给定一个长度为N的数组arr[] ，任务是找到数组的每个可能的无序对的成对和的异或。无序对总和定义如下 –
XOR of pairwise sum = (A[0] + A[1]) ^
(A[0] + A[2]) ^ ...(A[0] + A[N]) ^
(A[1] + A[2]) ^ ...(A[1] + A[N]) ^
.......
(A[N-1] + A[N])
Notice that after including A[0] and A[1]
as pairs, then A[1] and A[0] are not included.例子：
Input: arr[] = {1, 2}
Output: 3
Explanation:
There is only one unordered pair. That is (1, 2)
Input: arr[] = {1, 2, 3}
Output: 2
Explanation:
Unordered pairs of the numbers –
{(1, 2), (1, 3), (2, 3)}
XOR of unordered pairswise sum –
=> (1 + 2) ^ (1 + 3) ^ (2 + 3)
=> 3 ^ 4 ^ 5
=> 2
朴素的方法：这个想法是在两个循环的帮助下找到每个可能的无序对，并找到这些对的异或。
下面是上述方法的实现：
C++
// C++ implementation to find XOR of
// pairwise sum of every unordered
// pairs in an array
#include
using namespace std;
// Function to find XOR of pairwise
// sum of every unordered pairs
int xorOfSum(int a[], int n)
{
int answer = 0;
// Loop to choose every possible
// pairs in the array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++)
answer ^= (a[i] + a[j]);
}
return answer;
}
// Driver Code
int main()
{
int n = 3;
int A[n] = { 1, 2, 3 };
cout << xorOfSum(A, n);
return 0;
} Java
// Java implementation to find XOR of
// pairwise sum of every unordered
// pairs in an array
class GFG{
// Function to find XOR of pairwise
// sum of every unordered pairs
static int xorOfSum(int a[], int n)
{
int answer = 0;
// Loop to choose every possible
// pairs in the array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++)
answer ^= (a[i] + a[j]);
}
return answer;
}
// Driver Code
public static void main(String[] args)
{
int n = 3;
int A[] = { 1, 2, 3 };
System.out.print(xorOfSum(A, n));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation to find XOR of
# pairwise sum of every unordered
# pairs in an array
# Function to find XOR of pairwise
# sum of every unordered pairs
def xorOfSum(a, n):
answer = 0
# Loop to choose every possible
# pairs in the array
for i in range(n):
for j in range(i + 1, n):
answer ^= (a[i] + a[j])
return answer
# Driver Code
if __name__ == '__main__':
n = 3
A=[1, 2, 3]
print(xorOfSum(A, n))
# This code is contributed by mohit kumar 29C#
// C# implementation to find XOR of
// pairwise sum of every unordered
// pairs in an array
using System;
using System.Collections.Generic;
class GFG{
// Function to find XOR of pairwise
// sum of every unordered pairs
static int xorOfSum(int []a, int n)
{
int answer = 0;
// Loop to choose every possible
// pairs in the array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++)
answer ^= (a[i] + a[j]);
}
return answer;
}
// Driver Code
public static void Main(String[] args)
{
int n = 3;
int []A = { 1, 2, 3 };
Console.Write(xorOfSum(A, n));
}
}
// This code is contributed by PrinciRaj1992Javascript
C++
// C++ implementation to find XOR of
// pairwise sum of every unordered
// pairs in an array
#include
using namespace std;
// Function to find XOR of pairwise
// sum of every unordered pairs
int xorOfSum(int a[], int n)
{
int i, j, k;
// Sort the array
sort(a, a + n);
int ans = 0;
// Array elements are not greater
// than 1e7 so 27 bits are suffice
for (k = 0; k < 27; ++k) {
// Modded elements of array
vector b(n);
// Loop to find the modded
// elements of array
for (i = 0; i < n; i++)
b[i] = a[i] % (1 << (k + 1));
// Sort the modded array
sort(b.begin(), b.end());
int cnt = 0;
for (i = 0; i < n; i++) {
// finding the bound for j
// for given i using binary search
int l = lower_bound(b.begin() +
i + 1, b.end(),
(1 << k) - b[i]) - b.begin();
int r = lower_bound(b.begin() +
i + 1, b.end(), (1 << (k + 1)) -
b[i]) - b.begin();
// All the numbers in the range
// of indices can be added to the
// count to check the xor.
