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Problem solution · Go

Codeforces 1903D2 — Maximum And Queries (hard version)

Codeforces 1903D2 — Maximum And Queries (hard version): 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
73 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 1903D2 — Maximum And Queries (hard version), 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

  • 73 lines of Go from the credited upstream file 1903D2.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 1903D2 — Maximum And Queries (hard version) · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math/bits") // https://github.com/EndlessChengfunc cf1903D2(in io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	var n, q, v, k, total, maxVal int	Fscan(in, &n, &q) 	const w = 20	const u = 1 << w	cnt := [u]int{}	f := [u][w]int{}	for range n {		Fscan(in, &v)		total += v		maxVal = max(maxVal, v)		cnt[v]++		// 遍历 v 的每个 0		for j := uint32(u - 1 ^ v); j > 0; j &= j - 1 {			i := bits.TrailingZeros32(j)			f[v][i] += v & (1<<i - 1) // 累加 v 的低 i 位		}	} 	// 算完 SOS DP（从超集转移到当前状态）后：	// 二进制包含 s 的元素，有 cnt[s] 个	// 对于二进制包含 s 且第 i 位是 0 的元素，累加这些元素的低 i 位之和，即 f[s][i]	for i := range w {		for s := 0; s < u; s++ {			s |= 1 << i			cnt[s^1<<i] += cnt[s]			for j := range w {				f[s^1<<i][j] += f[s][j]			}		}	} 	for range q {		Fscan(in, &k)		avg := (total + k) / n		if avg >= maxVal {			Fprintln(out, avg)			continue		}		ans := 0		for i := w - 1; i >= 0; i-- {			// 现在我们要计算，让答案第 i 位是 1，代价是多少。也就是元素要包含 ans|1<<i			// 对于（在所有操作之前）已经包含 ans|1<<i 的元素，无需操作，代价是 0。设这样的元素有 cnt[ans|1<<i]] 个			// 其余 n-cnt[ans|1<<i]] 个元素呢？可以分为两类：			// 第一类是（在所有操作之前）包含 ans，但不包含 1<<i 的元素。这些元素要增大，比如从 001 到 100，需要 +3，而不是 +4			// 第二类是（在所有操作之前）不包含 ans 的元素。由于我们已经在之前的循环中增加了这些数，这些元素的低 i 位现在都是 0。这些元素要增大，比如从 000 到 100，直接 +4			// 我们可以先增加 n-cnt 个 2^i，再减去第一类元素多操作的次数，即第一类元素的低 i 位之和，记作 f[ans][i]			// 综上，预处理 cnt 和 f，就可以 O(1) 求出让答案第 i 位是 1 的代价：(n-cnt[ans|1<<i])<<i - f[ans][i]			cost := (n-cnt[ans|1<<i])<<i - f[ans][i]			if cost <= k {				k -= cost				ans |= 1 << i			}		}		Fprintln(out, ans)	}} //func main() { cf1903D2(bufio.NewReader(os.Stdin), os.Stdout) }

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