CPMagical CP Library

Problem solution · Go

Codeforces 1000F — One Occurrence

Codeforces 1000F — One Occurrence: 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 1000F — One Occurrence, 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 1000F.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 1000F — One Occurrence · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math"	"sort") // github.com/EndlessCheng/codeforces-gofunc CF1000F(_r io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	buf := make([]byte, 4096)	_i := len(buf)	rc := func() byte {		if _i == len(buf) {			_r.Read(buf)			_i = 0		}		b := buf[_i]		_i++		return b	}	ri := func() (x int) {		b := rc()		for ; '0' > b; b = rc() {		}		for ; '0' <= b; b = rc() {			x = x*10 + int(b&15)		}		return	}	type query struct{ bid, l, r, idx int } 	n := ri()	a := make([]int, n)	for i := range a {		a[i] = ri()	}	q := ri()	qs := make([]query, q)	blockSize := int(math.Round(math.Sqrt(float64(n))))	for i := range qs {		l := ri()		qs[i] = query{l / blockSize, l, ri() + 1, i}	}	sort.Slice(qs, func(i, j int) bool {		qi, qj := qs[i], qs[j]		if qi.bid != qj.bid {			return qi.bid < qj.bid		}		if qi.bid&1 == 0 {			return qi.r < qj.r		}		return qi.r > qj.r	}) 	cnt := [5e5 + 1]int{}	del := [5e5 + 1]int{} // 懒删除标记	s := []int{0} // 事先插入一个元素表示空栈，这样后面取栈元素时不需要判断空栈	update := func(i, d int) {		v := a[i-1]		if cnt[v] == 1 {			del[v]++ // 从栈中删除 v		}		cnt[v] += d		if cnt[v] == 1 {			// 把 v 插入栈			if del[v] > 0 {				del[v]--			} else {				s = append(s, v)			}		}	}	ans := make([]int, q)	l, r := 1, 1	for _, q := range qs {		for ; r < q.r; r++ {			update(r, 1)		}		for ; l < q.l; l++ {			update(l, -1)		}		for l > q.l {			l--			update(l, 1)		}		for r > q.r {			r--			update(r, -1)		}		// 获取栈顶元素		for del[s[len(s)-1]] > 0 {			del[s[len(s)-1]]--			s = s[:len(s)-1]		}		ans[q.idx] = s[len(s)-1]	}	for _, v := range ans {		Fprintln(out, v)	}} //func main() { CF1000F(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 ↗