Problem solution · Go

Codeforces 1428F — Fruit Sequences

Codeforces 1428F — Fruit Sequences: 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
38 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 1428F — Fruit Sequences, 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

  • 38 lines of Go from the credited upstream file 1428F.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.

Full codeCodeforces 1428F — Fruit Sequences · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://space.bilibili.com/206214func CF1428F(in io.Reader, out io.Writer) {	var n, ones int	var s string	Fscan(bufio.NewReader(in), &n, &s)	var ans, sum int64	last := make([]int, n+1)	for i := range last {		last[i] = -1	}	for i, c := range s {		if c == '1' {			// 1111011			// 11110111			//  | 从这里到 i 的左开右闭区间的(以 i 为右端点的)子串都多了 1			ones++			sum += int64(i - last[ones])		} else {			// 更新上一个连续 ones 个 1 的起始位置			for ; ones > 0; ones-- {				last[ones] = i - ones			}		}		ans += sum	}	Fprint(out, ans)} //func main() { CF1428F(os.Stdin, os.Stdout) } 

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