查找是否有可能制作一个包含给定数字“ 0”，“ 1”，“ 01”和“ 10”作为子序列的单宁酸的二进制字符串

给定四个整数l ， m ， x和y 。任务是检查是否可以创建由l 0 ， m 1 ， x“ 01”和y“ 10”作为子序列的二进制字符串。

例子：

Input: l = 3, m = 2, x = 4, y = 2
Output: Yes
Possible string is “00110”. It contains 3 0’s, 2 1’s,
4 “01” sub-sequences and 2 “10” sub-sequences.

Input: l = 3, m = 2, x = 4, y = 3
Output: No
No such binary string exists.

方法：可能的字符串始终为00…11…00…的形式。首先由一些零组成，然后由所有零组成，然后由剩余的零组成。
假设l1为1之前的零个数， l2为1后面的零个数，则等式为：

1. l1 + l2 = l (零总数)。
2. l1 * m = x (“ 01”子序列数)。
3. m * l2 = y (“ 10”个子序列的数量)。

从以上三个方程式中，我们得到x + y = l * m 。如果对于给定的值，此方程式失败，则该字符串不可能，否则打印Yes 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function that returns true if it is possible to
// make a binary string consisting of l 0's, m 1's,
// x "01" sub-sequences and y "10" sub-sequences
bool isPossible(int l, int m, int x, int y)
{
    if (l * m == x + y)
        return true;
  
    return false;
}
  
// Driver code
int main()
{
    int l = 3, m = 2, x = 4, y = 2;
  
    if (isPossible(l, m, x, y))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation of the approach 
class sol
{
      
// Function that returns true if it is possible to 
// make a binary string consisting of l 0's, m 1's, 
// x "01" sub-sequences and y "10" sub-sequences 
static boolean isPossible(int l, int m, int x, int y) 
{ 
    if (l * m == x + y) 
        return true; 
  
    return false; 
} 
  
// Driver code 
public static void main(String args[])
{ 
    int l = 3, m = 2, x = 4, y = 2; 
  
    if (isPossible(l, m, x, y)) 
        System.out.print("Yes"); 
    else
        System.out.print("No"); 
  
} 
}
  
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach 
  
# Function that returns true if it is possible to 
# make a binary string consisting of l 0's, m 1's, 
# x "01" sub-sequences and y "10" sub-sequences 
def isPossible(l, m, x, y) :
      
    if (l * m == x + y) :
        return True; 
  
    return False; 
  
# Driver code 
if __name__ == "__main__" : 
  
    l = 3; m = 2; x = 4; y = 2; 
  
    if (isPossible(l, m, x, y)) :
        print("Yes"); 
    else :
        print("No"); 
          
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach 
using System;
  
class sol
{
      
// Function that returns true if it is possible to 
// make a binary string consisting of l 0's, m 1's, 
// x "01" sub-sequences and y "10" sub-sequences 
static Boolean isPossible(int l, int m, int x, int y) 
{ 
    if (l * m == x + y) 
        return true; 
  
    return false; 
} 
  
// Driver code 
public static void Main(String []args)
{ 
    int l = 3, m = 2, x = 4, y = 2; 
  
    if (isPossible(l, m, x, y)) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
} 
}
  
// This code is contributed by Arnab Kundu

输出：

Yes