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Problem solution · Go

Codeforces 1C — Ancient Berland Circus

Codeforces 1C — Ancient Berland Circus: 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
39 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 1C — Ancient Berland Circus, 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

  • 39 lines of Go from the credited upstream file 1C.go.
  • The implementation keeps its working state in language-native values and containers.
  • 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 1C — Ancient Berland Circus · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io"	"math") // https://space.bilibili.com/206214type vecF struct{ x, y float64 } func (a vecF) sub(b vecF) vecF     { return vecF{a.x - b.x, a.y - b.y} }func (a vecF) det(b vecF) float64  { return a.x*b.y - a.y*b.x }func (a vecF) len2() float64       { return a.x*a.x + a.y*a.y }func (a vecF) dis2(b vecF) float64 { return a.sub(b).len2() } func CF1C(in io.Reader, out io.Writer) {	const eps = 1e-2 // 由于题目保证正多边形边数不超过 100，故 gcdf 的结果不会小于 2*Pi/100，这里简单地写成 1e-2 即可	gcdf := func(a, b float64) float64 {		for a > eps {			a, b = math.Mod(b, a), a		}		return b	} 	var a, b, c vecF	Fscan(in, &a.x, &a.y, &b.x, &b.y, &c.x, &c.y)	ab, ac := b.sub(a), c.sub(a)	ab2, ac2 := ab.len2(), ac.len2()	r2 := 0.25 * ab2 / ab.det(ac) * ac2 / ab.det(ac) * b.dis2(c) // 外接圆半径 r = abc/4S△abc	a1 := math.Acos(1 - ab2/2/r2) // 余弦定理	a2 := math.Acos(1 - ac2/2/r2)	a3 := 2*math.Pi - a1 - a2	t := gcdf(gcdf(a1, a2), a3)	Fprintf(out, "%.8f", math.Pi/t*r2*math.Sin(t))} //func main() { CF1C(os.Stdin, os.Stdout) }

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