生成一个数组，其中偶数和奇数和子数组的计数分别为E和O

给定三个整数N ， E和O。任务是找到一个大小为N的数组，以使sum的奇数和奇数的子数组数分别为E和O。
例子：

Input: N = 3, E = 2, O = 4 
Output: 0 1 0 
There are total 6 sub-arrays: {0}, {1}, {0}, {0, 1}, {1, 0}, {0, 1, 0}. 
Their sums are {0, 1, 0, 1, 1, 1} respectively. 
2 of them are even and 4 of them are odd.
Input: N = 3, E = 0, O = 6 
Output: -1 
 

天真的方法：使用位掩码来生成数组中0和1的所有组合。对于每个组合，我们计算偶数和奇数子数组的数量。如果它们等于给定值，那么它是正确的组合，我们打印
数组。
对于这种方法，将生成所有需要的集合$O(2^N)$对于每种组合，我们找到计算成本的子阵列数量$O(2^N * N^2)$ 。
高效的方法：众所周知，数组的PrefixSums。因此，我们将计算偶数PrefixSum和奇数PrefixSum的数量。如果我们以某种方式知道分别具有奇数和偶数奇偶校验的prefixSums的数量，我们可以相应地创建任何有效的数组，前提是oddPrefixSums和evenPrefixSums的总数为N + 1。
示例：如果我们有3个evenPrefixSums和2个奇数PrefixSums，我们可以创建一个数组[0，0，1，0]。诀窍是在放置(evenPrefixSums – 1)零之后放置唯一的1。所有剩余的prefixSums显然都是奇校验。
以下等式成立。

evenPrefixSums + oddPrefixSums = N + 1 
 

由于prefixSum_i – prefixSum_j有助于连续子数组的总和，因此两者都应具有不同的奇偶校验。因此，具有奇数奇偶性的连续子数组的数量将为C(evenPrefixSums，1)* C(oddPrefixSums，1)。这引起了另一个方程。

evenPrefixSums * oddPrefixSums = O 
 

我们可以形成一个二次方程式，并将其求解以获得相应的值。如果找不到任何有效值，则输出-1。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#include 
using namespace std;
 
// Function to generate and print the required array
void CreateArray(int N, int even, int odd)
{
    int temp = -1;
    int OddPreSums;
 
    // Find the number of odd prefix sums
    for (int i = 0; i <= N + 1; i++) {
        if (i * ((N + 1) - i) == odd) {
            temp = 0;
            OddPreSums = i;
            break;
        }
    }
 
    // If no odd prefix sum found
    if (temp == -1) {
        cout << temp << endl;
    }
    else {
 
        // Calculating the number of even prefix sums
        int EvenPreSums = (N + 1) - OddPreSums;
        int e = 1;
        int o = 0;
 
        // Stores the current prefix sum
        int CurrSum = 0;
        for (int i = 0; i < N; i++) {
 
            // If current prefix sum is even
            if (CurrSum % 2 == 0) {
 
                // Print 0 until e = EvenPreSums - 1
                if (e < EvenPreSums) {
                    e++;
                    cout << "0 ";
                }
                else {
                    o++;
 
                    // Print 1 when e = EvenPreSums
                    cout << "1 ";
                    CurrSum++;
                }
            }
            else {
                if (e < EvenPreSums) {
                    e++;
                    cout << "1 ";
                    CurrSum++;
                }
                else {
                    o++;
 
                    // Print 0 for rest of the values
                    cout << "0 ";
                }
            }
        }
        cout << endl;
    }
}
 
// Driver code
int main()
{
    int N = 15;
    int even = 60, odd = 60;
    CreateArray(N, even, odd);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG {
 
    // Function to generate and print the required array
    static void CreateArray(int N, int even, int odd)
    {
        int EvenPreSums = 1;
        int temp = -1;
        int OddPreSums = 0;
 
        // Find the number of odd prefix sums
        for (int i = 0; i <= N + 1; i++) {
            if (i * ((N + 1) - i) == odd) {
                temp = 0;
                OddPreSums = i;
                break;
            }
        }
 
        // If no odd prefix sum found
        if (temp == -1) {
            System.out.println(temp);
        }
        else {
 
            // Calculating the number of even prefix sums
 
            EvenPreSums = ((N + 1) - OddPreSums);
            int e = 1;
            int o = 0;
 
            // Stores the current prefix sum
            int CurrSum = 0;
            for (int i = 0; i < N; i++) {
 
                // If current prefix sum is even
                if (CurrSum % 2 == 0) {
 
                    // Print 0 until e = EvenPreSums - 1
                    if (e < EvenPreSums) {
                        e++;
                        System.out.print("0 ");
                    }
                    else {
                        o++;
 
                        // Print 1 when e = EvenPreSums
                        System.out.print("1 ");
                        CurrSum++;
                    }
                }
                else {
                    if (e < EvenPreSums) {
                        e++;
                        System.out.print("1 ");
                        CurrSum++;
                    }
                    else {
                        o++;
 
