BOJ 10830: 행렬 제곱

작성일

https://www.acmicpc.net/problem/10830

  • 2월동안 187문제 풀기(지난달 누적 87개) (42/187)
  • 행렬 곱셈방법 기억하기, 앞선 문제랑 동일한 pow 구현
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#include <bits/stdc++.h>
using namespace std;

long long N, B;

struct Mat {
  vector<vector<int>> arr;

	Mat() {
	  arr.resize(N, vector<int>(N));
  }

  Mat operator*(Mat& b) {
    Mat ans;
    for(int i = 0; i < N; i++)
      for(int j = 0; j < N; j++)
        for(int k = 0; k < N; k++) {
          int tmp = (arr[i][k] * b.arr[k][j]) % 1000;
          ans.arr[i][j] = (ans.arr[i][j] + tmp) % 1000;
        }
    return ans;
  }

  Mat& operator=(Mat b) {
    for(int i = 0; i < N; i++)
      for(int j = 0; j < N; j++)
          arr[i][j] = b.arr[i][j];
    return *this;
  }

  void print() {
    for(int i = 0; i < N; i++) {
      for(int j = 0; j < N; j++)
        cout << arr[i][j] << " ";
      cout << "\n";
    }
  }
};

Mat pow(Mat A, long long e) {
  Mat ret;
  for(int i = 0; i < N; i++)
    ret.arr[i][i] = 1;
  while(e) {
    if(e&1) ret = ret * A;
    e /= 2;
    A = A * A;
  }
  return ret;
}

int main() {
  ios::sync_with_stdio(false);
  cin >> N >> B;
  Mat A;
  for(int i = 0; i < N; i++)
    for(int j = 0; j < N; j++)
      cin >> A.arr[i][j];
  pow(A, B).print();
  return 0;
}