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Problem solution · Go

Codeforces 520E — Pluses everywhere

Codeforces 520E — Pluses everywhere: 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
53 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 520E — Pluses everywhere, 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

  • 53 lines of Go from the credited upstream file 520E.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 520E — Pluses everywhere · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF520E(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	const mod int64 = 1e9 + 7	const mx int = 1e5	F := [mx + 1]int64{1}	for i := 1; i <= mx; i++ {		F[i] = F[i-1] * int64(i) % mod	}	pow := func(x, n int64) (res int64) {		res = 1		for ; n > 0; n >>= 1 {			if n&1 > 0 {				res = res * x % mod			}			x = x * x % mod		}		return res	}	invF := [...]int64{mx: pow(F[mx], mod-2)}	for i := mx; i > 0; i-- {		invF[i-1] = invF[i] * int64(i) % mod	}	C := func(n, k int) int64 { return F[n] * invF[k] % mod * invF[n-k] % mod } 	var n, k int	var s string	Fscan(in, &n, &k, &s)	sum := make([]int64, n+1)	for i, b := range s {		sum[i+1] = sum[i] + int64(b&15)	}	ans := int64(0)	for i, p10 := 0, int64(1); i < n-k; i++ { // 枚举 10^i，计算 10^i 可以算到哪些数字的贡献上		if k > 0 {			ans = (ans + sum[n-i-1]*p10%mod*C(n-i-2, k-1)) % mod		}		ans = (ans + (sum[n-i]-sum[n-i-1])*p10%mod*C(n-i-1, k)) % mod // 作为最后一个数的部分，后面无法跟加号		p10 = p10 * 10 % mod	}	Fprint(out, ans)} //func main() { CF520E(os.Stdin, os.Stdout) }

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