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Problem solution · Go

Codeforces 1830C — Hyperregular Bracket Strings

Codeforces 1830C — Hyperregular Bracket Strings: 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
108 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 1830C — Hyperregular Bracket Strings, 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

  • 108 lines of Go from the credited upstream file 1830C.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 1830C — Hyperregular Bracket Strings · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math/rand"	"time") // https://space.bilibili.com/206214const mod30 = 998244353 func (c *comb30) pow(x int64, n int) (res int64) {	x %= mod30	res = 1	for ; n > 0; n >>= 1 {		if n&1 > 0 {			res = res * x % mod30		}		x = x * x % mod30	}	return} type comb30 struct{ _f, _invF []int64 } func newComb30(mx int) *comb30 {	c := &comb30{[]int64{1}, []int64{1}}	c._init(mx)	return c} func (c *comb30) _init(mx int) {	n := len(c._f)	c._f = append(make([]int64, 0, mx+1), c._f...)[:mx+1]	for i := n; i <= mx; i++ {		c._f[i] = c._f[i-1] * int64(i) % mod30	}	c._invF = append(make([]int64, 0, mx+1), c._invF...)[:mx+1]	c._invF[mx] = c.pow(c._f[mx], mod30-2)	for i := mx; i > n; i-- {		c._invF[i-1] = c._invF[i] * int64(i) % mod30	}} func (c *comb30) f(n int) int64 {	if n >= len(c._f) {		c._init(n * 2)	}	return c._f[n]} func (c *comb30) invF(n int) int64 {	if n >= len(c._f) {		c._init(n * 2)	}	return c._invF[n]} func (c *comb30) c(n, k int) int64 {	if k < 0 || k > n {		return 0	}	return c.f(n) * c.invF(k) % mod30 * c.invF(n-k) % mod30} func CF1830C(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	cm := newComb30(0)	rand.Seed(time.Now().UnixNano()) 	var T, n, k, l, r into:	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n, &k)		diff := make([]uint64, n)		for ; k > 0; k-- {			Fscan(in, &l, &r)			v := rand.Uint64()			diff[l-1] ^= v			if r < n {				diff[r] ^= v			}		}		cnt := map[uint64]int{}		s := uint64(0)		for _, d := range diff {			s ^= d			cnt[s]++		}		ans := int64(1)		for _, c := range cnt {			if c%2 > 0 {				Fprintln(out, 0)				continue o			}			// 卡特兰数 C(c/2)			ans = ans * (cm.c(c, c/2) - cm.c(c, c/2-1) + mod30) % mod30		}		Fprintln(out, ans)	}} //func main() { CF1830C(os.Stdin, os.Stdout) }

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