Educational Codeforces Round 100 (Rated for Div. 2) B.Find The Array（构造算法）

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本文介绍了一道编程题，要求在给定数组中找到满足特定条件的美丽数组：元素间相除余数为0或被除数除，且所有元素之差绝对和不超过总和的一半。给出了一个C++代码示例，展示了如何通过调整元素来构造美丽数组。

题目链接：https://codeforc.es/contest/1463/problem/B

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.

Example

input

4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3

output

3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3

分析

对于第二个要求，只要构造出1 x 1 x 1 x … 或者 x 1 x 1 x 1 … 这样的必然满足条件。
对于第三个要求，只要偶数位置的和和奇数位置的和我们取小的将他们变成 1 ，这样必然满足条件。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int a[57];
ll sum[2];
 
void solve()
{
	sum[0] = sum[1] = 0;
	int n,x;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		sum[i % 2] += a[i];
	}
	if(sum[0] < sum[1])
	{
		for(int i=1;i<=n;i++)
			if(i % 2 == 1) printf("%d ",a[i]);
			else printf("1 ");
	}
	else
	{
		for(int i=1;i<=n;i++)
			if(i % 2 == 0) printf("%d ",a[i]);
			else printf("1 ");
	}
	printf("\n");
}
 
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		solve();
	}
	return 0;
}