1 //A,B,C的排列
2 #include <iostream>
3 using namespace std;
4 void main(void){
5 char k=65;
6 int line,n=1;
7 cout<<"请输入你所需要的行数:";
8 cin>>line;
9
10 for(int a=1 ;a<=line;a++)
11 {
12 for(int b=1 ;b<=2*n-1;b++)
13 {cout<<k;}
14
15
16 k++;
17 n++;
18 cout<<endl;
19 }
20
21 }
1 //离散实验3作业源程序.cpp
2 #include<iostream>
3 using namespace std;
4
5 int main()
6 {
7 int m, n, i, j, k;
8 cout << "Please input m and n:" << endl;
9 cin >> m >> n;
10
11 int *arr = new int[n];
12 int **arr1 = new int *[n];
13 for (i = 0; i < n; i++)
14 arr1[i] = new int[n];
15 int *arr2 = new int[m];
16 int **arr3 = new int*[m];
17 for (i = 0; i < m; i++)
18 arr3[i] = new int[m];
19 int **arr4 = new int*[n];
20 for (i = 0; i < n; i++)
21 arr4[i] = new int[n];
22
23 cout << "Please input " << n << " numbers:" << endl;
24 for (i = 0; i < n; i++)
25 cin >> arr[i];
26 cout << "Please input " << n << " numbers:" << endl;
27 for (j = 0; j < m; j++)
28 cin >> arr2[j];
29
30 for (i = 0; i < n; i++)
31 for (j = 0; j < n; j++)
32 {
33 if (arr[i] > arr[j])
34 arr1[i][j] = 1;
35 else
36 arr1[i][j] = 0;
37 }
38 for (i = 0; i < m; i++)
39 for (j = 0; j < m; j++)
40 {
41 if (arr2[i] > arr2[j])
42 arr3[i][j] = 1;
43 else
44 arr3[i][j] = 0;
45 }
46 cout << "矩阵arr1的逆矩阵为：" << endl;
47 for (i = 0; i < n; i++)
48 {
49 for (j = 0; j < n; j++)
50 {
51 cout << arr1[j][i] << " ";
52 }
53 cout << endl;
54 }
55 cout << "矩阵arr3的逆矩阵为:" << endl;
56 for (i = 0; i < n; i++)
57 {
58 for (j = 0; j < n; j++)
59 {
60 cout << arr3[j][i] << " ";
61 }
62 cout << endl;
63 }
64
65 cout << "矩阵arr1和arr3的布尔乘积为：" << endl;
66 for (i = 0; i < n; i++)
67 for (j = 0; j < n; j++)
68 for (k = 0; k < m; k++)
69 {
70 if (arr1[i][k] * arr3[k][j] == 1)
71 arr4[i][j] = 1;
72 else
73 arr4[i][j] = 0;
74 }
75 for (i = 0; i < n; i++)
76 {
77 cout << endl;
78 for (j = 0; j < m; j++)
79 cout << arr4[i][j] << " ";
80 }
81 cout << endl;
82 cout << "矩阵arr1 Warshall算法实现程序:" << endl;
83
84 for (k = 0; k <n; k++)
85 for (i =0; i < n; i++)
86 {
87 if (arr1[i][k] == 1)
88 for (j = 0; j <m; j++)
89 {
90 if (arr1[k][j] == 1)
91 arr1[i][j] = 1;
92 }
93 }
94
95 for (k = 0; k < n; k++)
96 for (i = 0; i < n; i++)
97 {
98 if (arr3[i][k] == 1)
99 for (j = 0; j < m; j++)
100 {
101 if (arr3[k][j] == 1)
102 arr3[i][j] = 1;
103 }
104 }
105
106
107 for (i = 0; i < n; i++)
108 {
109 for (int j = 0; j < n; j++)
110 cout << arr1[i][j] << " ";
111 cout << endl;
112 }
113 cout << endl;
114 for (i = 0; i < n; i++)
115 {
116 for (int j = 0; j < n; j++)
117 cout << arr3[i][j] << " ";
118 cout << endl;
119 }
120 }
1 //参考代码.cpp
2 #include<iostream>
3
