给定长度的计数序列具有可以通过给定值生成的非负前缀和

给定两个整数M和X中，任务是找到可产生含有X的长度M的序列的-X使得它们各自的计数相等的数量和与所述前缀和高达所得到的序列的每个索引是非负面的。

例子：

Input: M = 4, X = 5
Output: 2
Explanation:
There are only 2 possible sequences that have all possible prefix sums non-negative: 

1. {+5, +5, -5, -5}
2. {+5, -5, +5, -5}

Input: M = 6, X = 2
Output: 5
Explanation:
There are only 5 possible sequences that have all possible prefix sums non-negative:

1. {+2, +2, +2, -2, -2, -2}
2. {+2, +2, -2, -2, +2, -2}
3. {+2, -2, +2, -2, +2, -2}
4. {+2, +2, -2, +2, -2, -2}
5. {+2, -2, +2, +2, -2, -2}

天真的方法：最简单的方法是使用给定的整数+ X和-X生成大小为M的所有可能排列，并找到形成的每种排列的前缀和，并对那些其前缀和数组仅包含非负元素的序列进行计数。在上述步骤之后，打印此序列的计数。

时间复杂度： O((M *(M！))/((M / 2)！) ² )
辅助空间： O(M)

高效的方法：想法是观察形成任何序列的模式 每个索引处出现的正X数始终大于或等于出现的负X数。这类似于加泰罗尼亚语数字的模式。在这种情况下，请检查在任何时候出现的正X的数量始终大于或等于出现的负X的数量，这是加泰罗尼亚语数字的形式。因此，任务是找到第N个加泰罗尼亚数字，其中N = M / 2 。

$K_{N} = \frac{\binom{2N}{N}}{N + 1}$

where, K_N is N^th the Catalan Number and 

$\binom{2N}{N}$

is the binomial coefficient. 
 

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Function to find the Binomial
// Coefficient C(n, r)
unsigned long int binCoff(unsigned int n,
                          unsigned int r)
{
    // Stores the value C(n, r)
    unsigned long int val = 1;
    int i;
  
    // Update C(n, r) = C(n, n - r)
    if (r > (n - r))
        r = (n - r);
  
    // Find C(n, r) iteratively
    for (i = 0; i < r; i++) {
        val *= (n - i);
        val /= (i + 1);
    }
  
    // Return the final value
    return val;
}
  
// Function to find number of sequence
// whose prefix sum at each index is
// always non-negative
void findWays(int M)
{
    // Find n
    int n = M / 2;
  
    unsigned long int a, b, ans;
  
    // Value of C(2n, n)
    a = binCoff(2 * n, n);
  
    // Catalan number
    b = a / (n + 1);
  
    // Print the answer
    cout << b;
}
  
// Driver Code
int main()
{
    // Given M and X
    int M = 4, X = 5;
  
    // Function Call
    findWays(M);
  
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
  
class GFG{
  
// Function to find the Binomial
// Coefficient C(n, r)
static long binCoff(long n, long r)
{
      
    // Stores the value C(n, r)
    long val = 1;
    int i;
  
    // Update C(n, r) = C(n, n - r)
    if (r > (n - r))
        r = (n - r);
  
    // Find C(n, r) iteratively
    for(i = 0; i < r; i++)
    {
        val *= (n - i);
        val /= (i + 1);
    }
  
    // Return the final value
    return val;
}
  
// Function to find number of sequence
// whose prefix sum at each index is
// always non-negative
static void findWays(int M)
{
      
    // Find n
    int n = M / 2;
  
    long a, b, ans;
  
    // Value of C(2n, n)
    a = binCoff(2 * n, n);
  
    // Catalan number
    b = a / (n + 1);
  
    // Print the answer
    System.out.print(b);
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Given M and X
    int M = 4, X = 5;
  
    // Function Call
    findWays(M);
}
}
  
// This code is contributed by akhilsaini

Python3

# Python3 program for the above approach
  
# Function to find the Binomial
# Coefficient C(n, r)
def binCoff(n, r):
  
    # Stores the value C(n, r)
    val = 1
  
    # Update C(n, r) = C(n, n - r)
    if (r > (n - r)):
        r = (n - r)
  
    # Find C(n, r) iteratively
    for i in range(0, r):
        val *= (n - i)
        val //= (i + 1)
  
    # Return the final value
    return val
  
# Function to find number of sequence
# whose prefix sum at each index is
# always non-negative
def findWays(M):
  
    # Find n
    n = M // 2
  
    # Value of C(2n, n)
    a = binCoff(2 * n, n)
  
    # Catalan number
    b = a // (n + 1)
  
    # Print the answer
    print(b)
  
# Driver Code
if __name__ == '__main__':
  
    # Given M and X
    M = 4
    X = 5
  
    # Function Call
    findWays(M)
  
# This code is contributed by akhilsaini

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the Binomial
// Coefficient C(n, r)
static long binCoff(long n, long r)
{
      
    // Stores the value C(n, r)
    long val = 1;
    int i;
  
    // Update C(n, r) = C(n, n - r)
    if (r > (n - r))
        r = (n - r);
  
    // Find C(n, r) iteratively
    for(i = 0; i < r; i++)
    {
        val *= (n - i);
        val /= (i + 1);
    }
  
    // Return the final value
    return val;
}
  
// Function to find number of sequence
// whose prefix sum at each index is
// always non-negative
static void findWays(int M, int X)
{
      
    // Find n
    int n = M / 2;
  
    long a, b;
  
    // Value of C(2n, n)
    a = binCoff(2 * n, n);
  
    // Catalan number
    b = a / (n + 1);
  
    // Print the answer
    Console.WriteLine(b);
}
  
// Driver Code
public static void Main()
{
      
    // Given M and X
    int M = 4;
    int X = 5;
  
    // Function Call
    findWays(M, X);
}
}
  
// This code is contributed by akhilsaini

输出：

2

时间复杂度： O(M)
辅助空间： O(1)