题目：https://acmore.cc/problem/LOCAL/1571#desc
解法一：二维数组（因为二维数组不能作为函数的返回值，所以处理起来比较麻烦，此时推荐使用vector表示二维数组，请看解法二）
1 #include <iostream>
2
3 using namespace std;
4
5 int a[30][30] = { 0 }; //初始矩阵
6 int b[30][30] = { 0 }; //结果矩阵
7
8 int getMultiSum(int a1[][30], int a2[][30], int m, int i, int j) //计算矩阵a1的第i行与矩阵a2的第j列相乘的结果，其中a1,a2是同为m阶的方阵
9 {
10 int result = 0;
11 for (int k = 0; k < m; ++k)
12 {
13 result += a1[i][k] * a2[k][j];
14 }
15 return result;
16 }
17
18
19 int main()
20 {
21 int N, M; //N表示矩阵的阶数，M表示幂
22 while (cin >> N >> M)
23 {
24 for (int i = 0; i < N; ++i)
25 for (int j = 0; j < N; ++j)
26 cin >> a[i][j];
27
28 switch (M)
29 {
30 case 0: //矩阵的零次方是单位矩阵
31 for (int i = 0; i < N; ++i)
32 b[i][i] = 1;
33 break;
34
35 case 1:
36 for (int i = 0; i < N; ++i)
37 for (int j = 0; j < N; ++j)
38 b[i][j] = a[i][j];
39
40 case 2:
41 for (int i = 0; i < N; ++i) //i表示原矩阵的行
42 {
43 for (int j = 0; j < N; ++j) //j表示原矩阵的列
44 {
45 b[i][j] = getMultiSum(a, a, N, i, j);
46 }
47
48 }
49 break;
50
51 case 3:
52 int tmp[30][30];
53 for (int i = 0; i < N; ++i) //i表示原矩阵的行
54 {
55 for (int j = 0; j < N; ++j) //j表示原矩阵的列
56 {
57 tmp[i][j] = getMultiSum(a, a, N, i, j);
58 }
59
60 }
61 for (int i = 0; i < N; ++i) //i表示原矩阵的行
62 {
63 for (int j = 0; j < N; ++j) //j表示原矩阵的列
64 {
65 b[i][j] = getMultiSum(a, tmp, N, i, j);
66 }
67
68 }
69 break;
70
71 case 4:
72 int tmp1[30][30];
73 for (int i = 0; i < N; ++i) //i表示原矩阵的行
74 {
75 for (int j = 0; j < N; ++j) //j表示原矩阵的列
76 {
77 tmp1[i][j] = getMultiSum(a, a, N, i, j);
78 }
79
80 }
81 int tmp2[30][30];
82 for (int i = 0; i < N; ++i) //i表示原矩阵的行
83 {
84 for (int j = 0; j < N; ++j) //j表示原矩阵的列
85 {
86 tmp2[i][j] = getMultiSum(a, tmp1, N, i, j);
87 }
88
89 }
90 for (int i = 0; i < N; ++i) //i表示原矩阵的行
91 {
92 for (int j = 0; j < N; ++j) //j表示原矩阵的列
93 {
94 b[i][j] = getMultiSum(a, tmp2, N, i, j);
95 }
96
97 }
98
99 break;
100
101 case 5:
102 int tmp3[30][30];
103 for (int i = 0; i < N; ++i) //i表示原矩阵的行
104 {
105 for (int j = 0; j < N; ++j) //j表示原矩阵的列
106 {
107 tmp3[i][j] = getMultiSum(a, a, N, i, j);
108 }
109
110 }
111 int tmp4[30][30];
112 for (int i = 0; i < N; ++i) //i表示原矩阵的行
113 {
114 for (int j = 0; j < N; ++j) //j表示原矩阵的列
115 {
116 tmp4[i][j] = getMultiSum(a, tmp3, N, i, j);
117 }
118
119 }
120 int tmp5[30][30];
121 for (int i = 0; i < N; ++i) //i表示原矩阵的行
122 {
123 for (int j = 0; j < N; ++j) //j表示原矩阵的列
124 {
125 tmp5[i][j] = getMultiSum(a, tmp4, N, i, j);
126 }
127
128 }
129 for (int i = 0; i < N; ++i) //i表示原矩阵的行
130 {
131 for (int j = 0; j < N; ++j) //j表示原矩阵的列
132 {
133 b[i][j] = getMultiSum(a, tmp5, N, i, j);
134 }
135
136 }
137
138
139 break;
140
141 }
142
143 for (int i = 0; i < N; ++i)
144 {
145 int j = 0;
146 for (; j < N-1; ++j)
147 {
148 cout << b[i][j] << ‘ ‘;
149 }
150 cout << b[i][j] << endl;
151 }
152 }
153
154 return 0;
155
156 }
解法二： 使用vector代替二维数组
1 #include <iostream>
2 #include <vector>
3
4 using namespace std;
5
6
7 //矩阵a1与a2相乘，其中a1和a2都是N阶方阵，返回值为vector表示的二维数组
8 vector<vector<int>> matrixMul(vector<vector<int>>a1, vector<vector<int>>a2, int N)
9 {
10 vector<vector<int>> result(30);
11 for (int i = 0; i < N; ++i)
12 for (int j = 0; j < N; ++j)
13 result[i].push_back(0); //把结果矩阵置为0
14
15
16 for (int i = 0; i < N; i++) //矩阵乘法的三重循环
17 {
18 for (int j = 0; j < N; j++)
19 {
20 for (int k = 0; k < N; k++)
21 {
22 result[i][j] += a1[i][k] * a2[k][j];
23 }
24
25 }
26 }
27
28 return result;
29
30 }
31
32
33 int main()
34 {
35 int N, M; //N表示矩阵的阶数，M表示幂
36 while (cin >> N >> M)
37 {
38 vector<vector<int>>a(30); //原矩阵
39 vector<vector<int>>b(30); //结果矩阵
40 int e;
41 for (int i = 0; i < N; ++i)
42 for (int j = 0; j < N; ++j)
43 {
44 cin >> e;
45 a[i].push_back(e); //输入原矩阵
46 }
47
48 for (int i = 0; i < N; ++i)
49 for (int j = 0; j < N; ++j)
50 b[i].push_back(0); //把结果矩阵初始化为0
51
52 if (M == 0)
53 {
54 for (int i = 0; i < N; ++i)
55 b[i][i] = 1; //矩阵的零次方为单位矩阵
56 }
57 else
58 {
59 b = a; //矩阵的一次方
60 if(M > 1) //矩阵的M次方（M大于1）
61 for (int i = 0; i < M - 1; ++i)
62 {
63 b = matrixMul(a, b, N);
64 }
65 }
66
67
68 for (int i = 0; i < N; ++i) //输出结果
69 {
70 int j = 0;
71 for (; j < N-1; ++j)
72 {
73 cout << b[i][j] << ‘ ‘;
74 }
75 cout << b[i][j] << endl;
76 }
77 }
78
79 return 0;
80
81 }
原文地址：https://www.cnblogs.com/FengZeng666/p/10355938.html
时间： 2024-08-28 04:19:43