Educational codeforces round 100 (rated for Div. 2) B. find the array solution

Front row thanks @Leonard.7 The idea given by the boss qwq

Without the help of the boss, I have this batch of vegetables A Don't drop this question qwq

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The original title is

You are given an array [a1,a2,…,an] such that 1≤ai≤109. Let S be the sum of all elements of the array a.

Let’s call an array b of n integers beautiful if:

1≤bi≤109 for each i from 1 to n;
for every pair of adjacent integers from the array (bi,bi+1), either bi divides bi+1, or bi+1 divides bi (or both);
2∑i=1n|ai−bi|≤S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.

Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.

Each test case consists of two lines. The first line contains one integer n (2≤n≤50).

The second line contains n integers a1,a2,…,an (1≤ai≤109).

Output
For each test case, print the beautiful array b1,b2,…,bn (1≤bi≤109) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.

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A simple explanation

Give an array a, The sum of each number of its array is S, Let you find an array b Meet the following conditions ：

• b[i] ∈ [1,1e9]
• 2*∑|a[i]-b[i]| <= S
• b[i] Can be b[i+1] Divide or be divided

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Here is a solution that basically presses the maximum .

In array 1,3,2,5,4 give an example , First, separate the array with odd and even numbers , namely
1 ,2, 4 and 3,5
The sum of one of the two separate arrays must be greater than S/2.
In this case, that is 1,2,4.
So if we want to ask for an array b, It's an array b The number in is 1,x,2,x,4.
The obvious thing is x=1. This question is proved .

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The following is a very inconvenient code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

const int N = 100;

using namespace std;
int main(){
    
	int n;
	cin >> n;
	for(int i = 0;i < n;i++){
    
		int x,a[N],b[N];
		long long sum = 0,tot = 0;
		cin >> x;
		for(int j = 0;j < x;j++){
    
			cin >> a[j];
			tot+= a[j];  // seek S
			if(j % 2==0) b[j] = 1;
			else b[j] = a[j];
			sum+=abs(a[j]-b[j]);  // Calculation ∑|a[i]-b[i]|
		}
		if(2*sum < tot)   // Judge from two situations , It's troublesome to write double here 
		 for(int j = 0;j < x;j++)cout << b[j] << " ";
		else{
      // The second case 
			for(int j = 0;j < x;j++){
    
			if(j%2!=0) b[j] = 1;
			else b[j] = a[j];
			}
			for(int j = 0;j < x;j++)cout << b[j] << " ";
		}
		cout << endl;
	}
	return 0;
}