给定一个数组arr []，其中包含前N个自然数的排列。在一个操作中，从阵列中删除一对(X，Y) ，然后将(X + Y + 1)/ 2插入阵列。任务是通过精确地执行给定的操作N – 1次和N – 1对，找到可以保留在数组中的最小数组元素。如果存在多个解决方案，则打印其中任何一种。
例子：
Input: arr[] = {1, 2, 3, 4}
Output: {2}, { (2, 4), (3, 3), (1, 3) }
Explanation:
Selecting a pair(arr[1], arr[3]) modifies the array arr[] = {1, 3, 3}
Selecting a pair(arr[1], arr[2]) modifies the array arr[] = {1, 3}
Selecting a pair(arr[0], arr[1]) modifies the array arr[] = {2}
Therefore, the smallest element left in array = {2} and the pairs that can be selected are {(2, 4), (3, 3), (1, 3)}.
Input: arr[] = {3, 2, 1}
Output: {2}, { (3, 2), (3, 1) }
方法：可以使用贪婪技术解决问题。请按照以下步骤解决问题：
- 初始化一个数组，例如， pairsArr []，以存储可以通过执行给定操作选择的所有对。
- 创建一个优先级队列，例如说pq，以将所有数组元素存储在优先级队列中。
- 横动PQ而留在PQ计数元件是大于1且在每个操作弹出顶部的两个元素PQ的(X，Y)，商店(X，Y)在pairsArr []，然后插入具有值的元素(X + Y更大+1)/ 2在pq中。
- 最后，打印留在pq和pairArr []中的元素的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to print the smallest element left
// in the array and the pairs by given operation
void smallestNumberLeftInPQ(int arr[], int N)
{
// Stores array elements and return
// the minimum element of arr[] in O(1)
priority_queue pq;
// Stores all the pairs that can be
// selected by the given operations
vector > pairsArr;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
pq.push(arr[i]);
}
// Traverse pq while count of elements
// left in pq greater than 1
while (pq.size() > 1) {
// Stores top element of pq
int X = pq.top();
// Pop top element of pq
pq.pop();
// Stores top element of pq
int Y = pq.top();
// Pop top element of pq
pq.pop();
// Insert (X + Y + 1) / 2 in pq
pq.push((X + Y + 1) / 2);
// Insert the pair (X, Y)
// in pairsArr[]
pairsArr.push_back({ X, Y });
}
// Print the element left in pq
// by performing the given operations
cout << "{" << pq.top() << "}, ";
// Stores count of elements
// in pairsArr[]
int sz = pairsArr.size();
// Print all the pairs that can
// be selected in given operations
for (int i = 0; i < sz; i++) {
// If i is the first
// index of pairsArr[]
if (i == 0) {
cout << "{ ";
}
// Print current pairs of pairsArr[]
cout << "(" << pairsArr[i].first
<< ", " << pairsArr[i].second << ")";
// If i is not the last index
// of pairsArr[]
if (i != sz - 1) {
cout << ", ";
}
// If i is the last index
// of pairsArr[]
if (i == sz - 1) {
cout << " }";
}
}
}
// Driver Code
int main()
{
int arr[] = { 3, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
smallestNumberLeftInPQ(arr, N);
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to print the smallest element left
// in the array and the pairs by given operation
static void smallestNumberLeftInPQ(int arr[], int N)
{
// Stores array elements and return
// the minimum element of arr[] in O(1)
PriorityQueue pq = new PriorityQueue<>((x, y) ->
Integer.compare(y, x));
// Stores all the pairs that can be
// selected by the given operations
Vector pairsArr = new Vector<>();
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
pq.add(arr[i]);
}
// Traverse pq while count of elements
// left in pq greater than 1
while (pq.size() > 1) {
// Stores top element of pq
int X = pq.peek();
// Pop top element of pq
pq.remove();
// Stores top element of pq
int Y = pq.peek();
// Pop top element of pq
pq.remove();
// Insert (X + Y + 1) / 2 in pq
pq.add((X + Y + 1) / 2);
// Insert the pair (X, Y)
// in pairsArr[]
pairsArr.add(new pair( X, Y ));
}
// Print the element left in pq
// by performing the given operations
System.out.print("{" + pq.peek()+ "}, ");
// Stores count of elements
// in pairsArr[]
int sz = pairsArr.size();
// Print all the pairs that can
// be selected in given operations
for (int i = 0; i < sz; i++) {
// If i is the first
// index of pairsArr[]
if (i == 0) {
System.out.print("{ ");
}
