Approach
Depth-first search
For Codeforces 1580B — Mathematics Curriculum, 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
- 74 lines of Go from the credited upstream file 1580B.go.
- The implementation visibly relies on sequence storage, 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)7 89func cf1580B(in io.Reader, out io.Writer) {10 var n, m, k, mod int11 Fscan(in, &n, &m, &k, &mod)12 const mx = 10013 F := [mx]int{1}14 for i := 1; i < mx; i++ {15 F[i] = F[i-1] * i % mod16 }17 C := [mx][mx]int{}18 for i := 0; i < mx; i++ {19 C[i][0] = 120 for j := 1; j <= i; j++ {21 C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod22 }23 }24 25 memo := make([][][]int, m)26 for i := range memo {27 memo[i] = make([][]int, n+1)28 for j := range memo[i] {29 memo[i][j] = make([]int, k+1)30 for p := range memo[i][j] {31 memo[i][j][p] = -132 }33 }34 }35 var dfs func(int, int, int) int36 dfs = func(dep, size, need int) int {37 if dep < 0 {38 if need > 0 {39 return 040 }41 return F[size] 42 }43 if size == 0 {44 return 145 }46 47 p := &memo[dep][size][need]48 if *p >= 0 {49 return *p50 }51 52 if dep == 0 { 53 need--54 }55 56 res := 057 for leftSz := range size {58 for leftNeed := max(need-(size-1-leftSz), 0); leftNeed <= min(leftSz, need); leftNeed++ {59 leftRes := dfs(dep-1, leftSz, leftNeed)60 if leftRes == 0 { 61 continue62 }63 rightRes := dfs(dep-1, size-1-leftSz, need-leftNeed)64 res = (res + C[size-1][leftSz]*leftRes%mod*rightRes) % mod65 }66 }67 *p = res68 return res69 }70 Fprint(out, dfs(m-1, n, k))71}72 7374