Educational Codeforces Round 100 (Rated for Div. 2) B. Find The Array 题解

前排感谢 @Leonard.7 大佬给的思路qwq

没有大佬的帮助我这个菜批A不掉这道题qwq

原题

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.

简单的解释

给一个数组a，其数组各个数的和为S，让你求一个数组b符合以下条件：

• b[i] ∈ [1,1e9]
• 2*∑|a[i]-b[i]| <= S
• b[i]可以被b[i+1]整除或被整除

这里是一个基本上压着最大值的解法。

以数组1,3,2,5,4举例，首先把这个数组以单双数分开，即
1 ,2, 4 和 3,5
被分开的这两个数组的其中一个的和必然大于S/2。
在这个例子中也就是1,2,4。
这样看来如果我们想要求出数组b，即数组b中的数为1,x,2,x,4。
显而易见的是x=1。此题得证。

以下是写的非常不简便的代码

#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];  //求S
			if(j % 2==0) b[j] = 1;
			else b[j] = a[j];
			sum+=abs(a[j]-b[j]);  //计算∑|a[i]-b[i]|
		}
		if(2*sum < tot)   //两种情况进行判断，这里又又双写麻烦了
		 for(int j = 0;j < x;j++)cout << b[j] << " ";
		else{
      //第二种情况
			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;
}