L1-048 矩阵 A 乘以 B

给定两个矩阵A和B，要求你计算它们的乘积矩阵A**B。需要注意的是，只有规模匹配的矩阵才可以相乘。即若A有R**a行、C**a列，B有R**b行、C**b列，则只有C**a与R**b相等时，两个矩阵才能相乘。

输入格式：

输入先后给出两个矩阵A和B。对于每个矩阵，首先在一行中给出其行数R和列数C，随后R行，每行给出C个整数，以 1 个空格分隔，且行首尾没有多余的空格。输入保证两个矩阵的R和C都是正数，并且所有整数的绝对值不超过 100。

输出格式：

若输入的两个矩阵的规模是匹配的，则按照输入的格式输出乘积矩阵A**B，否则输出Error: Ca != Rb，其中Ca是A的列数，Rb是B的行数。

输入样例 1：

tex

2 3
1 2 3
4 5 6
3 4
7 8 9 0
-1 -2 -3 -4
5 6 7 8

2 3
1 2 3
4 5 6
3 4
7 8 9 0
-1 -2 -3 -4
5 6 7 8

输出样例 1：

tex

2 4
20 22 24 16
53 58 63 28

2 4
20 22 24 16
53 58 63 28

输入样例 2：

输出样例 2：

Error: 2 != 3

Error: 2 != 3

Solution：

java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

public class Main {
    static int[][] a = new int[101][101];
    static int[][] b = new int[101][101];
    static int ra, ca, rb, cb;

    static int f(int r, int c) {
        int sum = 0;
        for (int i = 1; i <= ca; i++)
            sum += a[r][i] * b[i][c];
        return sum;

    }

    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        in.nextToken();
        ra = (int) in.nval;
        in.nextToken();
        ca = (int) in.nval;
        for (int i = 1; i <= ra; i++) {
            for (int j = 1; j <= ca; j++) {
                in.nextToken();
                a[i][j] = (int) in.nval;
            }
        }

        in.nextToken();
        rb = (int) in.nval;
        in.nextToken();
        cb = (int) in.nval;
        for (int i = 1; i <= rb; i++) {
            for (int j = 1; j <= cb; j++) {
                in.nextToken();
                b[i][j] = (int) in.nval;
            }
        }

        if (ca != rb) {
            System.out.printf("Error: %d != %d", ca, rb);
        } else {
            System.out.printf("%d %d\n", ra, cb);
            StringBuilder sb = new StringBuilder();
            for (int i = 1; i <= ra; i++) {
                for (int j = 1; j <= cb; j++) {
                    if (j > 1) sb.append(" ");
                    sb.append(f(i, j));
                }
                sb.append("\n");
            }
            System.out.print(sb);
        }
    }
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

public class Main {
    static int[][] a = new int[101][101];
    static int[][] b = new int[101][101];
    static int ra, ca, rb, cb;

    static int f(int r, int c) {
        int sum = 0;
        for (int i = 1; i <= ca; i++)
            sum += a[r][i] * b[i][c];
        return sum;

    }

    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        in.nextToken();
        ra = (int) in.nval;
        in.nextToken();
        ca = (int) in.nval;
        for (int i = 1; i <= ra; i++) {
            for (int j = 1; j <= ca; j++) {
                in.nextToken();
                a[i][j] = (int) in.nval;
            }
        }

        in.nextToken();
        rb = (int) in.nval;
        in.nextToken();
        cb = (int) in.nval;
        for (int i = 1; i <= rb; i++) {
            for (int j = 1; j <= cb; j++) {
                in.nextToken();
                b[i][j] = (int) in.nval;
            }
        }

        if (ca != rb) {
            System.out.printf("Error: %d != %d", ca, rb);
        } else {
            System.out.printf("%d %d\n", ra, cb);
            StringBuilder sb = new StringBuilder();
            for (int i = 1; i <= ra; i++) {
                for (int j = 1; j <= cb; j++) {
                    if (j > 1) sb.append(" ");
                    sb.append(f(i, j));
                }
                sb.append("\n");
            }
            System.out.print(sb);
        }
    }
}

L1-048 矩阵 A 乘以 B ​

输入格式： ​

输出格式： ​

输入样例 1： ​

输出样例 1： ​

输入样例 2： ​

输出样例 2： ​

Solution： ​