找到满足给定条件的大小为 N 的数组

给定三个整数N 、 S和K ，任务是创建一个由N 个正整数组成的数组，使得数组中任意两个连续元素的按位 OR 为奇数，并且恰好有K个子数组的总和等于S ，其中1 ≤ K ≤ N / 2 。

例子：

Input: N = 4, K = 2, S = 6 
Output: 6 7 6 7 
Here, there are exactly 2 subarray {6} and {6} 
whose sum is 6 and the bitwise OR of 
any adjacent elements is odd.
Input: N = 8, K = 3, S = 12 
Output: 12 13 12 13 12 13 13 13 
 

方法：

• 观察这里的模式{S, P, S, P, S, P, …, P, P, P, P} 。
• 这里P是一个奇数> S并且在每个S之后都有一个P 。已知与奇数的按位或总是奇数，因此确认每个相邻元素的按位或是奇数。
• 现在，在数组的上述模式中准确放置K个S。
• 除了S之外，所有元素（即 P）都大于 S，因此除了 K 个子数组之外，不可能有任何子数组的总和恰好为S。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Utility function to print the
// contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Function to generate and print
// the required array
void findArray(int n, int k, int s)
{
 
    // Initially all the positions are empty
    int vis[n] = { 0 };
 
    // To store the count of positions
    // i such that arr[i] = s
    int cnt = 0;
 
    // To store the final array elements
    int arr[n];
 
    for (int i = 0; i < n && cnt < k; i += 2) {
 
        // Set arr[i] = s and the gap between
        // them is exactly 2 so in for loop
        // we use i += 2
        arr[i] = s;
 
        // Mark the i'th position as visited
        // as we put arr[i] = s
        vis[i] = 1;
 
        // Increment the count
        cnt++;
    }
    int val = s;
 
    // Finding the next odd number after s
    if (s % 2 == 0)
        val++;
    else
        val = val + 2;
 
    for (int i = 0; i < n; i++) {
        if (vis[i] == 0) {
 
            // If the i'th position is not visited
            // it means we did not put any value
            // at position i so we put 1 now
            arr[i] = val;
        }
    }
 
    // Print the final array
    printArr(arr, n);
}
 
// Driver code
int main()
{
    int n = 8, k = 3, s = 12;
 
    findArray(n, k, s);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Utility function to print the
    // contents of an array
    static void printArr(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
     
    // Function to generate and print
    // the required array
    static void findArray(int n, int k, int s)
    {
     
        // Initially all the positions are empty
        int vis[] = new int[n] ;
     
        // To store the count of positions
        // i such that arr[i] = s
        int cnt = 0;
     
        // To store the final array elements
        int arr[] = new int[n];
     
        for (int i = 0; i < n && cnt < k; i += 2)
        {
     
            // Set arr[i] = s and the gap between
            // them is exactly 2 so in for loop
            // we use i += 2
            arr[i] = s;
     
            // Mark the i'th position as visited
            // as we put arr[i] = s
            vis[i] = 1;
     
            // Increment the count
            cnt++;
        }
        int val = s;
     
        // Finding the next odd number after s
        if (s % 2 == 0)
            val++;
        else
            val = val + 2;
     
        for (int i = 0; i < n; i++)
        {
            if (vis[i] == 0)
            {
     
                // If the i'th position is not visited
                // it means we did not put any value
                // at position i so we put 1 now
                arr[i] = val;
            }
        }
     
        // Print the final array
        printArr(arr, n);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 8, k = 3, s = 12;
     
        findArray(n, k, s);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Utility function to print the
# contents of an array
def printArr(arr, n) :
 
    for i in range(n) :
        print(arr[i], end= " ");
 
# Function to generate and print
# the required array
def findArray(n, k, s) :
 
    # Initially all the positions are empty
    vis = [0] * n;
 
    # To store the count of positions
    # i such that arr[i] = s
    cnt = 0;
 
    # To store the final array elements
    arr = [0] * n;
    i = 0;
     
    while (i < n and cnt < k) :
         
        # Set arr[i] = s and the gap between
        # them is exactly 2 so in for loop
        # we use i += 2
        arr[i] = s;
 
        # Mark the i'th position as visited
        # as we put arr[i] = s
        vis[i] = 1;
 
        # Increment the count
        cnt += 1;
        i += 2;
    val = s;
     
    # Finding the next odd number after s
    if (s % 2 == 0) :
        val += 1;
    else :
        val = val + 2;
 
    for i in range(n) :
        if (vis[i] == 0) :
 
            # If the i'th position is not visited
            # it means we did not put any value
            # at position i so we put 1 now
            arr[i] = val;
 
    # Print the final array
    printArr(arr, n);
 
# Driver code
if __name__ == "__main__" :
 
    n = 8; k = 3; s = 12;
 
    findArray(n, k, s);
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Utility function to print the
    // contents of an array
    static void printArr(int []arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
     
    // Function to generate and print
    // the required array
    static void findArray(int n, int k, int s)
    {
     
        // Initially all the positions are empty
        int []vis = new int[n] ;
     
        // To store the count of positions
        // i such that arr[i] = s
        int cnt = 0;
     
        // To store the final array elements
        int []arr = new int[n];
     
        for (int i = 0; i < n && cnt < k; i += 2)
        {
     
            // Set arr[i] = s and the gap between
            // them is exactly 2 so in for loop
            // we use i += 2
            arr[i] = s;
     
            // Mark the i'th position as visited
            // as we put arr[i] = s
            vis[i] = 1;
     
            // Increment the count
            cnt++;
        }
        int val = s;
     
        // Finding the next odd number after s
        if (s % 2 == 0)
            val++;
        else
            val = val + 2;
     
        for (int i = 0; i < n; i++)
        {
            if (vis[i] == 0)
            {
     
                // If the i'th position is not visited
                // it means we did not put any value
                // at position i so we put 1 now
                arr[i] = val;
            }
        }
     
        // Print the final array
        printArr(arr, n);
    }
     
    // Driver code
    public static void Main()
    {
        int n = 8, k = 3, s = 12;
     
        findArray(n, k, s);
    }
}
 
// This code is contributed by AnkitRai01

Javascript

// javascript implementation of the approach    
// Utility function to prvar the
    // contents of an array
    function printArr(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }
 
    // Function to generate and print
    // the required array
    function findArray(n , k , s) {
 
        // Initially all the positions are empty
        var vis = Array(n).fill(0);
 
        // To store the count of positions
        // i such that arr[i] = s
        var cnt = 0;
 
        // To store the final array elements
        var arr = Array(n).fill(0);
 
        for (i = 0; i < n && cnt < k; i += 2) {
 
            // Set arr[i] = s and the gap between
            // them is exactly 2 so in for loop
            // we use i += 2
            arr[i] = s;
 
            // Mark the i'th position as visited
            // as we put arr[i] = s
            vis[i] = 1;
 
            // Increment the count
            cnt++;
        }
        var val = s;
 
        // Finding the next odd number after s
        if (s % 2 == 0)
            val++;
        else
            val = val + 2;
 
        for (i = 0; i < n; i++) {
            if (vis[i] == 0) {
 
                // If the i'th position is not visited
                // it means we did not put any value
                // at position i so we put 1 now
                arr[i] = val;
            }
        }
 
        // Prvar the final array
        printArr(arr, n);
    }
 
    // Driver code
     
        var n = 8, k = 3, s = 12;
 
        findArray(n, k, s);
 
// This code contributed by umadevi9616

输出：

12 13 12 13 12 13 13 13

时间复杂度： O(N)

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