CPMagical CP Library

Problem solution · Go

Codeforces 1461F — Mathematical Expression

Codeforces 1461F — Mathematical Expression: 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
117 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 1461F — Mathematical Expression, 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

  • 117 lines of Go from the credited upstream file 1461F.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 1461F — Mathematical Expression · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	"bytes"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF1461F(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	output := func(a []byte, op string) {		Fprint(out, a[0])		for _, v := range a[1:] {			Fprint(out, op, v)		}	} 	var n, st int	Fscan(in, &n)	a := make([]byte, n)	for i := range a {		Fscan(in, &a[i])	}	var s string	Fscan(in, &s)	if s == "+-" || s == "-+" {		s = "+"	} 	if len(s) == 1 {		output(a, s)		return	} 	if s == "*-" || s == "-*" {		posZ := bytes.IndexByte(a, 0)		if posZ <= 0 {			output(a, "*")		} else {			output(a[:posZ], "*")			Fprint(out, "-")			output(a[posZ:], "*")		}		return	} 	pos := []int{}   // 保存大于 1 的数的位置	a = append(a, 0) // 方便处理没有 0 的情况	// 按 0 分割成若干段，分别处理	for i, v := range a {		if v > 1 {			pos = append(pos, i)			continue		}		if v == 1 {			if len(pos) == 0 { // 一段的开头				if i > 0 {					Fprint(out, "+")				}				Fprint(out, 1)			}			continue		}		if m := len(pos); m > 0 {			addPos := 0			// 粗略的比较：若乘积的最小值（将所有大于 1 的数视作 2）比加法的最大值大，就说明没有取加号的机会，否则二进制枚举，看看加号放哪能得到最大值			if m < 20 && 1<<m <= 9*(i-st) {				max := int64(0)				for sub := 0; sub < 1<<(m-1); sub++ {					sum, mul := int64(0), int64(a[pos[0]])					for i := 1; i < m; i++ {						if sub>>(i-1)&1 > 0 {							sum += mul + int64(pos[i]-pos[i-1]-1)							mul = int64(a[pos[i]])						} else {							mul *= int64(a[pos[i]])						}						if sum+mul > max {							max, addPos = sum+mul, sub						}					}				}			}			if pos[0] > 0 {				Fprint(out, "+")			}			Fprint(out, a[pos[0]])			for i := 1; i < m; i++ {				op := "*"				if addPos>>(i-1)&1 > 0 {					op = "+"				}				for _, v := range a[pos[i-1]+1 : pos[i]+1] {					Fprint(out, op, v)				}			}			for _, v := range a[pos[m-1]+1 : i] {				Fprint(out, "+", v)			}		}		if i < n {			if i > 0 {				Fprint(out, "+")			}			Fprint(out, 0)		}		pos = nil		st = i + 1	}} //func main() { CF1461F(os.Stdin, os.Stdout) }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