提示：
Codeforces Round #618 (Div. 1)-----A. Anu Has a Function

题意：

定义一个函数f(x , y) = f(x | y) - y
给定一个长度为n的序列a，给序列重新排序，使得f(f…f(f(a1,a2),a3),…an−1),an)的值最大
n(1≤n≤10^5)， ai(0≤ai≤10^9)

解题思路：

上图是计算f(11,6)的过程，可以发现运算结果对于 “11” 的二进制表示就是如果该位在"6"的二进制表示中对应的值是1那么这位就是0，否则不变，然后我们就可以得出f(x,y) = (x | y) - y = x & (~y)

由题中所要求的f(f…f(f(a1,a2),a3),…an−1),an)化简为a1 & (~a2) & (~a3) & (~a4) & … & (~an)，要使得这个式子最大，只需要找到一个a1和剩余的式子与最大就ok，假设a1为数组中的第k个，则

最后记录令函数值最大的k，最后a1与ak交换输出序列就ok，时间复杂度为O(n)

解题代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int N = 1e5 + 10;
int n, k, mx = -0x3f3f3f3f, tp;
int sum1[N], sum2[N], a[N];int main()
{scanf("%d",&n);for(int i = 1 ; i <= n ; ++i){scanf("%d",&a[i]);sum1[i] = sum1[i-1] | a[i];}for(int i = n ; i >= 1 ; --i)sum2[i] = sum2[i+1] | a[i];for(int i = 1 ; i <= n ; ++i)if((tp = a[i] & ~(sum1[i-1] | sum2[i+1])) > mx)mx = tp, k = i;printf("%d",a[k]);for(int i = 1 ; i <= n ; ++i)if(i != k)printf(" %d",a[i]);printf("\n");return 0;
}

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Codeforces Round #618 (Div. 1)-----A. Anu Has a Function

题意：

解题思路：

解题代码：

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