题干：
You are given a set Y of n distinct positive integers y1, y2, ..., yn.
Set X of n distinct positive integers x1, x2, ..., xn is said to generate set Y if one can transform X to Y by applying some number of the following two operation to integers in X:
- Take any integer xi and multiply it by two, i.e. replace xi with 2·xi.
- Take any integer xi, multiply it by two and add one, i.e. replace xi with 2·xi + 1.
Note that integers in X are not required to be distinct after each operation.
Two sets of distinct integers X and Y are equal if they are equal as sets. In other words, if we write elements of the sets in the array in the increasing order, these arrays would be equal.
Note, that any set of integers (or its permutation) generates itself.
You are given a set Y and have to find a set X that generates Y and the maximum element of X is mininum possible.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50 000) — the number of elements in Y.
The second line contains n integers y1, ..., yn (1 ≤ yi ≤ 109), that are guaranteed to be distinct.
Output
Print n integers — set of distinct integers that generate Y and the maximum element of which is minimum possible. If there are several such sets, print any of them.
Examples
Input
5 1 2 3 4 5
Output
4 5 2 3 1
Input
6 15 14 3 13 1 12
Output
12 13 14 7 3 1
Input
6 9 7 13 17 5 11
Output
4 5 2 6 3 1
题目大意：
有一个原数列a，你可以对任何一个数字进行操作，令x变为2*x 或2*x+1。现在给你一个目标数列b，让你求一个原数列，原数列可能有多个，要求你求的这个原数列最大的一个元素最小。注意原数列和目标数列都必须满足每个元素都不相同，但是在变化的过程中（也就是2*x,2*x+1这种变换过程中）不需要满足这个条件。
输入：第一行一个数n，第二行n个数代表b数组。
解题报告：
将数字之间的变换画出来发现是一棵二叉树，所以每个bi变小的过程，是唯一的，因为父节点是唯一的。发现从最小的数开始贪心或从大的数开始贪心都不是很合适，所以二分上界，然后从大到小贪心，让大的数字找地方放下，能放下则放下，给小的数字腾出尽可能大的空间，不难证明这种贪心的正确性。代码是勉勉强强的O(n*lognlogn)的复杂度。
当然这题可以不用二分，直接拿个优先队列也可，这样复杂度可以做到O(nlogn)。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<unordered_set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 50000 + 6;
int n,a[MAX],mx;
vector<int> ans,tmp;
unordered_set<int> ss;
bool ok(int up) {tmp.clear();ss.clear();//从大到小进行安排位置for(int x,i = 1; i<=n; i++) {x = a[i];while(x >= 1) {if(x<=up && ss.find(x) == ss.end()) {ss.insert(x);tmp.pb(x);break;}if(x&1) x=(x-1)/2;else x=x/2;}}return (int)tmp.size() == n;
}
int main()
{
// cout << log2(50000)*log2(50000)*log2(50000)*50000<<endl;cin >> n;for(int i = 1; i<=n; i++) scanf("%d",a+i),mx=max(mx,a[i]);sort(a+1,a+n+1); reverse(a+1,a+n+1);int l = 1,r = mx,mid;while(l<=r) {mid=(l+r)>>1;if(ok(mid)) r = mid-1,ans=tmp;else l = mid+1;}for(auto x : ans) printf("%d ",x);return 0 ;
}
