Codeforces Round #618 (Div. 2)C、Anu Has a Function_codeforces 618div2 c-优快云博客

本文探讨了一种特殊函数f(x,y)的性质，并提出了一种排序策略，以使由数组元素通过此函数计算得到的最终值达到最大。通过分析二进制位，文章详细解释了如何确定最优排序，确保具有唯一高位置位的元素优先排列。

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C. Anu Has a Function
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Anu has created her own function f: f(x,y)=(x|y)−y where | denotes the bitwise OR operation. For example, f(11,6)=(11|6)−6=15−6=9. It can be proved that for any nonnegative numbers x and y value of f(x,y) is also nonnegative.

She would like to research more about this function and has created multiple problems for herself. But she isn’t able to solve all of them and needs your help. Here is one of these problems.

A value of an array [a1,a2,…,an] is defined as f(f(…f(f(a1,a2),a3),…an−1),an) (see notes). You are given an array with not necessarily distinct elements. How should you reorder its elements so that the value of the array is maximal possible?

Input
The first line contains a single integer n (1≤n≤105).

The second line contains n integers a1,a2,…,an (0≤ai≤109). Elements of the array are not guaranteed to be different.

Output
Output n integers, the reordering of the array with maximum value. If there are multiple answers, print any.

Examples
inputCopy
4
4 0 11 6
outputCopy
11 6 4 0
inputCopy
1
13
outputCopy
13
Note
In the first testcase, value of the array [11,6,4,0] is f(f(f(11,6),4),0)=f(f(9,4),0)=f(9,0)=9.

[11,4,0,6] is also a valid answer.

题意：给了n个数让给他排序后按照题中的公式最大

思路：很明显当前这个数只对当前有影响，对于每个数二进制来看都是32位，从最高位开始去看每一位中，该二进制位置位1的个数是不是只有1个只有一个的话，就对答案有贡献，放到前面即可，其他的随便排序都行
举个例子 32 32 3 排序后应该是3 32 32 答案为3

#include<bits/stdc++.h>
using namespace std;
int a[100005];
int c[100005][32];
int v[100005];
int main()
{
    int n;cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        for(int j=0;j<=31;j++)
        {
            c[i][j]=(a[i]>>j)&1;
        }
    }
    int k=0;
    vector<int> q;
    for(int i=31;i>=0;i--)
    {
        int sum=0,w=0;
        for(int j=1;j<=n;j++){

            if(c[j][i]&1)
            {
                sum++;
                w=j;
            }
        }
        if(sum==1&&!v[w]) q.push_back(a[w]),v[w]=1;
    }
    for(int i=0;i<q.size();i++) cout<<q[i]<<" ";
    for(int i=1;i<=n;i++){
        if(!v[i]) cout<<a[i]<<" ";
    }

   return 0;
}