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Problem solution · Go

Codeforces 93D — Flags

Codeforces 93D — Flags: 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
95 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 93D — Flags, 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

  • 95 lines of Go from the credited upstream file 93D.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 93D — Flags · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengconst mod93 = 1_000_000_007 type matrix93 [][]int func newMatrix93(n, m int) matrix93 {	a := make(matrix93, n)	for i := range a {		a[i] = make([]int, m)	}	return a} func (a matrix93) mul(b matrix93) matrix93 {	c := newMatrix93(len(a), len(b[0]))	for i, row := range a {		for k, x := range row {			if x == 0 {				continue			}			for j, y := range b[k] {				c[i][j] = (c[i][j] + x*y) % mod93			}		}	}	return c} func (a matrix93) powMul(n int, f0 matrix93) matrix93 {	res := f0	for ; n > 0; n /= 2 {		if n%2 > 0 {			res = a.mul(res)		}		a = a.mul(a)	}	return res} func f93(n int) int {	if n == 0 {		return 0	}	if n == 1 {		return 4	} 	m := newMatrix93(17, 17)	f2 := newMatrix93(17, 1)	f2[16][0] = 4 // 长度等于 1 的方案数	// 0123 = 白黑红黄	for i := range 4 {		for j := range 4 {			if j == i || i+j == 3 {				continue			}			for k := range 4 {				if k != j && j+k != 3 && (j != 0 || i+k != 3) {					m[i*4+j][j*4+k] = 1					m[16][j*4+k]++ // 计算 m 矩阵 j*4+k 列之和				}			}			f2[i*4+j][0] = 1			f2[16][0]++ // 长度等于 2 的方案数		}	}	m[16][16] = 1 	return m.powMul(n-2, f2)[16][0]} // (f(n) - 回文方案数) / 2 + 回文方案数 = (f(n) + 回文方案数) / 2// 回文方案数：长度恰好等于偶数不可能（中间两个颜色相同），此时 <= n 变成 <= n-1// n 为奇数时，根据左半就能确定右半，所以只有前 ceil(n/2) 个颜色需要计算方案数func solve93(n int) int {	const inv2 = (mod93 + 1) / 2	return (f93(n) + f93((n+1)/2)) * inv2} func cf93D(in io.Reader, out io.Writer) {	var l, r int	Fscan(in, &l, &r)	ans := solve93(r) - solve93(l-1)	Fprintln(out, (ans%mod93+mod93)%mod93)} //func main() { cf93D(os.Stdin, os.Stdout) }

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