[BaekJoon] 2740번 - 행렬 곱셈 [Java][C++]

문제 링크

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1. 문제 풀이

두 행렬의 곱셈을 구현하는 문제로 2차원 배열로 행렬을 다루면 간단하게 해결할 수 있다.

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2. 코드

1. 풀이 [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));
        StringBuilder sb = new StringBuilder();
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());

        int[][] a = new int[n][m];
        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < m; j++) {
                a[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        st = new StringTokenizer(br.readLine());
        m = Integer.parseInt(st.nextToken());
        int t = Integer.parseInt(st.nextToken());

        int[][] b = new int[m][t];
        for (int i = 0; i < m; i++) {
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < t; j++) {
                b[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        int[][] res = multiply(a, b, n, m, t);
        for (int i = 0; i < res.length; i++) {
            for (int j = 0; j < res[i].length; j++) {
                sb.append(res[i][j]).append(" ");
            }
            sb.append("\n");
        }

        System.out.print(sb);
    }

    static int[][] multiply(int[][] a, int[][] b, int n, int m, int t) {
        int[][] res = new int[n][t];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < t; j++) {
                for (int k = 0; k < m; k++) {
                    res[i][j] += a[i][k] * b[k][j];
                }
            }
        }

        return res;
    }
}

2. 풀이 [C++]

#include <bits/stdc++.h>
using namespace std;

int a[100][100];
int b[100][100];
int res[100][100];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, k;
    cin >> n >> m;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> a[i][j];
        }
    }

    cin >> m >> k;

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < k; j++) {
            cin >> b[i][j];
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < k; j++) {
            for (int t = 0; t < m; t++) {
                res[i][j] += a[i][t] * b[t][j];
            }
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < k; j++) {
            cout << res[i][j] << ' ';
        }
        cout << '\n';
    }
}

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