Problem Statement

Suppose that A is a list of n numbers {A1,A2,A3,…,An} and B={B1,B2,B3,..,Bn} is a permutation of these numbers, we say B is K-Manipulative if and only if:

M(B)=minimum(B1⊕B2,B2⊕B3,B3⊕B4,…,Bn−1⊕Bn,Bn⊕B1) is not less than 2K , where ⊕ represents the XOR operator.

You are given A . Find the largest K such that there exists a K-manipulative permutation B .

Input:

The first line is an integer N . The second line contains N space separated integers - A1 A2 … An .

Output:
The largest possible K , or −1 if there is no solution.

Constraints:

• 1<n<=100
• 0≤Ai≤109,where i∈[1,n]

Sample Input #00

3
13 3 10

Sample Output #00

2

解题思路：Bi^Bi+1 >= 2^K当且仅当(Bi>>K) != (Bi+1>>K),对于上述的排列如果M(B)=K,则Bi>>K相同的数最多不能超过的n/2，这个分为奇数和偶数两种情况，画画图就很明显了。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <functional>
using namespace std;
map<int, int> myMap;
int arr[110];int main() {int n;int ans = -1;scanf("%d", &n);for(int i = 1; i <= n; ++i) {scanf("%d", &arr[i]);}for(int k = 30; k >= 0; --k) {myMap.clear();for(int i = 1; i <= n; ++i) {myMap[arr[i]>>k]++;}bool ok = true;for(map<int,int>::iterator iter = myMap.begin(); iter != myMap.end(); ++iter) {if(iter->second > n/2) {ok = false;break;}}if(ok) {ans = k;break;}}printf("%d\n", ans);return 0;
}