1486C2(in input86, Q func(req86) resp86) (gs guess86) {	q := func(l, r int) int { return Q(req86{l, r}).v }	n := in.n	ans := 0	defer func() { gs.ans = ans }() 	p := q(1, n)	l, r := 1, n+1	ll, rr := l, r	for {		if r-l == 2 {			ans = l + r - 1 - q(l, r-1)			return		} 		if r-l == 3 {			p, pl, pr := q(l, r-1), q(l, l+1), q(l+1, l+2)			if pl == pr {				if p > pl {					ans = pl - 1					return				}				ans = pl + 1				return			}			if pl == p {				ans = 2*l + 1 - p				return			}			ans = 2*l + 3 - p			return		} 		m := (l + r) >> 1		if l == ll && r == rr {			if p < m {				if pp := q(l, m-1); pp == p {					r, rr = m, m				} else {					l, ll = m, p				}			} else {				if pp := q(m, r-1); pp == p {					l, ll = m, m				} else {					r, rr = m, p+1				}			}		} else if r == rr {			if pp := q(ll, m-1); pp == p {				r, rr = m, m			} else {				l = m			}		} else {			if pp := q(m, rr-1); pp == p {				l, ll = m, m			} else {				r = m			}		}	}} func ioq86() {	in := os.Stdin	out := os.Stdout 	Q := func(req req86) (resp resp86) {		Fprintln(out, "?", req.l, req.r)		//out.Flush()		Fscan(in, &resp.v)		return	} 	d := input86{}	Fscan(in, &d.n)	gs := CF1486C2(d, Q)	ans := gs.ans	Fprintln(out, "!", ans)	//out.Flush()} //func main() { ioq86() }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