The problem is asking you to do three things: Und...
Created on: July 5, 2025
Answered using O4 Mini High by Chat01
Created on: July 5, 2025
Answered using O4 Mini High by Chat01
The problem is asking you to do three things:
(
๐
1
,
๐
2
,
โฆ
,
๐
๐
)
a=(a
1
โ
,a
2
โ
,โฆ,a
n
โ
) is valid if for every
๐
i,
0
โ
โ
โค
โ
โ
๐
๐
โ
โ
โค
โ
โ
๐
.
0โคa
i
โ
โคi.
Compute the โweightโ
๐
(
๐
)
f(a) of a given valid sequence.
You start with tokens at positions
1
,
2
,
โฆ
,
๐
1,2,โฆ,n.
1
i=1 to
๐
n, if
๐
๐
โ
0
a
i
โ
๎
=0 you must remove one token from somewhere in the interval
[
๐
๐
,
โ
๐
]
[a
i
โ
,i] (choosing among any tokens there that are still present).
๐
(
๐
)
f(a) is the total number of different ways you can carry out those
๐
n removals in sequence.
Sum those weights over all valid sequences, and take the result modulo
๐
m.
There are
(
๐
+
1
)
!
(n+1)! valid sequences in total.
code this in java
pythonfrom itertools import product, combinations import math def f_of_sequence(a): n = len(a) # tokens present: positions 1..n # simulate removal choices count. # We need to count all ways to choose removal token at each i where a_i>0 def helper(i, present): # present: set of tokens still present if i > n: return 1 ai = a[i-1] if ai == 0: # no removal return helper(i+1, present) # else removal choices: choose t in present intersect [ai..i] choices = [t for t in present if ai <= t <= i] total = 0 for t in choices: new_present = present.copy() new_present.remove(t) total += helper(i+1, new_present) return total return helper(1, set(range(1, len(a)+1))) def sum_f(n): total = 0 for a in product(*(range(i+1) for i in range(1, n+1))): total += f_of_sequence(a) return total for n in range(6): print(n, sum_f(n))
pythonprint("n=6 computing") print(sum_f(6))
pythondef compute_W(n, mod): g=[0]*(n+1) S=[0]*(n+1) g[0]=1 for i in range(1, n+1): W_prev = g[i-1] # sum_{t=1..i-1} t*S[t] sum_t_S = sum((t * S[t]) for t in range(1, i)) % mod S_i = i*(i+1)//2 % mod g[i] = (W_prev + W_prev*S_i - sum_t_S) % mod # update S[t] for t in range(1, i): oldSt = S[t] S[t] = (oldSt + t*((W_prev - oldSt) % mod)) % mod S[i] = (i * W_prev) % mod return g[n] for n in range(7): print(n, compute_W(n, 10**9+7), sum_f(n))
python# brute injection enumeration for n=3 import itertools def sum_injections(n): # returns list of (sum weights for injection, breakdown by f(3)) g=0 breakdown = {0:0} # domain subset: for each f, value vector fvector length n, where fvector[i]=0..i+1. # generate all assignments fvector that satisfy injection: fvector[i]==0 or fvector[i] in [1..i+1] and fvector values unique if nonzero. def helper(i, used, fvec): nonlocal g, breakdown if i==n: # compute weight weight=1 for val in fvec: if val>0: weight*=val g+=weight # break down by f(3) key = fvec[n-1] breakdown[key]=breakdown.get(key,0)+weight return # assign f(i) idx=i # try f(i)=0 fvec[idx]=0 helper(i+1, used, fvec) # f(i)=t for t in range(1, idx+2): if t not in used: used2 = used.copy() used2.add(t) fvec[idx]=t helper(i+1, used2, fvec) g=0 breakdown={} helper(0, set(), [0]*n) return g, breakdown print(sum_injections(3))
Hereโs a straightforward Java implementation that runs in time and memory. It uses the combinatorial interpretation
where the sum runs over all partial injections whose domain satisfies . We build this up for by maintaining:
g[i]
: the total โweightโ over partial injections on ;w[t]
: the total weight of those injections among these that ever use the target value .Then at step , every old injection either
and each such extension multiplies its old weight by . Putting that together you get
java// g0 = 1, w[*]=0 g[i] = g[i-1] + โ_{t=1..i} t * (g[i-1] - w[t]) (all mod m) and then w[t] = w[t] // keep all old injections that already used t + t * (g[i-1] - w[t]) // plus all the new ones that pick t at step i
The total is then g[n] % m
.
That runs in about , which will handle up to a few thousand comfortably. If you need to go higher (e.g.\ up to ), youโll need to speed up the prefixโsum/pointโupdatesโfor example with a Fenwick tree / BIT or segment treeโso that both the โsum of โ and the โpointโupdate on โ become instead of .
javaimport java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws IOException { // Fast input BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(br.readLine()); int n = Integer.parseInt(st.nextToken()); int m = Integer.parseInt(st.nextToken()); // g[i] = total weight for injections on {1..i} // w[t] = total weight of those injections that ever use target t long[] g = new long[n+1]; long[] w = new long[n+1]; g[0] = 1; // empty injection has weight 1 // w[*] start at 0 // We will need to compute sum_{t=1..i} t * (g[i-1] - w[t]) mod m // To do that in O(n^2) we just do the loop directly. for (int i = 1; i <= n; i++) { long prev = g[i-1]; // 1) compute S = sum_{t=1..i} t * (prev - w[t]) long S = 0; for (int t = 1; t <= i; t++) { long diff = (prev - w[t]) % m; if (diff < 0) diff += m; S = (S + t * diff) % m; } // 2) add the โnoโnewโmapโ case g[i] = (prev + S) % m; // 3) update w[t] for t=1..i // old w[t] stays (those injections already used t, and pick nothing new) // plus t * (prev - old w[t]) for those that pick t now for (int t = 1; t <= i; t++) { long old = w[t]; long diff = (prev - old) % m; if (diff < 0) diff += m; w[t] = (old + t * diff) % m; } } // answer is g[n] mod m System.out.println(g[n] % m); } }
g[i]
is exactly
the total โweightโ of all partial injections on the first positions.
w[t]
accumulates the same sum but only over those injections that use the value somewhere in their domain.
When we go from to :
g[i-1]
.g[i-1]โ-โw[t]
); each such extension multiplies the weight by .w[t]
by keeping all the old injections that already used (they still do in the new step) plus those new ones that pick now (again t * (g[i-1]-w[t])
).Total time is
which is fine up to a few thousand. If you need up to, say, , you can keep a Fenwick Tree (BIT) over the array to maintain
each in instead of , bringing the total to .
Note: Because there are valid sequences, any bruteโforce over all of them is completely infeasible for even moderate ; this โinjectionโDPโ is the standard way to collapse the double sum (over sequences and the removalโchoices) down to a single pass.