离散实验程序代码3

//A,B,C的排列
#include <iostream>
using namespace std;
void main(void){
    char k=65;
    int line,n=1;
    cout<<"请输入你所需要的行数:";
    cin>>line;
    
    for(int a=1 ;a<=line;a++)
    {
        for(int b=1 ;b<=2*n-1;b++)
        {cout<<k;}
            
            
            k++;
            n++;
            cout<<endl;
        }
            
    }

//离散实验3作业源程序.cpp
#include<iostream>
using namespace std;

int main()
{
    int m, n, i, j, k;
    cout << "Please input m and n:" << endl;
    cin >> m >> n;

    int *arr = new int[n];
    int **arr1 = new int *[n];
    for (i = 0; i < n; i++)
        arr1[i] = new int[n];
    int *arr2 = new int[m];
    int **arr3 = new int*[m];
    for (i = 0; i < m; i++)
        arr3[i] = new int[m];
    int **arr4 = new int*[n];
    for (i = 0; i < n; i++)
        arr4[i] = new int[n];

    cout << "Please input " << n << " numbers:" << endl;
    for (i = 0; i < n; i++)
        cin >> arr[i];
    cout << "Please input " << n << " numbers:" << endl;
    for (j = 0; j < m; j++)
        cin >> arr2[j];

    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            if (arr[i] > arr[j])
                arr1[i][j] = 1;
            else
                arr1[i][j] = 0;
        }
        for (i = 0; i < m; i++)
            for (j = 0; j < m; j++)
            {
                if (arr2[i] > arr2[j])
                    arr3[i][j] = 1;
                else
                    arr3[i][j] = 0;
            }
            cout << "矩阵arr1的逆矩阵为：" << endl;
            for (i = 0; i < n; i++)
            {
                for (j = 0; j < n; j++)
                {
                    cout << arr1[j][i] << " ";
                }
                cout << endl;
            }
            cout << "矩阵arr3的逆矩阵为:" << endl;
            for (i = 0; i < n; i++)
            {
                for (j = 0; j < n; j++)
                {
                    cout << arr3[j][i] << " ";
                }
                cout << endl;
            }

            cout << "矩阵arr1和arr3的布尔乘积为：" << endl;
            for (i = 0; i < n; i++)
                for (j = 0; j < n; j++)
                    for (k = 0; k < m; k++)
                    {
                        if (arr1[i][k] * arr3[k][j] == 1)
                            arr4[i][j] = 1;
                        else
                            arr4[i][j] = 0;
                    }
                    for (i = 0; i < n; i++)
                    {
                        cout << endl;
                        for (j = 0; j < m; j++)
                            cout << arr4[i][j] << " ";
                    }
                    cout << endl;
                    cout << "矩阵arr1 Warshall算法实现程序:" << endl;
                
                    for (k = 0; k <n; k++)
                        for (i =0; i < n; i++)
                        {
                            if (arr1[i][k] == 1)
                                for (j = 0; j <m; j++)
                                {
                                    if (arr1[k][j] == 1)
                                        arr1[i][j] = 1;
                                }
                        }
                        
                        for (k = 0; k < n; k++)
                            for (i = 0; i < n; i++)
                            {
                                if (arr3[i][k] == 1)
                                    for (j = 0; j < m; j++)
                                    {
                                        if (arr3[k][j] == 1)
                                            arr3[i][j] = 1;
                                    }
                            }


                    for (i = 0; i < n; i++)
                    {
                        for (int j = 0; j < n; j++)
                            cout << arr1[i][j] << " ";
                        cout << endl;
                    }
                    cout << endl;
                    for (i = 0; i < n; i++)
                    {
                        for (int j = 0; j < n; j++)
                            cout << arr3[i][j] << " ";
                        cout << endl;
                    }
}

//参考代码.cpp
#include<iostream>

using namespace std;
void strMatrix(int *a1, int **a2, int n);
void output(int **a2, int n);
int main()
{
    int n, i, c, d;
    cout << "How many numbers do you want to input:" << endl;
    cin >> n;
    int *arr1 = new int[n];
    int **arr2 = new int*[n];
    for (i = 0; i < n; i++)
        arr2[i] = new int[n];
    cout << "Please input the numbers:" << endl;
    for (i = 0; i < n; i++)
        cin >> arr1[i];
    cout << "Please input the ralations:" << endl;
    cin >> c;
    cin >> d;
    strMatrix(arr1, arr2, n);
    output(arr2, n);
}
void strMatrix(int *a1, int **a2, int n)
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if (a1[i] % a1[j] == 0)
                a2[i][j] = 1;
            else
                a2[i][j] = 0;
        }
}
void output(int **a2, int n)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
            cout << a2[i][j] << " ";
        cout << endl;
    }
}

void zfx(int a[M][M], int n)
{
    int i = 0, vale1 = 1, vale2 = 1;/*vale1确定关系的自反性,vale2
                                    判断关系的反自反性*/
    while (i < n)
    {
        if (a[i][i] == 1) vale2 = 0;
        else vale1 = 0;
        i++;
    }
    if (vale1 == 1) printf("zfx
");
    if (vale2 == 1) printf("fzfx
");
    if (!(vale1) && !(vale2)) printf("wzfxywfzfx
");
}
void dcx(int a[M][M], int n)
{
    int i, j, vale3 = 1, vale4 = 1;/*vale3确定关系的对称性,vale4
                                   判断关系的反对称性*/
    for (i = 0; i < n; i++)
        for (j = 0; j < i; j++)
            if (a[i][j] == a[j][i])
                vale4 = 0;
            else vale3 = 0;
            if (vale3 == 1) printf("dcx
");
            if (vale4 == 1) printf("fdcx
");
            if (!(vale3) && !(vale4)) printf("wdcx wfdcx
");
} void cdx1(int a[M][M], int n)/*用MM ⊆2 算法来判断传递 性*/
{
    int b[M][M], i, j, k, vale = 1;
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            b[i][j] = 0;
            for (k = 0; k < n; k++)
                b[i][j] += a[i][k] * a[k][j];/*计算 2M */
        }
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                if (b[i][j] != 0)
                    b[i][j] = 1;
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                if (b[i][j] == 1)
                    if (a[i][j] != b[i][j])
                        vale = 0;
        if (vale)
            printf("ycdx
");
        else   
            printf("wcdx
");
}
void cdx2(int a[M][M], int n)/*用warshall算法来求传递闭
                             包数组b */
{
    int b[M][M], i, j, k, vale = 1;
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
            b{ [i][j] = a[i][j];
    for (k = 0; k < n; k++)
        i{ (f a[i][j] && (a[i][k] || a[j][k]))
        b[i][k] = 1;
    }
    }
    for (i = 0; i < n; i++)     /* 判断数组a和b相等 */
        for (j = 0; j < n; j++)
            (iaf[i][j] != b[i][j])
        {
            vale = 0; break;
        }
        if (vale == 1) printf("cdx
");
        else printf("wcdx
");
}