按照给定的约束构造一个二进制字符串

给定三个整数A ， B和X。任务是构造一个二进制字符串str ，该字符串的正好是A的0和B的1，前提是必须至少有X个索引，这样str [i]！= str [i + 1] 。输入的信息总是有一个有效的解决方案。
例子：

Input: A = 2, B = 2, X = 1 
Output: 1100 
There are two 0’s and two 1’s and one (=X) index such that s[i] != s[i+1] (i.e. i = 1)
Input: A = 4, B = 3, X = 2 
Output: 0111000 
 

方法：

• 将x除以2并将其存储在变量d中。
• 检查d是否为偶数，并且d / 2！= a ，如果条件为true，则打印0并将d和a减1 。
• 从1循环到d并打印10 ，最后更新a = a – d和b = b – d 。
• 最后根据a和b的值打印其余的0和1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
int constructBinString(int a, int b, int x)
{
    int d, i;
 
    // Divide index value by 2 and store
    // it into d
    d = x / 2;
 
    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a) {
        d--;
        cout << 0;
        a--;
    }
 
    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        cout << "10";
 
    // subtract d from a and b
    a = a - d;
    b = b - d;
 
    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++) {
        cout << "1";
    }
 
    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++) {
        cout << "0";
    }
}
 
// Driver code
int main()
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
static void constructBinString(int a, int b, int x)
{
    int d, i;
 
    // Divide index value by 2 and store
    // it into d
    d = x / 2;
 
    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a)
    {
        d--;
        System.out.print("0");
        a--;
    }
 
    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        System.out.print("10");
 
    // subtract d from a and b
    a = a - d;
    b = b - d;
 
    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++)
    {
        System.out.print("1");
    }
 
    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++)
    {
        System.out.print("0");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
}
}
 
// This code is contributed
// by Mukul Singh

Python3

# Python3 implementation of the above approach
 
# Function to print a binary string which
# has 'a' number of 0's, 'b' number of 1's
# and there are at least 'x' indices such
# that s[i] != s[i+1]
def constructBinString(a, b, x):
 
    # Divide index value by 2 and
    # store it into d
    d = x // 2
 
    # If index value x is even and
    # x/2 is not equal to a
    if x % 2 == 0 and x // 2 != a:
        d -= 1
        print("0", end = "")
        a -= 1
 
    # Loop for d for each d print 10
    for i in range(d):
        print("10", end = "")
 
    # subtract d from a and b
    a = a - d
    b = b - d
 
    # Loop for b to print remaining 1's
    for i in range(b):
        print("1", end = "")
     
    # Loop for a to print remaining 0's
    for i in range(a):
        print("0", end = "")
 
# Driver Code
if __name__ == "__main__":
 
    a, b, x = 4, 3, 2
    constructBinString(a, b, x)
 
# This code is contributed by Rituraj_Jain

C#

// C# implementation of the approach
using System;
 
class GFG
{
// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
static void constructBinString(int a, int b, int x)
{
    int d, i;
 
    // Divide index value by 2 and store
    // it into d
    d = x / 2;
 
    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a)
    {
        d--;
        Console.Write("0");
        a--;
    }
 
    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        Console.Write("10");
 
    // subtract d from a and b
    a = a - d;
    b = b - d;
 
    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++)
    {
        Console.Write("1");
    }
 
    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++)
    {
        Console.Write("0");
    }
}
 
// Driver code
public static void Main()
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
}
}
 
// This code is contributed
// by Akanksha Rai

Javascript

 
    // Javascript implementation of the approach
     
    // Function to print a binary string which has
    // 'a' number of 0's, 'b' number of 1's and there are
    // at least 'x' indices such that s[i] != s[i+1]
    function constructBinString(a, b, x)
    {
        let d, i;
 
        // Divide index value by 2 and store
        // it into d
        d = parseInt(x / 2, 10);
 
        // If index value x is even and
        // x/2 is not equal to a
        if (x % 2 == 0 && parseInt(x / 2, 10) != a)
        {
            d--;
            document.write("0");
            a--;
        }
 
        // Loop for d for each d print 10
        for (i = 0; i < d; i++)
            document.write("10");
 
        // subtract d from a and b
        a = a - d;
        b = b - d;
 
        // Loop for b to print remaining 1's
        for (i = 0; i < b; i++)
        {
            document.write("1");
        }
 
        // Loop for a to print remaining 0's
        for (i = 0; i < a; i++)
        {
            document.write("0");
        }
    }
     
    let a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
     

输出：

0111000

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