检查是否可以创建一个矩阵，使得每一行都有A 1，每列都有B 1

给定四个整数N，M，A，B ，其中N为行数，M为列数，任务是检查是否有可能创建尺寸为N x M的二进制矩阵，使得每一行都有A数为1s ，每列的B数为1s 。如果有任何这样的矩阵，则打印它，否则打印“ -1” 。

例子：

Input: N = 3, M = 6, A = 2, B = 1
Output: 
1 1 0 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
Explanation:
Every row has A ones i.e. 2 and every column has B ones i.e., 1.

Input: N = 2, M = 2, A = 2, B = 1
Output: No
Explanation:
It is not possible to create such a 2 x 2 matrix in which every row has 2 ones and every column has 1 ones because of the following two observations:
1. For every row place two ones because of which we will never be able to have one 1 in every column.
1 1         
1 1

2. For every column place one 1 because of which we can never have 2 ones in every row.         
1 0         
0 1 

方法：这样做的目的是要观察到，由于每一行应具有精确的A 1s ，每列应具有精确的B 1s ，因此所有行A * N的个数应等于所有列B *的1s M．因此，当且仅当A * N = B * M时，才存在期望的矩阵。下图是插图：

1. 发现任意数量的0 使得(d * N)％M == 0，其中A％B是A除以B的余数。
2. 在所需矩阵的第一行中，将其插入位置[1，A] 。
3. 与^第i – 1行一样，在第i行中放入那些，但将d循环右移。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Number of rows
const int n = 3;
  
// Number of columns
const int m = 6;
  
// Function that prints the matrix
// if it exists
void printMatrix(int arr[][m],
                 string ans)
{
    if (ans == "No")
        cout << "No\n";
    else {
        // Print if matrix exists
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                cout << arr[i][j] << " ";
            cout << '\n';
        }
    }
}
  
// Function to check if it is possible
// to create a matrix such that every
// row has A 1s & every column has B 1s
void createMatrix(int a, int b)
{
    int matrix[n][m], row[n], col[m];
  
    // Initialize all matrix
    // entries equal to 0
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            matrix[i][j] = 0;
        }
    }
  
    // Initialize the number of
    // ones required in every row
    for (int i = 0; i < n; i++)
        row[i] = a;
  
    // Initialize the number of
    // ones required in each column
    for (int i = 0; i < m; i++)
        col[i] = b;
  
    int l = 0, d = 0;
  
    // Check if the total number of
    // ones required in every row is
    // not equal to total number of
    // ones required in every column
    // then print No
    if (n * a != m * b)
        printMatrix(matrix, "No");
  
    else {
  
        for (int i = 0; i < n; i++) {
            int j;
            if (l == m)
                l = 0;
  
            for (j = l; j < m; j++) {
  
                if (row[i] > 0 && col[j] > 0) {
  
                    // Fill a one if there is a
                    // place to be filled
                    matrix[i][j] = 1;
  
                    // Decrease the number of
                    // ones required in ith row
                    row[i]--;
  
                    // Decrease the number of
                    // ones required in jth column
                    col[j]--;
                    d = j;
                }
            }
            l = d + 1;
            if (row[i] != 0) {
  
                for (j = 0; j < m; j++) {
  
                    if (row[i] > 0 && col[j] > 0) {
  
                        // Fill a one if there is
                        // a place to be filled
                        matrix[i][j] = 1;
  
                        // Decrease the number of 1s
                        // required in ith row
                        row[i]--;
                        // Decrease the number of 1s
                        // required in jth column
                        col[j]--;
                        l = j + 1;
                    }
                    // Break the loop if no place
                    // is left for ones to filled
                    if (row[i] == 0)
                        break;
                }
            }
        }
  
        // Function call to print the matrix
        printMatrix(matrix, "Yes");
    }
}
  
// Driver Code
int main()
{
  
    // Number of ones required
    // in every row
    int a = 2;
  
