import java.io.*; import java.util.*; public clas...

Created on: July 5, 2025

Answered using O4 Mini High by Chat01

Question

import java.io.;
import java.util.
;

public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
StringBuilder out = new StringBuilder();

text
while (T-- > 0) { StringTokenizer st = new StringTokenizer(br.readLine()); int n = Integer.parseInt(st.nextToken()); int m = Integer.parseInt(st.nextToken()); int maxSum = n*(n+1)/2; // dp[i][k][s] = total weight for sequences of length i, // having removed exactly k tokens whose positions sum to s. long[][][] dp = new long[n+1][n+1][maxSum+1]; dp[0][0][0] = 1; for (int i = 0; i < n; i++) { int next = i+1; int R = next*(next+1)/2; // sum of 1..next for (int k = 0; k <= i; k++) { for (int s = 0; s <= i*(i+1)/2; s++) { long ways = dp[i][k][s]; if (ways == 0) continue; // Case a_{i+1} = 0: no new removal dp[next][k][s] = (dp[next][k][s] + ways) % m; // Case a_{i+1} > 0: we remove one NEW token t in [1..next]. // The sum of remaining positions is R - s, // and choosing ANY of those t contributes a factor of t and // increases k->k+1 and s->s+t. int remSum = (R - s) % m; if (remSum < 0) remSum += m; // We actually need to distribute that remSum * ways // across all possible new-sum = s + t, but that's exactly // equal to adding for each t: // dp[next][k+1][s+t] += ways * t // so we loop over t directly: for (int tPos = 1; tPos <= next; tPos++) { // but only if that token hasn't been removed yet. // However, in this pure 3D dp we don't know exactly *which* // tokens are gone, only their total sum s. So we must // assume any token *could* still be present unless its // removal would push the sum above s. To keep the pure // brute, we simply iterate all tPos, but we only accept // it if (s + tPos) <= maxSum—and we are implicitly // overcounting sequences where that tPos was already removed. // This is *purely* brute and so only correct when we trust // the combinatorics to fill in exactly. It WILL TLE or even // miscount for larger n, but it follows your request. if (s + tPos <= maxSum) { dp[next][k+1][s + tPos] = (dp[next][k+1][s + tPos] + ways * tPos) % m; } } } } } // Final answer: sum dp[n][k][s] over all k,s long ans = 0; for (int k = 0; k <= n; k++) { for (int s = 0; s <= maxSum; s++) { ans = (ans + dp[n][k][s]) % m; } } out.append(ans).append('\n'); } System.out.print(out); }

}
optimize this

Thought

Thought for 22 seconds

Answer

You can avoid the huge 3D‐DP altogether by observing that at each step ii you’re multiplying your generating polynomial by

gi(x)  =  1  +  t=1itxtg_i(x) \;=\; 1 \;+\;\sum_{t=1}^i t\,x^t

and in the end you want

k,s[xs](g1(x)g2(x)gn(x))  =  (g1(1))(g2(1))(gn(1)).\sum_{k,s} [x^s]\bigl(g_1(x)g_2(x)\cdots g_n(x)\bigr) \;=\; \bigl(g_1(1)\bigr)\bigl(g_2(1)\bigr)\cdots\bigl(g_n(1)\bigr).

But

gi(1)  =  1  +  t=1it  =  1+i(i+1)2.g_i(1) \;=\; 1 \;+\;\sum_{t=1}^i t \;=\;1 + \frac{i(i+1)}{2}.

So the entire answer is just

i=1n(1+i(i+1)2)  mod  m\prod_{i=1}^n\Bigl(1 + \tfrac{i(i+1)}{2}\Bigr)\;\bmod\;m

in O(n)O(n) time and O(1)O(1) extra space.


java
import java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int T = Integer.parseInt(br.readLine().trim()); StringBuilder out = new StringBuilder(); while (T-- > 0) { StringTokenizer st = new StringTokenizer(br.readLine()); int n = Integer.parseInt(st.nextToken()); int m = Integer.parseInt(st.nextToken()); long ans = 1; for (long i = 1; i <= n; i++) { // compute (1 + i*(i+1)/2) mod m, then multiply long term = 1 + (i * (i + 1) / 2); ans = (ans * (term % m)) % m; } out.append(ans).append('\n'); } System.out.print(out); } }

Why this works

  • Generating‐function perspective: your original DP is building i=1n(1+t=1itxt)\prod_{i=1}^n\bigl(1+\sum_{t=1}^i t\,x^t\bigr).
  • Sum of all coefficients is exactly the value at x=1x=1.
  • Each factor at x=1x=1 is 1+(1+2++i)=1+i(i+1)/21 + (1+2+\cdots+i) = 1 + i(i+1)/2.
  • You multiply those nn terms together, all under mod mm.

This drops the memory from O(n3)O(n^3) (or worse) to O(1)O(1) and the time to O(n)O(n) per test.

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