根据给定条件可能的二进制字符串计数

给定两个整数N和M ，其中N表示‘0’的计数， M表示‘1’的计数，以及一个整数K ，任务是找到以下两个可以生成的二进制字符串的最大数量类型：

• 一个字符串可以由K 个“ 0 ”和一个“ 1 ”组成。
• 一个字符串可以由K 个“ 1 ”和一个“ 0 ”组成。

例子：

Input: N = 4, M = 4, K = 2 
Output: 6 
Explanation: 
Count of ‘0‘s = 4 
Count of ‘1‘s = 4 
Possible ways to combine 0’s and 1’s under given constraints are {“001”, “001”} or {“001”, “110”} or {“110”, “110”} 
Therefore, at most 2 combinations exists in a selection. 
Each combination can be arranged in K + 1 ways, i.e. “001” can be rearranged to form “010, “100” as well. 
Therefore, the maximum possible strings that can be generated is 3 * 2 = 6

Input: N = 101, M = 231, K = 15 
Output: 320 

方法：
请按照以下步骤解决问题：

• 考虑以下三个条件来生成二进制字符串的最大可能组合：
  • 组合数不能超过N 。
  • 组合数不能超过M 。
  • 组合数不能超过(A+B)/(K + 1) 。
• 因此，最大可能的组合是min(A, B, (A + B)/ (K + 1)) 。
• 因此，可以生成的最大可能字符串为(K + 1) * min(A, B, (A + B)/ (K + 1)) 。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to generate maximum
// possible strings that can be generated
long long countStrings(long long A,
                    long long B,
                    long long K)
{
 
    long long X = (A + B) / (K + 1);
 
    // Maximum possible strings
    return (min(A, min(B, X)) * (K + 1));
}
int main()
{
 
    long long N = 101, M = 231, K = 15;
    cout << countStrings(N, M, K);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to generate maximum
// possible strings that can be generated
static long countStrings(long A, long B,
                        long K)
{
    long X = (A + B) / (K + 1);
 
    // Maximum possible strings
    return (Math.min(A, Math.min(B, X)) *
                                (K + 1));
}
 
// Driver Code
public static void main (String[] args)
{
    long N = 101, M = 231, K = 15;
     
    System.out.print(countStrings(N, M, K));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement 
# the above approach 
  
# Function to generate maximum 
# possible strings that can be
# generated 
def countStrings(A, B, K): 
   
    X = (A + B) // (K + 1) 
   
    # Maximum possible strings 
    return (min(A, min(B, X)) * (K + 1)) 
 
# Driver code
N, M, K = 101, 231, 15
 
print(countStrings(N, M, K))
 
# This code is contributed divyeshrabadiya07

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to generate maximum
// possible strings that can be generated
static long countStrings(long A, long B,
                        long K)
{
    long X = (A + B) / (K + 1);
 
    // Maximum possible strings
    return (Math.Min(A, Math.Min(B, X)) *
                                (K + 1));
}
 
// Driver Code
public static void Main (string[] args)
{
    long N = 101, M = 231, K = 15;
     
    Console.Write(countStrings(N, M, K));
}
}
 
// This code is contributed by rock_cool

Javascript

// JavaScript Program to implement
// the above approach
 
// Function to generate maximum
// possible strings that can be generated
function countStrings(A, B, K)
{
 
    let X = Math.floor((A + B) / (K + 1));
 
    // Maximum possible strings
    return (Math.min(A, Math.min(B, X)) * (K + 1));
}
 
 
    let N = 101, M = 231, K = 15;
    document.write(countStrings(N, M, K));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

320

时间复杂度： O(1)
辅助空间： O(1)

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