给定一个正整数N ，对于大小为N且具有[0，N – 1]范围内的元素的任何置换，任务是计算所有元素在其各自位置的按位XOR之和。
For Example: For the permutation {3, 4, 2, 1, 0}, sum = (0^3 + 1^4 + 2^2 + 3^1 + 4^0) = 2.
例子：
Input: N = 3
Output: 6
Explanation: For the permutations {1, 2, 0} and {2, 0, 1}, the sum is 6.
Input: N = 2
Output: 2
方法：解决此问题的想法是递归生成整数[0，N – 1]的所有可能排列，并计算每个整数的分数，然后在所形成的所有排列中找到最大分数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate the score
int calcScr(vectorarr){
// Stores the possible score for
// the current permutation
int ans = 0;
// Traverse the permutation array
for(int i = 0; i < arr.size(); i++)
ans += (i ^ arr[i]);
// Return the final score
return ans;
}
// Function to generate all the possible
// permutation and get the max score
int getMax(vector arr, int ans, vector chosen, int N)
{
// If arr[] length is equal to N
// process the permutation
if (arr.size() == N){
ans = max(ans, calcScr(arr));
return ans;
}
// Generating the permutations
for (int i = 0; i < N; i++)
{
// If the current element is
// chosen
if(chosen[i])
continue;
// Mark the current element
// as true
chosen[i] = true;
arr.push_back(i);
// Recursively call for next
// possible permutation
ans = getMax(arr, ans, chosen, N);
// Backtracking
chosen[i] = false;
arr.pop_back();
}
// Return the ans
return ans;
}
// Driver Code
int main()
{
int N = 2;
// Stores the permutation
vector arr;
// To display the result
int ans = -1;
vectorchosen(N,false);
ans = getMax(arr, ans, chosen, N);
cout << ans << endl;
}
// This code is contributed by bgangwar59. Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to calculate the score
static int calcScr(ArrayListarr)
{
// Stores the possible score for
// the current permutation
int ans = 0;
// Traverse the permutation array
for(int i = 0; i < arr.size(); i++)
ans += (i ^ arr.get(i));
// Return the final score
return ans;
}
// Function to generate all the possible
// permutation and get the max score
static int getMax(ArrayList arr, int ans, ArrayList chosen, int N)
{
// If arr[] length is equal to N
// process the permutation
if (arr.size() == N)
{
ans = Math.max(ans, calcScr(arr));
return ans;
}
// Generating the permutations
for (int i = 0; i < N; i++)
{
// If the current element is
// chosen
if(chosen.get(i))
continue;
// Mark the current element
// as true
chosen.set(i, true);
arr.add(i);
// Recursively call for next
// possible permutation
ans = getMax(arr, ans, chosen, N);
// Backtracking
chosen.set(i, false);
arr.remove(arr.size()-1);
}
// Return the ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
int N = 2;
// Stores the permutation
ArrayList arr = new ArrayList();
// To display the result
int ans = -1;
ArrayList chosen = new ArrayList(Collections.nCopies(N, false));
ans = getMax(arr, ans, chosen, N);
System.out.print(ans +"\n");
}
}
// This code is contributed by 29AjayKumar Python3
# Python program for the above approach
# Function to generate all the possible
# permutation and get the max score
def getMax(arr, ans, chosen, N):
# If arr[] length is equal to N
# process the permutation
if len(arr) == N:
ans = max(ans, calcScr(arr))
return ans
# Generating the permutations
for i in range(N):
# If the current element is
# chosen
if chosen[i]:
continue
# Mark the current element
# as true
chosen[i] = True
arr.append(i)
# Recursively call for next
# possible permutation
ans = getMax(arr, ans, chosen, N)
# Backtracking
chosen[i] = False
arr.pop()
# Return the ans
return ans
# Function to calculate the score
def calcScr(arr):
# Stores the possible score for
# the current permutation
ans = 0
# Traverse the permutation array
for i in range(len(arr)):
ans += (i ^ arr[i])
# Return the final score
return ans
# Driver Code
N = 2
# Stores the permutation
arr = []
# To display the result
ans = -1
chosen = [False for i in range(N)]
ans = getMax(arr, ans, chosen, N)
print(ans)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to calculate the score
static int calcScr(Listarr)
{
// Stores the possible score for
// the current permutation
int ans = 0;
// Traverse the permutation array
for(int i = 0; i < arr.Count; i++)
ans += (i ^ arr[i]);
// Return the readonly score
return ans;
}
// Function to generate all the possible
// permutation and get the max score
static int getMax(List arr, int ans, List chosen, int N)
{
// If []arr length is equal to N
// process the permutation
if (arr.Count == N)
{
ans = Math.Max(ans, calcScr(arr));
return ans;
}
// Generating the permutations
for (int i = 0; i < N; i++)
{
// If the current element is
// chosen
if(chosen[i])
continue;
// Mark the current element
// as true
chosen[i] = true;
arr.Add(i);
// Recursively call for next
// possible permutation
ans = getMax(arr, ans, chosen, N);
// Backtracking
chosen[i] = false;
arr.Remove(arr.Count-1);
}
// Return the ans
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int N = 2;
// Stores the permutation
List arr = new List();
// To display the result
int ans = -1;
List chosen = new List(N);
for(int i = 0; i < N; i++)
chosen.Add(false);
ans = getMax(arr, ans, chosen, N);
Console.Write(ans +"\n");
}
}
// This code is contributed by shikhasingrajput 输出：
2时间复杂度： O(N * N！)
辅助空间： O(N)