给定一个由N 个整数和一个整数K组成的数组arr[] ,任务是通过遍历数组并在数组末尾添加arr[i] / K , K次来找到可能的数组元素的总和, 如果arr[i]可被K整除。否则,停止遍历。
例子:
Input: arr[] = {4, 6, 8, 2}, K = 2
Output: 44
Explanation:
The following operations are performed:
- For arr[0](= 4): arr[0](= 4) is divisible by 2, therefore append 4/2 = 2, 2 numbers of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2}.
- For arr[1](= 6): arr[1](= 6) is divisible by 2, therefore append 6/2 = 3, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3}.
- For arr[2](= 8): arr[2](= 8) is divisible by 2, therefore append 8/2 = 4, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4}.
- For arr[3](= 2): arr[3](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1}.
- For arr[4](= 2): arr[4](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1}.
- For arr[5](= 2): arr[5](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1}.
After completing the above steps, the sum of the array elements is = 4 + 6 + 8 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 1 + 1 + 1 + 1 + 1 + 1 = 44.
Input: arr[] = {4, 6, 8, 9}, K = 2
Output: 45
朴素方法:解决给定问题的最简单方法是遍历给定数组并在数组末尾添加值(arr[i]/K) K次。完成上述步骤后,打印数组元素的总和。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
vector v;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
v.push_back(arr[i]);
}
// Traverse the vector
for (int i = 0;
i < v.size(); i++) {
// If v[i] is divisible by K
if (v[i] % K == 0) {
long long x = v[i] / K;
// Iterate over the range
// [0, K]
for (int j = 0; j < K; j++) {
// Update v
v.push_back(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for (int i = 0; i < v.size(); i++)
sum = sum + v[i];
// Return the sum of the updated array
return sum;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << sum(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
ArrayList v = new ArrayList<>();
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
v.add(arr[i]);
}
// Traverse the vector
for(int i = 0; i < v.size(); i++)
{
// If v[i] is divisible by K
if (v.get(i) % K == 0)
{
int x = v.get(i) / K;
// Iterate over the range
// [0, K]
for(int j = 0; j < K; j++)
{
// Update v
v.add(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for(int i = 0; i < v.size(); i++)
sum = sum + v.get(i);
// Return the sum of the updated array
return sum;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = arr.length;
System.out.println(sum(arr, N, K));
}
}
// This code is contributed by Kingash
Python3
# Python3 program for the above approach
# Function to calculate sum of array
# elements after adding arr[i] / K
# to the end of the array if arr[i]
# is divisible by K
def summ(arr, N, K):
# Stores the sum of the array
sum = 4
v = [i for i in arr]
# Traverse the vector
for i in range(len(v)):
# If v[i] is divisible by K
if (v[i] % K == 0):
x = v[i] // K
# Iterate over the range
# [0, K]
for j in range(K):
# Update v
v.append(x)
# Otherwise
else:
break
# Traverse the vector v
for i in range(len(v)):
sum = sum + v[i]
# Return the sum of the updated array
return sum
# Driver Code
if __name__ == '__main__':
arr = [ 4, 6, 8, 2 ]
K = 2
N = len(arr)
print(summ(arr, N, K))
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int[] arr, int N, int K)
{
// Stores the sum of the array
int sum = 0;
List v = new List();
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
v.Add(arr[i]);
}
// Traverse the vector
for (int i = 0; i < v.Count; i++) {
// If v[i] is divisible by K
if (v[i] % K == 0) {
int x = v[i] / K;
// Iterate over the range
// [0, K]
for (int j = 0; j < K; j++) {
// Update v
v.Add(x);
}
}
// Otherwise
else
break;
}
// Traverse the vector v
for (int i = 0; i < v.Count; i++)
sum = sum + v[i];
// Return the sum of the updated array
return sum;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 4, 6, 8, 2 };
int K = 2;
int N = arr.Length;
Console.WriteLine(sum(arr, N, K));
}
}
// This code is contributed by ukasp.
