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Problem solution · Go

Codeforces 1007B — Pave the Parallelepiped

Codeforces 1007B — Pave the Parallelepiped: 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
82 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 1007B — Pave the Parallelepiped, 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

  • 82 lines of Go from the credited upstream file 1007B.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 1007B — Pave the Parallelepiped · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF1007B(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	gcd := func(a, b int) int {		for a != 0 {			a, b = b%a, a		}		return b	}	const MX int = 1e5	d := [MX + 1]int{}	for i := 1; i <= MX; i++ {		for j := i; j <= MX; j += i {			d[j]++		}	}	const mx = 140	C := [mx][4]int64{}	C[0][0] = 1	for i := 1; i < mx; i++ {		C[i][0] = 1		for j := 1; j < 4; j++ {			C[i][j] = C[i-1][j-1] + C[i-1][j]		}	}	perm3 := [][]int{{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 0, 1}, {2, 1, 0}} 	var T, a, b, c int	x := [8]int{}	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &a, &b, &c)		a, b, c, ab, ac, bc, abc := d[a], d[b], d[c], d[gcd(a, b)], d[gcd(a, c)], d[gcd(b, c)], d[gcd(gcd(a, b), c)]		x[1] = a - ab - ac + abc // 某个类型的因子个数		x[2] = b - ab - bc + abc		x[4] = c - ac - bc + abc		x[3] = ab - abc		x[5] = ac - abc		x[6] = bc - abc		x[7] = abc 		ans := int64(0)		t := [3]int{}		for t[0] = 1; t[0] < 8; t[0]++ {			for t[1] = t[0]; t[1] < 8; t[1]++ {				for t[2] = t[1]; t[2] < 8; t[2]++ {					for _, p := range perm3 {						if t[p[0]]&1 > 0 && t[p[1]]&2 > 0 && t[p[2]]&4 > 0 {							goto calc						}					}					continue				calc:					c := [8]int{}					c[t[0]]++					c[t[1]]++					c[t[2]]++					res := int64(1)					for i := 1; i < 8; i++ {						if c[i] > 0 {							res *= C[x[i]+c[i]-1][c[i]] // 重复选择						}					}					ans += res				}			}		}		Fprintln(out, ans)	}} //func main() { CF1007B(os.Stdin, os.Stdout) }

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