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Problem solution · Go

Codeforces 2045A — Scrambled Scrabble

Codeforces 2045A — Scrambled Scrabble: 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
42 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 2045A — Scrambled Scrabble, 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

  • 42 lines of Go from the credited upstream file 2045A.go.
  • The implementation keeps its working state in language-native values and containers.
  • 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 2045A — Scrambled Scrabble · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengfunc cf2045A(in io.Reader, out io.Writer) {	var s string	Fscan(in, &s)	n := len(s)	cnt := ['Z' + 1]int{}	for _, b := range s {		cnt[b]++	} 	a := cnt['A'] + cnt['E'] + cnt['I'] + cnt['O'] + cnt['U']	y := cnt['Y']	b := n - a - y	ng := min(cnt['N'], cnt['G']) 	if (a+y)*2 <= ng { // 元音少，甚至比 NG 还少		Fprint(out, (a+y)*5) // 只用 NG 辅音	} else if (a+y)*2 <= b-ng { // 元音少，即使把 NG 合并（减少辅音个数）仍然少		Fprint(out, (a+y)*3+ng) // 所有 NG 全部用上	} else if (b+y)/2 <= a { // 辅音少		res := (b + y) / 2 * 3		if (b+y)%2 > 0 && ng > 0 {			res++ // 多出的一个辅音可以是 N 或者 G，合并到 NG 中		}		Fprint(out, res)	} else {		// 如果没有 NG，那么答案一定是 3 的倍数		// 如果只有一个 NG，那么当 n 是 3k+2 时，一定会多出一个字母，例如 AYNGG		// 其余情况可以用 NG 和 Y 灵活调整，答案是 n		Fprint(out, n-max(n%3-ng, 0))	}} //func main() { cf2045A(bufio.NewReader(os.Stdin), os.Stdout) }

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