cnt += r - l;
l = lower_bound(b.begin() + i + 1,
b.end(), (1 << (k + 1)) +
(1 << k) - b[i]) - b.begin();
cnt += n - l;
}
// Remainder of cnt * kth power
// of 2 added to the xor value
ans += (cnt % 2) * 1LL * (1 << k);
}
return ans;
}
// Driver Code
int main()
{
int n = 3;
int A[n] = { 1, 2, 3 };
cout << xorOfSum(A, n);
return 0;
} 输出：
2有效的方法：
- 为了获得ķ我们在所有对和数看到最后的异或值的个位，即第k个位他们设置与否。如果有，甚至一些有第K个位设置，则第k个位对，他们的XOR为零别的一个。
- 为了找到设置了第K 位的对和的计数，我们注意到我们可以将所有数组元素乘以 2 (K+1) 。这是因为 X 和 Y 属于输入数组并且Sum = X + Y 。然后 X + Y 可以加起来设置它们的第K 位，这意味着 Sum >= 2 K 。它也可以被观察到，它们可以具有从另外一个遗留，这使得数字在范围[2（K + 1），2（K + 1）+ 2 K）具有其第K位未设置。所以，我们只关心第K和（K + 1）的所有号码的个位检查K的XOR次位。
- 执行模运算后，对于设置第 k 位的 sum，其值将在范围内 – [2 K , 2 (K+1) ) U [2 (K+1) + 2 K , Max-Value-Sum -可以采取 ]。
- 要查找所述范围内的数字，请创建另一个包含 arr[] 修改后的数组元素的数组 B，并对它们进行排序。那么 Sum 可以假设为 Sum = B i + B j 。最后，使用二分查找（C++ 内置的lower_bound）找到j 的最大边界。修复 i 并且由于数组已排序，因此找到满足给定条件的最后一个 j，并且可以将索引范围内的所有数字添加到计数中以检查异或。
下面是上述方法的实现：
C++
// C++ implementation to find XOR of
// pairwise sum of every unordered
// pairs in an array
#include
using namespace std;
// Function to find XOR of pairwise
// sum of every unordered pairs
int xorOfSum(int a[], int n)
{
int i, j, k;
// Sort the array
sort(a, a + n);
int ans = 0;
// Array elements are not greater
// than 1e7 so 27 bits are suffice
for (k = 0; k < 27; ++k) {
// Modded elements of array
vector b(n);
// Loop to find the modded
// elements of array
for (i = 0; i < n; i++)
b[i] = a[i] % (1 << (k + 1));
// Sort the modded array
sort(b.begin(), b.end());
int cnt = 0;
for (i = 0; i < n; i++) {
// finding the bound for j
// for given i using binary search
int l = lower_bound(b.begin() +
i + 1, b.end(),
(1 << k) - b[i]) - b.begin();
int r = lower_bound(b.begin() +
i + 1, b.end(), (1 << (k + 1)) -
b[i]) - b.begin();
// All the numbers in the range
// of indices can be added to the
// count to check the xor.
cnt += r - l;
l = lower_bound(b.begin() + i + 1,
b.end(), (1 << (k + 1)) +
(1 << k) - b[i]) - b.begin();
cnt += n - l;
}
// Remainder of cnt * kth power
// of 2 added to the xor value
ans += (cnt % 2) * 1LL * (1 << k);
}
return ans;
}
// Driver Code
int main()
{
int n = 3;
int A[n] = { 1, 2, 3 };
cout << xorOfSum(A, n);
return 0;
}
输出：
2性能分析：
- 时间复杂度： O(N * log(max(A))*log(N)
最外层循环运行 log(max(A)) 次，对于每个循环，我们创建和排序数组 b ，它由N 个元素组成，因此复杂度为O(N*log(N)*log(max(A)))