给定一个大小为N的数组arr[] ，每个由N / 2 个偶数和奇数组成，以及一个整数K ，任务是找到一对不同奇偶校验模K的数组元素之和的最大余数。
例子：
Input: arr[] = {3, 2, 4, 11, 6, 7}, K = 7
Output: 6
Explanation:
Sum of a pair of array elements = 2 + 11
Sum % K = 13 % 7 = 6.
Therefore, the maximum remainder possible is 6.
Input: arr[] = {8, 11, 17, 16}, K = 13
Output: 12
处理方法：按照以下步骤解决问题：
- 初始化一个 HashSet，比如even ，以存储所有偶数数组元素。
- 初始化一个 TreeSet，比如说奇数，以存储所有奇数数组元素。
- 初始化一个变量，比如max_rem，以存储可能的最大余数。
- 遍历 HashSet 并为每个元素找到它的补码并在集合奇数中搜索它，该集合小于等于其补码。
- 用元素的总和更新max_rem ，它是补码。
- 打印最大余数，即max_rem 的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum
// remainder of sum of a pair
// of array elements modulo K
void maxRemainder(int A[], int N, int K)
{
// Stores all even numbers
unordered_set even;
// Stores all odd numbers
set odd;
// Segregate remainders of even
// and odd numbers in respective sets
for(int i = 0; i < N; i++)
{
int num = A[i];
if (num % 2 == 0)
even.insert(num % K);
else
odd.insert(num % K);
}
// Stores the maximum
// remainder obtained
int max_rem = 0;
// Find the complement of remainder
// of each even number in odd set
for(int x : even)
{
// Find the complement
// of remiander x
int y = K - 1 - x;
auto it = odd.upper_bound(y);
if (it != odd.begin())
{
it--;
max_rem = max(max_rem, x + *it);
}
}
// Print the answer
cout << max_rem;
}
// Driver code
int main()
{
// Given array
int arr[] = { 3, 2, 4, 11, 6, 7 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Given value of K
int K = 7;
maxRemainder(arr, N, K);
return 0;
}
// This code is contributed by Kingash Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the maximum
// remainder of sum of a pair
// of array elements modulo K
static void maxRemainder(int A[],
int N, int K)
{
// Stores all even numbers
HashSet even
= new HashSet<>();
// Stores all odd numbers
TreeSet odd
= new TreeSet<>();
// Segregate remainders of even
// and odd numbers in respective sets
for (int num : A) {
if (num % 2 == 0)
even.add(num % K);
else
odd.add(num % K);
}
// Stores the maximum
// remainder obtained
int max_rem = 0;
// Find the complement of remainder
// of each even number in odd set
for (int x : even) {
// Find the complement
// of remiander x
int y = K - 1 - x;
if (odd.floor(y) != null)
max_rem
= Math.max(
max_rem,
x + odd.floor(y));
}
// Print the answer
System.out.print(max_rem);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 3, 2, 4, 11, 6, 7 };
// Size of the array
int N = arr.length;
// Given value of K
int K = 7;
maxRemainder(arr, N, K);
}
} Python3
# Python3 program for the above approach
from bisect import bisect_left
# Function to find the maximum
# remainder of sum of a pair
# of array elements modulo K
def maxRemainder(A, N, K):
# Stores all even numbers
even = {}
# Stores all odd numbers
odd = {}
# Segregate remainders of even
# and odd numbers in respective sets
for i in range(N):
num = A[i]
if (num % 2 == 0):
even[num % K] = 1
else:
odd[num % K] = 1
# Stores the maximum
# remainder obtained
max_rem = 0
# Find the complement of remainder
# of each even number in odd set
for x in even:
# Find the complement
# of remiander x
y = K - 1 - x
od = list(odd.keys())
it = bisect_left(od, y)
if (it != 0):
max_rem = max(max_rem, x + od[it])
# Print the answer
print (max_rem)
# Driver code
if __name__ == '__main__':
# Given array
arr = [3, 2, 4, 11, 6, 7]
# Size of the array
N = len(arr)
# Given value of K
K = 7
maxRemainder(arr, N, K)
# This code is contributed by mohit kumar 29输出：
6时间复杂度： O(N * logN)
辅助空间： O(N)
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