// Print current pairs of pairsArr[]
System.out.print("(" + pairsArr.get(i).first
+ ", " + pairsArr.get(i).second+ ")");
// If i is not the last index
// of pairsArr[]
if (i != sz - 1) {
System.out.print(", ");
}
// If i is the last index
// of pairsArr[]
if (i == sz - 1) {
System.out.print(" }");
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, 2, 1 };
int N = arr.length;
smallestNumberLeftInPQ(arr, N);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to implement
# the above approach
# Function to prthe smallest
# element left in the array
# and the pairs by given operation
def smallestNumberLeftInPQ(arr, N):
# Stores array elements and
# return the minimum element
# of arr[] in O(1)
pq = []
# Stores all the pairs that can
# be selected by the given operations
pairsArr = []
# Traverse the array arr[]
for i in range(N):
pq.append(arr[i])
pq = sorted(pq)
# Traverse pq while count of
# elements left in pq greater
# than 1
while (len(pq) > 1):
# Stores top element of pq
X = pq[-1]
del pq[-1]
# Stores top element of pq
Y = pq[-1]
# Pop top element of pq
del pq[-1]
# Insert (X + Y + 1) / 2
# in pq
pq.append((X + Y + 1) // 2)
# Insert the pair (X, Y)
# in pairsArr[]
pairsArr.append([X, Y])
pq = sorted(pq)
# Print element left in pq
# by performing the given
# operations
print("{", pq[-1], "}, ",
end = "")
# Stores count of elements
# in pairsArr[]
sz = len(pairsArr)
# Print the pairs that can
# be selected in given operations
for i in range(sz):
# If i is the first
# index of pairsArr[]
if (i == 0):
print("{ ", end = "")
# Print pairs of pairsArr[]
print("(", pairsArr[i][0], ",",
pairsArr[i][1], ")", end = "")
# If i is not the last index
# of pairsArr[]
if (i != sz - 1):
print(end = ", ")
# If i is the last index
# of pairsArr[]
if (i == sz - 1):
print(end = " }")
# Driver Code
if __name__ == '__main__':
arr = [3, 2, 1]
N = len(arr)
smallestNumberLeftInPQ(arr, N)
# This code is contributed by Mohit Kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to print the smallest element left
// in the array and the pairs by given operation
static void smallestNumberLeftInPQ(int[] arr,
int N)
{
// Stores array elements and return
// the minimum element of []arr in O(1)
List pq = new List();
// Stores all the pairs that can be
// selected by the given operations
List pairsArr = new List();
// Traverse the array []arr
for(int i = 0; i < N; i++)
{
pq.Add(arr[i]);
}
pq.Sort();
pq.Reverse();
// Traverse pq while count of elements
// left in pq greater than 1
while (pq.Count > 1)
{
// Stores top element of pq
int X = pq[0];
// Pop top element of pq
pq.RemoveAt(0);
// Stores top element of pq
int Y = pq[0];
// Pop top element of pq
pq.RemoveAt(0);
// Insert (X + Y + 1) / 2 in pq
pq.Add((X + Y + 1) / 2);
pq.Sort();
pq.Reverse();
// Insert the pair (X, Y)
// in pairsArr[]
pairsArr.Add(new pair(X, Y));
}
// Print the element left in pq
// by performing the given operations
Console.Write("{" + pq[0] + "}, ");
// Stores count of elements
// in pairsArr[]
int sz = pairsArr.Count;
// Print all the pairs that can
// be selected in given operations
for(int i = 0; i < sz; i++)
{
// If i is the first
// index of pairsArr[]
if (i == 0)
{
Console.Write("{ ");
}
// Print current pairs of pairsArr[]
Console.Write("(" + pairsArr[i].first + ", " +
pairsArr[i].second + ")");
// If i is not the last index
// of pairsArr[]
if (i != sz - 1)
{
Console.Write(", ");
}
// If i is the last index
// of pairsArr[]
if (i == sz - 1)
{
Console.Write(" }");
}
}
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 3, 2, 1 };
int N = arr.Length;
smallestNumberLeftInPQ(arr, N);
}
}
// This code is contributed by aashish1995 输出：
{2}, { (3, 2), (3, 1) }时间复杂度： O(N * log(N))
辅助空间： O(N)