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Problem solution · Go

Codeforces 1616H — Keep XOR Low

Codeforces 1616H — Keep XOR Low: a Go solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Depth-first search
Source
EndlessCheng Codeforces Go
Length
83 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

Depth-first search

For Codeforces 1616H — Keep XOR Low, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 83 lines of Go from the credited upstream file 1616H.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 1616H — Keep XOR Low · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengfunc cf1616H(in io.Reader, out io.Writer) {	type node struct {		son [2]*node		sz  int	}	size := func(o *node) int {		if o == nil {			return 0		}		return o.sz	}	root := &node{} 	var n, x, v int	Fscan(in, &n, &x)	for range n {		Fscan(in, &v)		o := root		for i := 29; i >= 0; i-- {			b := v >> i & 1			if o.son[b] == nil {				o.son[b] = &node{}			}			o = o.son[b]			o.sz++		}	} 	const mod = 998244353	pow2 := make([]int, n+1)	pow2[0] = 1	for i := 1; i <= n; i++ {		pow2[i] = pow2[i-1] * 2 % mod	} 	var dfs func(*node, *node, int) int	dfs = func(p, q *node, d int) int {		if p == nil || q == nil {			return pow2[size(p)+size(q)]		}		if d < 0 {			if p == q {				return pow2[p.sz]			}			return pow2[p.sz+q.sz]		} 		bit := x >> d & 1		// 用数学归纳法可以证明，对于固定的 p.son[i]，第二个参数是固定的		// 所以相当于在 dfs(p.son[i])，这等价于 dfs 一棵二叉树		res1 := dfs(p.son[0], q.son[bit], d-1)		res2 := dfs(p.son[1], q.son[bit^1], d-1) 		if bit > 0 {			if p == q {				return res1			}			return res1 * res2 % mod		} 		if p == q {			return (res1 + res2 - 1) % mod		}		// 注意我们算的是子序列（子集）的个数，不是数对的个数		// bit == 0 时，res1 和 res2 不能有交，所以是加法而不是乘法		// 考虑上面是 1，这里是 0 的情形，那么同一组子树内部可以随意选，这解释了末两行代码		return (res1 + res2 - 1 +			(pow2[size(p.son[0])]-1)*(pow2[size(p.son[1])]-1) +			(pow2[size(q.son[0])]-1)*(pow2[size(q.son[1])]-1)) % mod	}	Fprint(out, (dfs(root, root, 29)+mod*2-1)%mod)} //func main() { debug.SetGCPercent(-1); cf1616H(bufio.NewReader(os.Stdin), os.Stdout) }

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