import java.util.*; public class Main { stati...

생성일: 2025년 7월 13일

답변에 사용된 모델 O4 Mini High by Chat01

질문

import java.util.*;

public class Main {
static final int MOD = (int)1e9+7;

text
static class Mat { long a00, a01, a10, a11; Mat(long a00, long a01, long a10, long a11) { this.a00 = a00; this.a01 = a01; this.a10 = a10; this.a11 = a11; } Mat mul(Mat o) { return new Mat( (a00 * o.a00 + a01 * o.a10) % MOD, (a00 * o.a01 + a01 * o.a11) % MOD, (a10 * o.a00 + a11 * o.a10) % MOD, (a10 * o.a01 + a11 * o.a11) % MOD ); } } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int Q = sc.nextInt(); Mat[] base = new Mat[Math.max(0, N-1)]; for (int i = 0; i < N; i++) { long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); if (i < N-1) { base[i] = new Mat(r % MOD, w % MOD, w % MOD, b % MOD); } } long xorSum = 0; while (Q-- > 0) { int type = sc.nextInt(); if (type == 0) { int u = sc.nextInt(); int v = sc.nextInt(); int c = sc.nextInt(); if (u > v) { int tmp = u; u = v; v = tmp; } Mat cur = new Mat(1, 0, 0, 1); for (int k = u; k < v; k++) { cur = cur.mul(base[k-1]); } long ways = (c == 0 ? cur.a00 : cur.a11); xorSum^ = ways; } else { int i = type; long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); if (i >= 1 && i <= N-1) { base[i-1] = new Mat(r % MOD, w % MOD, w % MOD, b % MOD); } } } System.out.println(xorSum); sc.close(); }

}
this code is passing 8/12 tc but giving wrong answer in 4 tc

사고

Debugging matrix exponentiation

The user is working with Java code for matrix exponentiation and says they have incorrect results for 4 out of 12 tests, though no timeouts. They mention using a naive product method within a loop to process each query. This could be inefficient since it’s O(N * Q) in the worst case and may cause timeouts or integer overflows. Since the claim isn't about timing but wrong answers, the issue might lie in how the matrix multiplication is handled or perhaps rounding.

Reviewing code logic

The user has shared Java code that performs matrix exponentiation, uses matrix multiplication, and handles two types of queries. One type (type == 0) processes a range of matrices, calculates a result, and XORs it. The other updates a matrix value. The XOR approach makes sense, but there might be potential issues if matrix multiplication accumulates inaccuracies due to integer overflow. It could also be slow due to the repeated multiplications within the loop for each query, and the matrix sizes and number of queries are likely factors in the performance problems.

Identifying input handling bugs

The user’s code seems to incorrectly read input for updates. Specifically, for update queries, the code treats type as the index, but type==1 should indicate an update, followed by the index i. So, the bug is that the code reads type as the index, leading to wrong updates. Additionally, matrix operations should be optimized for performance using a segment tree, as the current approach is inefficient for large inputs. The incorrect results might be due to faulty input handling logic, not timeouts. Fixing input parsing should resolve the issue.

Fixing input handling and

Thought for 16 seconds

답변

The core issue isn’t modular arithmetic or overflow—it’s in how you’re reading the input:

  1. You’re reading one too many “edge”-triples.
    The problem statement gives you exactly N−1 triples (r, b, w), one per edge, but your loop runs i=0..N−1, so you actually consume the first query’s parameters as that last “useless” triple.

  2. You never read the update index.
    On an update you do:

    java
    int type = sc.nextInt(); // … } else { int i = type; // WRONG: `type` is just the code (1), not the edge-index! long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); base[i-1] =; }

    But the input for an update is actually 1 idx r b w, so you need to do:

    java
    int code = sc.nextInt(); if (code == 0) { … query … } else { int idx = sc.nextInt(); // <-- read the edge-index long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); base[idx-1] = new Mat(r%MOD, w%MOD, w%MOD, b%MOD); }

With those two fixes your input will line up correctly and you won’t silently “eat” part of the queries. Here’s a drop-in corrected version of your main:

java
import java.util.*; public class Main { static final int MOD = (int)1e9+7; static class Mat { long a00, a01, a10, a11; Mat(long a00, long a01, long a10, long a11) { this.a00 = a00; this.a01 = a01; this.a10 = a10; this.a11 = a11; } Mat mul(Mat o) { return new Mat( (a00*o.a00 + a01*o.a10) % MOD, (a00*o.a01 + a01*o.a11) % MOD, (a10*o.a00 + a11*o.a10) % MOD, (a10*o.a01 + a11*o.a11) % MOD ); } } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int Q = sc.nextInt(); // read exactly N-1 triples Mat[] base = new Mat[N-1]; for (int i = 0; i < N-1; i++) { long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); base[i] = new Mat(r % MOD, w % MOD, w % MOD, b % MOD); } long xorSum = 0; while (Q-- > 0) { int type = sc.nextInt(); if (type == 0) { // a range‐query int u = sc.nextInt(), v = sc.nextInt(), c = sc.nextInt(); if (u > v) { int t=u; u=v; v=t; } Mat cur = new Mat(1,0,0,1); for (int k = u; k < v; k++) { cur = cur.mul(base[k-1]); } long ways = (c == 0 ? cur.a00 : cur.a11); xorSum ^= ways; } else { // an update: read the actual index int idx = sc.nextInt(); // <-- fix! long r = sc.nextLong(); long b = sc.nextLong(); long w = sc.nextLong(); if (idx >= 1 && idx < N) { base[idx-1] = new Mat(r % MOD, w % MOD, w % MOD, b % MOD); } } } System.out.println(xorSum); sc.close(); } }

With these changes:

  • You only consume the N−1 edge specifications.
  • You correctly read the update’s idx (instead of mis-using type).

That should eliminate the “mysterious” misalignment that was wrecking 4 of your 12 tests.

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