Approach
Depth-first search
For Codeforces 1245F — Daniel and Spring Cleaning, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.
- Define the state carried into one recursive or stack frame.
- Mark or choose the current state before exploring children.
- Combine child results or undo the choice when the branch finishes.
Code notes
- 62 lines of Go from the credited upstream file 1245F.go.
- The implementation visibly relies on cached states.
- No explicit loop blocks detected, together with recursive traversal.
Complexity
Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
1package main2 3import (4 . "fmt"5 "io"6 "math/bits"7)8 910func cf1245F(in io.Reader, out io.Writer) {11 var T, l, r int12 for Fscan(in, &T); T > 0; T-- {13 Fscan(in, &l, &r)14 type args struct {15 p int16 a, b, c, d bool17 }18 memo := map[args]int{}19 var dfs func(int, bool, bool, bool, bool) int20 dfs = func(p int, limHigh1, limLow1, limHigh2, limLow2 bool) (res int) {21 if p < 0 {22 return 123 }24 t := args{p, limHigh1, limLow1, limHigh2, limLow2}25 if v, ok := memo[t]; ok {26 return v27 }28 29 hi1 := 130 if limHigh1 {31 hi1 = r >> p & 132 }33 lo1 := 034 if limLow1 {35 lo1 = l >> p & 136 }37 38 hi2 := 139 if limHigh2 {40 hi2 = r >> p & 141 }42 lo2 := 043 if limLow2 {44 lo2 = l >> p & 145 }46 47 for i := lo1; i <= hi1; i++ {48 for j := lo2; j <= hi2; j++ {49 if i == 0 || j == 0 {50 res += dfs(p-1, limHigh1 && i == hi1, limLow1 && i == lo1, limHigh2 && j == hi2, limLow2 && j == lo2)51 }52 }53 }54 memo[t] = res55 return56 }57 Fprintln(out, dfs(bits.Len(uint(r))-1, true, true, true, true))58 }59}60 6162