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Problem solution · Go

Codeforces 1710C — XOR Triangle

Codeforces 1710C — XOR Triangle: a Go solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Dynamic programming
Source
EndlessCheng Codeforces Go
Length
44 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

Dynamic programming

For Codeforces 1710C — XOR Triangle, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 44 lines of Go from the credited upstream file 1710C.go.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 1710C — XOR Triangle · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengfunc cf1710C(in io.Reader, out io.Writer) {	const mod = 998244353	var s string	Fscan(in, &s)	dp := make([][2][2][2][2][2][2]int, len(s))	var f func(p int, x, y, z byte, limX, limY, limZ bool) int	f = func(p int, x, y, z byte, limX, limY, limZ bool) (res int) {		if p == len(s) {			return int(x & y & z)		}		t := &dp[p][x][y][z][b2i10(limX)][b2i10(limY)][b2i10(limZ)]		if *t > 0 {			return *t - 1		}		var upX, upY, upZ byte = 1, 1, 1		if limX { upX = s[p] - '0' }		if limY { upY = s[p] - '0' }		if limZ { upZ = s[p] - '0' }		for i := range upX + 1 {			for j := range upY + 1 {				for k := range upZ + 1 {					res += f(p+1, x|(i^j)&(j^k), y|(i^k)&(j^k), z|(i^k)&(i^j), 						limX && i == upX, limY && j == upY, limZ && k == upZ)				}			}		}		res %= mod		*t = res + 1		return	}	Fprint(out, f(0, 0, 0, 0, true, true, true))} //func main() { cf1710C(bufio.NewReader(os.Stdin), os.Stdout) }func b2i10(b bool)uint8{if b{return 1};return 0}

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