                        // Print 0 for rest of the values
                        System.out.print("0 ");
                    }
                }
            }
            System.out.println();
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int N = 15;
        int even = 60, odd = 60;
        CreateArray(N, even, odd);
    }
}
 
// This code is contributed by akt_mit

Python3

# Python 3 implementation of the approach
 
# Function to generate and print
# the required array
def CreateArray(N, even, odd):
    temp = -1
     
    # Find the number of odd prefix sums
    for i in range(N + 2):
        if (i * ((N + 1) - i) == odd):
            temp = 0
            OddPreSums = i
            break
 
    # If no odd prefix sum found
    if (temp == -1):
        print(temp)
    else:
         
        # Calculating the number
        # of even prefix sums
        EvenPreSums = (N + 1) - OddPreSums
        e = 1
        o = 0
 
        # Stores the current prefix sum
        CurrSum = 0
        for i in range(N):
             
            # If current prefix sum is even
            if (CurrSum % 2 == 0):
                 
                # Print 0 until e = EvenPreSums - 1
                if (e < EvenPreSums):
                    e += 1
                    print("0 ", end = "")
                else:
                    o += 1
 
                    # Print 1 when e = EvenPreSums
                    print("1 ", end = "")
                    CurrSum += 1
     
            else:
                if (e < EvenPreSums):
                    e += 1
                    print("1 ")
                    CurrSum += 1
                else:
                    o += 1
 
                    # Print 0 for rest of the values
                    print("0 ", end = "")
        print("\n", end = "")
 
# Driver code
if __name__ == '__main__':
    N = 15
    even = 60
    odd = 60
    CreateArray(N, even, odd)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to generate and print the required array
    static void CreateArray(int N, int even, int odd)
    {
        int EvenPreSums = 1;
        int temp = -1;
        int OddPreSums = 0;
 
        // Find the number of odd prefix sums
        for (int i = 0; i <= N + 1; i++) {
            if (i * ((N + 1) - i) == odd) {
                temp = 0;
                OddPreSums = i;
                break;
            }
        }
 
        // If no odd prefix sum found
        if (temp == -1) {
            Console.WriteLine(temp);
        }
        else {
 
            // Calculating the number of even prefix sums
 
            EvenPreSums = ((N + 1) - OddPreSums);
            int e = 1;
            int o = 0;
 
            // Stores the current prefix sum
            int CurrSum = 0;
            for (int i = 0; i < N; i++) {
 
                // If current prefix sum is even
                if (CurrSum % 2 == 0) {
 
                    // Print 0 until e = EvenPreSums - 1
                    if (e < EvenPreSums) {
                        e++;
                        Console.Write("0 ");
                    }
                    else {
                        o++;
 
                        // Print 1 when e = EvenPreSums
                        Console.Write("1 ");
                        CurrSum++;
                    }
                }
                else {
                    if (e < EvenPreSums) {
                        e++;
                        Console.Write("1 ");
                        CurrSum++;
                    }
                    else {
                        o++;
 
                        // Print 0 for rest of the values
                        Console.Write("0 ");
                    }
                }
            }
            Console.WriteLine();
        }
    }
 
    // Driver code
    static public void Main()
    {
        int N = 15;
        int even = 60, odd = 60;
        CreateArray(N, even, odd);
    }
}
 
// This code is contributed by Tushil

Javascript

    // Javascript implementation of the approach
     
    // Function to generate and print the required array
    function CreateArray(N, even, odd)
    {
        let EvenPreSums = 1;
        let temp = -1;
        let OddPreSums = 0;
   
        // Find the number of odd prefix sums
        for (let i = 0; i <= N + 1; i++) {
            if (i * ((N + 1) - i) == odd) {
                temp = 0;
                OddPreSums = i;
                break;
            }
        }
   
        // If no odd prefix sum found
        if (temp == -1) {
            document.write(temp);
        }
        else {
   
            // Calculating the number of even prefix sums
   
            EvenPreSums = ((N + 1) - OddPreSums);
            let e = 1;
            let o = 0;
   
            // Stores the current prefix sum
            let CurrSum = 0;
            for (let i = 0; i < N; i++) {
   
                // If current prefix sum is even
                if (CurrSum % 2 == 0) {
   
                    // Print 0 until e = EvenPreSums - 1
                    if (e < EvenPreSums) {
                        e++;
                        document.write("0 ");
                    }
                    else {
                        o++;
   
                        // Print 1 when e = EvenPreSums
                        document.write("1 ");
                        CurrSum++;
                    }
                }
                else {
                    if (e < EvenPreSums) {
                        e++;
                        document.write("1 ");
                        CurrSum++;
                    }
                    else {
                        o++;
   
                        // Print 0 for rest of the values
                        document.write("0 ");
                    }
                }
            }
            document.write();
        }
    }
     
    let N = 15;
    let even = 60, odd = 60;
    CreateArray(N, even, odd);
     

输出：

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