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Problem solution · Go

Codeforces 1292B — Aroma's Search

Codeforces 1292B — Aroma's Search: 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
66 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 1292B — Aroma's Search, 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

  • 66 lines of Go from the credited upstream file 1292B.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks 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 1292B — Aroma's Search · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF1292B(_r io.Reader, _w io.Writer) {	type point struct{ x, y int64 }	abs := func(x int64) int64 {		if x < 0 {			return -x		}		return x	}	dis := func(a, b point) int64 { return abs(a.x-b.x) + abs(a.y-b.y) } 	var ax, ay, bx, by, tt int64	var p, st point	Fscan(_r, &p.x, &p.y, &ax, &ay, &bx, &by, &st.x, &st.y, &tt)	ps := []point{}	// 没注意范围 WA 了一发	for ; p.x < 9e16 && p.y < 9e16; p.x, p.y = ax*p.x+bx, ay*p.y+by {		ps = append(ps, p)	} 	ans := 0	for i, p := range ps {		t := tt		// 注：因为算的不是欧几里得距离，这样写复杂了，只需要算两段距离就行，但是这样写适用性更广。		if d := dis(st, p); d <= t {			t -= d			c := 1			for j := i; j > 0; j-- {				if d := dis(ps[j], ps[j-1]); d <= t {					t -= d					c++				} else {					break				}			}			if i+1 < len(ps) {				if d := dis(ps[0], ps[i+1]); d <= t {					t -= d					c++					for j := i + 2; j < len(ps); j++ {						if d := dis(ps[j], ps[j-1]); d <= t {							t -= d							c++						} else {							break						}					}				}			}			if c > ans {				ans = c			}		}	}	Fprint(_w, ans)} //func main() { CF1292B(os.Stdin, os.Stdout) }

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