題意

給出一個長度為 $n$的序列，初始序列為 $1,2,3,$

亂搞題解

很明顯最後的GCD序列長度是 $n$

正規題解

可以發現一開始需要刪除掉全是奇數的數字 $cnt$個，輸出 $cnt$個1，之後剩下的都是 $2,4,6,8,\dots$的序列，這相當於 $1,2,3,4,\dots$乘一個2，所以問題還是和前面一樣，刪除下標為奇數的數字 $cnt$個，然後輸出最前面的那個數字 $cnt$個。
然後不斷將問題縮小，當 $n \leq 3$時要特殊處理一下。

亂搞程式碼

#include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b) {
	return b == 0 ? a: gcd(b,a%b);
}
int main() {
	int n;
	scanf("%d", &n);
	int cnt = 0;
	if(n <= 3) {
		if(n == 3)
			cout <<"1 1 3" << endl;
		if(n == 2)
			cout << "1 2" << endl;
		if(n == 1)
			cout << 1 << endl;
		return 0;
	}
	int last;
	int sum = 0;
	int p = 1;
	sum++;
	cout << p << " ";
	last = p;
	int f = 2;
	int t = n;
	while((n-p)/f > 0) {
		int cnt = (n-p)/f;
		if(n&1) n--;
		for(int i = 0; i < cnt; ++i)
			cout << p <<" ";
		sum += cnt;
		last = p;
		p *= 2; f *= 2;
		if(sum < t-1) {
			sum++;
			cout << p <<" ";
			last = p;
		}
	}
	cout << n/last*last << endl;
	return 0;
}

正規程式碼

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e6 + 6;

int seq[maxn];
int ans[maxn];
int ptr = 0;

void solve(int n, int mul){
	if(n == 1){ans[ptr++] = mul; return;}
	if(n == 2){ans[ptr++] = mul; ans[ptr++] = mul * 2; return;}
	if(n == 3){ans[ptr++] = mul; ans[ptr++] = mul; ans[ptr++] = mul * 3; return;}
	for(int i = 0; i < n; i++)if(seq[i]&1)ans[ptr++] = mul;
	for(int i = 0; i < n/2; i++)seq[i] = seq[2*i + 1]/2;
	solve(n/2, mul * 2);
}

int main(){
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++)seq[i] = i + 1;
	solve(n, 1);
	for(int i = 0; i < n; i++)printf("%d ", ans[i]);
	return 0;
}