# Generate an array B[] from the given array A[] which satisfies the given conditions

Given an array **A[]** of **N** integers such that **A[0] + A[1] + A[2] + … A[N – 1] = 0**. The task is to generate an array **B[]** such that **B[i]** is either **⌊A[i] / 2⌋** or **⌈A[i] / 2⌉** for all valid **i** and **B[0] + B[1] + B[2] + … + B[N – 1] = 0**.

**Examples:**

Input:A[] = {1, 2, -5, 3, -1}

Output:0 1 -2 1 0

Input:A[] = {3, -5, -7, 9, 2, -2}

Output:1 -2 -4 5 1 -1

**Approach:** For even integers, it is safe to assume that **B[i]** will be **A[i] / 2** but for odd integers, to maintain the sum equal to zero, take the ceil of exactly half of odd integers and floor of exactly other half odd integers. Since Odd – Odd = Even and Even – Even = Even and 0 is also Even, it can be said that **A[]** will always contain even number of odd integers so that the sum can be **0**. So for a valid input, there will always be an answer.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Utility function to print ` `// the array elements ` `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` `generateArr(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// To switch the ceil and floor ` ` ` `// function alternatively ` ` ` `bool` `flip = ` `true` `; ` ` ` ` ` `// For every element of the array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// If the number is odd then print the ceil ` ` ` `// or floor value after division by 2 ` ` ` `if` `(arr[i] & 1) { ` ` ` ` ` `// Use the ceil and floor alternatively ` ` ` `if` `(flip ^= ` `true` `) ` ` ` `cout << ` `ceil` `((` `float` `)(arr[i]) / 2.0) << ` `" "` `; ` ` ` `else` ` ` `cout << ` `floor` `((` `float` `)(arr[i]) / 2.0) << ` `" "` `; ` ` ` `} ` ` ` ` ` `// If arr[i] is even then it will ` ` ` `// be completely divisible by 2 ` ` ` `else` `{ ` ` ` `cout << arr[i] / 2 << ` `" "` `; ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 3, -5, -7, 9, 2, -2 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); ` ` ` ` ` `generateArr(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 implementation of the approach ` `from` `math ` `import` `ceil, floor ` ` ` `# Utility function to print ` `# the array elements ` `def` `printArr(arr, n): ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `print` `(arr[i], end ` `=` `" "` `) ` ` ` `# Function to generate and print ` `# the required array ` `def` `generateArr(arr, n): ` ` ` ` ` `# To switch the ceil and floor ` ` ` `# function alternatively ` ` ` `flip ` `=` `True` ` ` ` ` `# For every element of the array ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# If the number is odd then prthe ceil ` ` ` `# or floor value after division by 2 ` ` ` `if` `(arr[i] & ` `1` `): ` ` ` ` ` `# Use the ceil and floor alternatively ` ` ` `flip ^` `=` `True` ` ` `if` `(flip): ` ` ` `print` `(` `int` `(ceil((arr[i]) ` `/` `2` `)), ` ` ` `end ` `=` `" "` `) ` ` ` `else` `: ` ` ` `print` `(` `int` `(floor((arr[i]) ` `/` `2` `)), ` ` ` `end ` `=` `" "` `) ` ` ` ` ` `# If arr[i] is even then it will ` ` ` `# be completely divisible by 2 ` ` ` `else` `: ` ` ` `print` `(` `int` `(arr[i] ` `/` `2` `), end ` `=` `" "` `) ` ` ` `# Driver code ` `arr ` `=` `[` `3` `, ` `-` `5` `, ` `-` `7` `, ` `9` `, ` `2` `, ` `-` `2` `] ` `n ` `=` `len` `(arr) ` ` ` `generateArr(arr, n) ` ` ` `# This code is contributed by Mohit Kumar ` |

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**Output:**

1 -2 -4 5 1 -1

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