    // Number of ones required
    // in every column
    int b = 1;
  
    // Function call
    createMatrix(a, b);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Number of rows
static int n = 3;
  
// Number of columns
static int m = 6;
  
// Function that prints the matrix
// if it exists
static void printMatrix(int arr[][],
                        String ans)
{
    if (ans == "No")
        System.out.print("No\n");
    else 
    {
          
        // Print if matrix exists
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
                System.out.print(arr[i][j] + " ");
            System.out.println();
        }
    }
}
  
// Function to check if it is possible
// to create a matrix such that every
// row has A 1s & every column has B 1s
static void createMatrix(int a, int b)
{
    int [][]matrix = new int[n][m];
    int []row = new int[n]; 
    int []col = new int[m];
  
    // Initialize all matrix
    // entries equal to 0
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            matrix[i][j] = 0;
        }
    }
  
    // Initialize the number of
    // ones required in every row
    for(int i = 0; i < n; i++)
        row[i] = a;
  
    // Initialize the number of
    // ones required in each column
    for(int i = 0; i < m; i++)
        col[i] = b;
  
    int l = 0, d = 0;
  
    // Check if the total number of
    // ones required in every row is
    // not equal to total number of
    // ones required in every column
    // then print No
    if (n * a != m * b)
        printMatrix(matrix, "No");
  
    else 
    {
        for(int i = 0; i < n; i++)
        {
            int j;
            if (l == m)
                l = 0;
  
            for(j = l; j < m; j++)
            {
                if (row[i] > 0 && col[j] > 0)
                {
                      
                    // Fill a one if there is a
                    // place to be filled
                    matrix[i][j] = 1;
  
                    // Decrease the number of
                    // ones required in ith row
                    row[i]--;
  
                    // Decrease the number of
                    // ones required in jth column
                    col[j]--;
                    d = j;
                }
            }
            l = d + 1;
            if (row[i] != 0)
            {
                for(j = 0; j < m; j++)
                {
                    if (row[i] > 0 && col[j] > 0) 
                    {
                          
                        // Fill a one if there is
                        // a place to be filled
                        matrix[i][j] = 1;
  
                        // Decrease the number of 1s
                        // required in ith row
                        row[i]--;
                        // Decrease the number of 1s
                        // required in jth column
                        col[j]--;
                        l = j + 1;
                    }
                      
                    // Break the loop if no place
                    // is left for ones to filled
                    if (row[i] == 0)
                        break;
                }
            }
        }
  
        // Function call to print the matrix
        printMatrix(matrix, "Yes");
    }
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Number of ones required
    // in every row
    int a = 2;
  
    // Number of ones required
    // in every column
    int b = 1;
  
    // Function call
    createMatrix(a, b);
}
}
  
// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
  
# Number of rows
n = 3
  
# Number of columns
m = 6
  
# Function that prints the matrix
# if it exists
def printMatrix(arr, ans):
      
    if (ans == "No"):
        print("No")
      
    else:
          
        # Print if matrix exists
        for i in range(n):
            for j in range(m):
                print(arr[i][j], end = " ")
                  
            print()
  
# Function to check if it is possible
# to create a matrix such that every
# row has A 1s & every column has B 1s
def createMatrix(a, b):
      
    matrix = [[0 for i in range(m)]
                 for i in range(n)]
    row = [a for i in range(n)]
    col = [b for i in range(m)]
  
    l = 0
    d = 0
  
    # Check if the total number of
    # ones required in every row is
    # not equal to total number of
    # ones required in every column
    # then prNo
    if (n * a != m * b):
        printMatrix(matrix, "No")
  
    else:
        for i in range(n):
            j = 0
              
            if (l == m):
                l = 0
  
            for j in range(l, m):
                if (row[i] > 0 and col[j] > 0):
  