4 using namespace std;
5 void strMatrix(int *a1, int **a2, int n);
6 void output(int **a2, int n);
7 int main()
8 {
9 int n, i, c, d;
10 cout << "How many numbers do you want to input:" << endl;
11 cin >> n;
12 int *arr1 = new int[n];
13 int **arr2 = new int*[n];
14 for (i = 0; i < n; i++)
15 arr2[i] = new int[n];
16 cout << "Please input the numbers:" << endl;
17 for (i = 0; i < n; i++)
18 cin >> arr1[i];
19 cout << "Please input the ralations:" << endl;
20 cin >> c;
21 cin >> d;
22 strMatrix(arr1, arr2, n);
23 output(arr2, n);
24 }
25 void strMatrix(int *a1, int **a2, int n)
26 {
27 for (int i = 0; i < n; i++)
28 for (int j = 0; j < n; j++)
29 {
30 if (a1[i] % a1[j] == 0)
31 a2[i][j] = 1;
32 else
33 a2[i][j] = 0;
34 }
35 }
36 void output(int **a2, int n)
37 {
38 for (int i = 0; i < n; i++)
39 {
40 for (int j = 0; j < n; j++)
41 cout << a2[i][j] << " ";
42 cout << endl;
43 }
44 }
45
46 void zfx(int a[M][M], int n)
47 {
48 int i = 0, vale1 = 1, vale2 = 1;/*vale1确定关系的自反性,vale2
49 判断关系的反自反性*/
50 while (i < n)
51 {
52 if (a[i][i] == 1) vale2 = 0;
53 else vale1 = 0;
54 i++;
55 }
56 if (vale1 == 1) printf("zfx\n");
57 if (vale2 == 1) printf("fzfx\n");
58 if (!(vale1) && !(vale2)) printf("wzfxywfzfx\n");
59 }
60 void dcx(int a[M][M], int n)
61 {
62 int i, j, vale3 = 1, vale4 = 1;/*vale3确定关系的对称性,vale4
63 判断关系的反对称性*/
64 for (i = 0; i < n; i++)
65 for (j = 0; j < i; j++)
66 if (a[i][j] == a[j][i])
67 vale4 = 0;
68 else vale3 = 0;
69 if (vale3 == 1) printf("dcx\n");
70 if (vale4 == 1) printf("fdcx\n");
71 if (!(vale3) && !(vale4)) printf("wdcx wfdcx\n");
72 } void cdx1(int a[M][M], int n)/*用MM ⊆2 算法来判断传递 性*/
73 {
74 int b[M][M], i, j, k, vale = 1;
75 for (i = 0; i < n; i++)
76 for (j = 0; j < n; j++)
77 {
78 b[i][j] = 0;
79 for (k = 0; k < n; k++)
80 b[i][j] += a[i][k] * a[k][j];/*计算 2M */
81 }
82 for (i = 0; i < n; i++)
83 for (j = 0; j < n; j++)
84 if (b[i][j] != 0)
85 b[i][j] = 1;
86 for (i = 0; i < n; i++)
87 for (j = 0; j < n; j++)
88 if (b[i][j] == 1)
89 if (a[i][j] != b[i][j])
90 vale = 0;
91 if (vale)
92 printf("ycdx\n");
93 else
94 printf("wcdx\n");
95 }
96 void cdx2(int a[M][M], int n)/*用warshall算法来求传递闭
97 包数组b */
98 {
99 int b[M][M], i, j, k, vale = 1;
100 for (i = 0; i < n; i++)
101 for (j = 0; j < n; j++)
102 b{ [i][j] = a[i][j];
103 for (k = 0; k < n; k++)
104 i{ (f a[i][j] && (a[i][k] || a[j][k]))
105 b[i][k] = 1;
106 }
107 }
108 for (i = 0; i < n; i++) /* 判断数组a和b相等 */
109 for (j = 0; j < n; j++)
110 (iaf[i][j] != b[i][j])
111 {
112 vale = 0; break;
113 }
114 if (vale == 1) printf("cdx\n");
115 else printf("wcdx\n");
116 }
