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Problem solution · Go

Codeforces 986F — Oppa Funcan Style Remastered

Codeforces 986F — Oppa Funcan Style Remastered: a Go solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Stack-based processing
Source
EndlessCheng Codeforces Go
Length
136 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

Stack-based processing

For Codeforces 986F — Oppa Funcan Style Remastered, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 136 lines of Go from the credited upstream file 986F.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

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 986F — Oppa Funcan Style Remastered · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	"container/heap"	. "fmt"	"io"	"math") // https://space.bilibili.com/206214func CF986F(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	const mx = 31622776	primes := make([]int, 0, 1951957)	np := [mx + 1]bool{}	for i := 2; i <= mx; i++ {		if !np[i] {			primes = append(primes, i)		}		for _, p := range primes {			if p*i > mx {				break			}			np[p*i] = true			if i%p == 0 {				break			}		}	}	pow := func(x, n, mod int64) int64 {		x %= mod		res := int64(1)		for ; n > 0; n >>= 1 {			if n&1 > 0 {				res = res * x % mod			}			x = x * x % mod		}		return res	} 	var T int	var n, k int64	Fscan(in, &T)	type query struct {		n int64		i int	}	qs := map[int64][]query{}	for i := 0; i < T; i++ {		Fscan(in, &n, &k)		qs[k] = append(qs[k], query{n, i})	}	ans := make([]bool, T)	for k, qs := range qs {		ps := []int64{}		x := k		for _, p := range primes {			p := int64(p)			if p > x {				break			}			if x%p == 0 {				for x /= p; x%p == 0; x /= p {				}				ps = append(ps, p)			}		}		if x > 1 {			ps = append(ps, x)		}		if len(ps) == 0 { // k = 1		} else if len(ps) == 1 {			for _, q := range qs {				ans[q.i] = q.n%ps[0] == 0			}		} else if len(ps) == 2 {			x, y := ps[0], ps[1]			for _, q := range qs {				t := q.n % x * pow(y, x-2, x) % x				ans[q.i] = t*y <= q.n			}		} else {			dis := make([]int64, ps[0])			for i := range dis {				dis[i] = math.MaxInt64			}			dis[0] = 0			h := hp86{{}}			for len(h) > 0 {				top := h.pop()				v := top.v				if top.dis > dis[v] {					continue				}				for _, p := range ps[1:] {					w := (int64(v) + p) % ps[0]					if newD := dis[v] + p; newD < dis[w] {						dis[w] = newD						h.push(pair86{int(w), newD})					}				}			}			for _, q := range qs {				ans[q.i] = dis[q.n%ps[0]] <= q.n			}		}	}	for _, b := range ans {		if b {			Fprintln(out, "YES")		} else {			Fprintln(out, "NO")		}	}} //func main() { CF986F(os.Stdin, os.Stdout) } type pair86 struct {	v   int	dis int64}type hp86 []pair86 func (h hp86) Len() int              { return len(h) }func (h hp86) Less(i, j int) bool    { return h[i].dis < h[j].dis }func (h hp86) Swap(i, j int)         { h[i], h[j] = h[j], h[i] }func (h *hp86) Push(v interface{})   { *h = append(*h, v.(pair86)) }func (h *hp86) Pop() (v interface{}) { a := *h; *h, v = a[:len(a)-1], a[len(a)-1]; return }func (h *hp86) push(v pair86)        { heap.Push(h, v) }func (h *hp86) pop() pair86          { return heap.Pop(h).(pair86) }

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