给定一个整数N和的N的所有非负功率为S = {N 0,N 1,N 2,N 3,…}组,安排在增加子集和的顺序S的所有非空子集。任务是从该顺序中找到第K个子集的最大和最小元素之和。
例子:
Input: N = 4, K = 3
Output: 5
Explanation:
S = {1, 4, 16, 64, … }.
Therefore, the non-empty subsets arranged in increasing order of their sum = {{1}, {4}, {1, 4}, {16}, {1, 16}, {4, 16}, {1, 4, 16}………}.
So the elements of the subset at Kth(3rd) subset are {1, 4}. So the sum of the greatest and smallest element of this subset = (4 + 1) = 5.
Input: N = 3, K = 4
Output: 18
Explanation:
S = {1, 3, 9, 27, 81, …}.
Therefore, the non-empty subsets arranged in increasing order of their sum = {{1}, {3}, {1, 3}, {9}, {1, 9}, {3, 9}, {1, 3, 9} ……..}.
So the element in the subset at 4th position is {9}. So the sum of greatest and smallest element of this subset = (9 + 9) = 18.
方法:此方法基于功率设定的概念。请按照以下步骤解决问题:
- 以相反的顺序生成整数K的对应二进制表达式(即子集的位置),以保持子集中元素的和不断增加。
- 然后,根据lst []列表中连续位置上出现的位,计算子集中相应位置是否存在元素。
- 迭代lst [] ,如果lst [i]为0 ,则忽略该位,否则( N i )* lst [i]为 在第K个子集num []中。
- 然后,通过在位置0和最后位置取num []的元素来计算总和。
- 完成上述步骤后,打印结果总和。
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 sum of greatest
// and smallest element of Kth Subset
void sumofExtremes(int N, int K)
{
// Stores the binary equivalent
// of k in inverted order
list lst;
while (K > 0)
{
lst.push_back(K % 2);
K = K / 2;
}
// Stores the kth subset
list num;
int x = 0;
// Iterate over the list
for(auto element : lst)
{
// If the element is non-zero
if (element != 0)
{
int a = pow(N, x);
a = a * (element);
num.push_back(a);
}
x++;
}
// Update S to length of num
int s = num.size();
// If length of the subset is 1
if (s == 1)
// Print twice of that element
cout << (2 * num.front()) << "\n";
// Print the sum of first and
// last element
else
cout << num.front() + num.back();
}
// Driver Code
int main()
{
// Given number N
int N = 4;
// Given position K
int K = 6;
sumofExtremes(N, K);
}
// This code is contributed by akhilsaini
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to find the sum of greatest
// and smallest element of Kth Subset
public static void sumofExtremes(int N, int K)
{
// Stores the binary equivalent
// of k in inverted order
List lst = new ArrayList();
while (K > 0)
{
lst.add(K % 2);
K = K / 2;
}
// Stores the kth subset
List num = new ArrayList();
int x = 0;
// Iterate over the list
for(int element : lst)
{
// If the element is non-zero
if (element != 0)
{
int a = (int)Math.pow(N, x);
a = a * (element);
num.add(a);
}
x++;
}
// Update S to length of num
int s = num.size();
// If length of the subset is 1
if (s == 1)
// Print twice of that element
System.out.println(2 * num.get(0));
// Print the sum of first and
// last element
else
System.out.println(num.get(0) +
num.get(s - 1));
}
// Driver Code
public static void main(String[] args)
{
// Given number N
int N = 4;
// Given position K
int K = 6;
// Function call
sumofExtremes(N, K);
}
}
// This code is contributed by akhilsaini
Python3
# Python3 program for the above approach
# Function to find the sum of greatest
# and smallest element of Kth Subset
def sumofExtremes(N, K):
# Stores the binary equivalent
# of k in inverted order
lst = []
while K > 0:
lst.append(K % 2)
K = K//2
# Stores the kth subset
num = []
# Iterate over the list
for i in range(0, len(lst)):
# If the element is non-zero
if(lst[i] != 0):
a = N**i
a = a * lst[i]
num.append(a)
# Update S to length of num
s = len(num)
# If length of the subset is 1
if(s == 1):
# Print twice of that element
print(2 * num[0])
# Print the sum of first and
# last element
else:
print(num[0] + num[s - 1])
# Driver Code
# Given number N
N = 4
# Given position K
K = 6
# Function Call
sumofExtremes(N, K)
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the sum of greatest
// and smallest element of Kth Subset
public static void sumofExtremes(int N, int K)
{
// Stores the binary equivalent
// of k in inverted order
List lst = new List();
while (K > 0)
{
lst.Add(K % 2);
K = K / 2;
}
// Stores the kth subset
List num = new List();
// Iterate over the list
for(int i = 0; i < lst.Count; i++)
{
// If the element is non-zero
if (lst[i] != 0)
{
int a = (int)Math.Pow(N, i);
a = a * (lst[i]);
num.Add(a);
}
}
// Update S to length of num
int s = num.Count;
// If length of the subset is 1
if (s == 1)
// Print twice of that element
Console.WriteLine(2 * num[0]);
// Print the sum of first and
// last element
else
Console.WriteLine(num[0] + num[s - 1]);
}
// Driver Code
static public void Main()
{
// Given number N
int N = 4;
// Given position K
int K = 6;
// Function call
sumofExtremes(N, K);
}
}
// This code is contributed by akhilsaini
20
时间复杂度: O(log K)
辅助空间: O(N)