                    # Fill a one if there is a
                    # place to be filled
                    matrix[i][j] = 1
  
                    # Decrease the number of
                    # ones required in ith row
                    row[i] -= 1
  
                    # Decrease the number of
                    # ones required in jth column
                    col[j] -= 1
                    d = j
  
            l = d + 1
            if (row[i] != 0):
                for j in range(m):
                    if (row[i] > 0 and col[j] > 0):
  
                        # Fill a one if there is
                        # a place to be filled
                        matrix[i][j] = 1
  
                        # Decrease the number of 1s
                        # required in ith row
                        row[i] -= 1
                          
                        # Decrease the number of 1s
                        # required in jth column
                        col[j] -= 1
                        l = j + 1
  
                    # Break the loop if no place
                    # is left for ones to filled
                    if (row[i] == 0):
                        break
  
        # Function call to prthe matrix
        printMatrix(matrix, "Yes")
  
# Driver Code
if __name__ == '__main__':
  
    # Number of ones required
    # in every row
    a = 2
  
    # Number of ones required
    # in every column
    b = 1
  
    # Function call
    createMatrix(a, b)
  
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
  
// Number of rows
static int n = 3;
  
// Number of columns
static int m = 6;
  
// Function that prints the matrix
// if it exists
static void printMatrix(int [,]arr,
                        String ans)
{
    if (ans == "No")
        Console.Write("No\n");
    else 
    {
          
        // Print if matrix exists
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
                Console.Write(arr[i, j] + " ");
            Console.WriteLine();
        }
    }
}
  
// Function to check if it is possible
// to create a matrix such that every
// row has A 1s & every column has B 1s
static void createMatrix(int a, int b)
{
    int [,]matrix = new int[n, m];
    int []row = new int[n]; 
    int []col = new int[m];
  
    // Initialize all matrix
    // entries equal to 0
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            matrix[i, j] = 0;
        }
    }
  
    // Initialize the number of
    // ones required in every row
    for(int i = 0; i < n; i++)
        row[i] = a;
  
    // Initialize the number of
    // ones required in each column
    for(int i = 0; i < m; i++)
        col[i] = b;
  
    int l = 0, d = 0;
  
    // Check if the total number of
    // ones required in every row is
    // not equal to total number of
    // ones required in every column
    // then print No
    if (n * a != m * b)
        printMatrix(matrix, "No");
  
    else 
    {
        for(int i = 0; i < n; i++)
        {
            int j;
            if (l == m)
                l = 0;
  
            for(j = l; j < m; j++)
            {
                if (row[i] > 0 && col[j] > 0)
                {
                      
                    // Fill a one if there is a
                    // place to be filled
                    matrix[i,j] = 1;
  
                    // Decrease the number of
                    // ones required in ith row
                    row[i]--;
  
                    // Decrease the number of
                    // ones required in jth column
                    col[j]--;
                    d = j;
                }
            }
            l = d + 1;
            if (row[i] != 0)
            {
                for(j = 0; j < m; j++)
                {
                    if (row[i] > 0 && col[j] > 0) 
                    {
                          
                        // Fill a one if there is
                        // a place to be filled
                        matrix[i,j] = 1;
  
                        // Decrease the number of 1s
                        // required in ith row
                        row[i]--;
                        // Decrease the number of 1s
                        // required in jth column
                        col[j]--;
                        l = j + 1;
                    }
                      
                    // Break the loop if no place
                    // is left for ones to filled
                    if (row[i] == 0)
                        break;
                }
            }
        }
  
        // Function call to print the matrix
        printMatrix(matrix, "Yes");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Number of ones required
    // in every row
    int a = 2;
  
    // Number of ones required
    // in every column
    int b = 1;
  
    // Function call
    createMatrix(a, b);
}
}
  
// This code is contributed by gauravrajput1

输出：

1 1 0 0 0 0 
0 0 1 1 0 0 
0 0 0 0 1 1

时间复杂度： O(N * M)
辅助空间： O(N * M)