Javascript
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
vector v;
// Traverse the array
for (int i = 0; i < N; i++) {
v.push_back(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = 0;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.push_back(v[i] / K);
// Otherwise, set flag as true
else {
flag = 1;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << sum(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
ArrayList v = new ArrayList();
// Traverse the array
for (int i = 0; i < N; i++) {
v.add(arr[i]);
}
// Stores if the operation
// should be formed or not
boolean flag = false;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v.get(i) % K == 0)
v.add(v.get(i) / K);
// Otherwise, set flag as true
else {
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v.get(i % N);
}
// Return the resultant sum
return sum;
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = arr.length;
System.out.println(sum(arr, N, K));
}
}
// This code is contributed by Dharanendra L V.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int []arr, int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
List v = new List();
// Traverse the array
for(int i = 0; i < N; i++)
{
v.Add(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = false;
// Traverse the vector V
for(int i = 0; i < v.Count; i++)
{
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.Add(v[i] / K);
// Otherwise, set flag as true
else
{
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
static void Main()
{
int[] arr = { 4, 6, 8, 2 };
int K = 2;
int N = arr.Length;
Console.WriteLine(sum(arr, N, K));
}
}
// This code is contributed by SoumikMondal
Javascript
输出:
44
时间复杂度: O(N * K * log N)
辅助空间: O(M),M 是数组的最大元素。
高效的方法:上述方法也可以基于以下观察进行优化:
- 如果arr[i]可被K整除,则添加arr[i] / K , K次将总和增加arr[i] 。
- 因此,这个想法是只在向量的末尾推送arr[i] / K一次。
请按照以下步骤解决问题:
- 初始化一个变量,比如sum为0 ,它存储所有数组元素 array A[]的总和。
- 初始化一个数组,比如A[]并将所有数组元素arr[] 存储在A[] 中。
- 初始化一个变量,比如flag为0 ,它存储元素是否要添加到数组的末尾。
- 遍历数组A[]并执行以下步骤:
- 如果值标志为0并且A[i]可被K整除,则将A[i]推入V的末尾。
- 否则,将flag的值更新为1 。
- 将总和的值增加V[i % N] 。
- 完成上述步骤后,打印总和的值作为结果总和。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
vector v;
// Traverse the array
for (int i = 0; i < N; i++) {
v.push_back(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = 0;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.push_back(v[i] / K);
// Otherwise, set flag as true
else {
flag = 1;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
int main()
{
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << sum(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
ArrayList v = new ArrayList();
// Traverse the array
for (int i = 0; i < N; i++) {
v.add(arr[i]);
}
// Stores if the operation
// should be formed or not
boolean flag = false;
// Traverse the vector V
for (int i = 0; i < v.size(); i++) {
// If flag is false and if
// v[i] is divisible by K
if (!flag && v.get(i) % K == 0)
v.add(v.get(i) / K);
// Otherwise, set flag as true
else {
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v.get(i % N);
}
// Return the resultant sum
return sum;
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 4, 6, 8, 2 };
int K = 2;
int N = arr.length;
System.out.println(sum(arr, N, K));
}
}
// This code is contributed by Dharanendra L V.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int []arr, int N, int K)
{
// Stores the sum of the array
int sum = 0;
// Stores the array elements
List v = new List();
// Traverse the array
for(int i = 0; i < N; i++)
{
v.Add(arr[i]);
}
// Stores if the operation
// should be formed or not
bool flag = false;
// Traverse the vector V
for(int i = 0; i < v.Count; i++)
{
// If flag is false and if
// v[i] is divisible by K
if (!flag && v[i] % K == 0)
v.Add(v[i] / K);
// Otherwise, set flag as true
else
{
flag = true;
}
// Increment the sum by v[i % N]
sum = sum + v[i % N];
}
// Return the resultant sum
return sum;
}
// Driver Code
static void Main()
{
int[] arr = { 4, 6, 8, 2 };
int K = 2;
int N = arr.Length;
Console.WriteLine(sum(arr, N, K));
}
}
// This code is contributed by SoumikMondal
Javascript
输出:
44
时间复杂度: O(N * log N)
辅助空间: O(N * log N)
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