给定N个整数和整数K的数组arr [] ，任务是找到一个长度为K的子数组，该数组的元素之和等于任意数量的阶乘。如果不存在这样的子数组，则打印“ -1” 。
例子：
Input: arr[] = {23, 45, 2, 4, 6, 9, 3, 32}, K = 5
Output: 2 4 6 9 3
Explanation:
Subarray {2, 4, 6, 9, 3} with sum 24 (= 4!) satisfies the required condition.
Input: arr[] = {23, 45, 2, 4, 6, 9, 3, 32}, K = 3
Output: -1
Explanation:
No such subarray of length K (= 3) exists.
天真的方法：解决该问题的最简单方法是计算所有长度为K的子数组的和，并检查这些和中的任何一个是否为任意数量的阶乘。如果发现对于任何子数组为true，则打印该子数组。否则，打印“ -1” 。
时间复杂度： O(N * K)
辅助空间： O(1)
高效方法：为优化上述方法，其思想是使用滑动窗口技术计算长度为K的所有子数组的总和，然后检查总和是否为阶乘。步骤如下：
- 计算前K个数组元素的总和，并将总和存储在变量中，例如sum 。
- 然后遍历剩余的数组，并通过从先前的子数组中减去第一个元素并添加当前的数组元素，不断更新sum以获得大小为K的当前子数组的总和。
- 要检查总和是否是一个阶乘，请将该总和除以2、3，以此类推，直到无法再对其求和。如果数字减少为1，则总和是数字的阶乘。
- 如果以上步骤中的总和是一个数字的阶乘，则存储该子数组的开始和结束索引以打印该子数组。
- 完成上述步骤后，如果找不到此类子数组，请打印“ -1” 。否则，打印存储开始索引和结束索引的子数组。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a number
// is factorial of a number or not
int isFactorial(int n)
{
int i = 2;
while (n != 1) {
// If n is not a factorial
if (n % i != 0) {
return 0;
}
n /= i;
i++;
}
return i - 1;
}
// Function to return the index of
// the valid subarray
pair sumFactorial(
vector arr, int K)
{
int i, sum = 0, ans;
// Calaculate the sum of
// first subarray of length K
for (i = 0; i < K; i++) {
sum += arr[i];
}
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If sum of first K length subarray
// is factorial of a number
if (ans != 0) {
return make_pair(ans, 0);
}
// Find the number formed from the
// subarray which is a factorial
for (int j = i; j < arr.size(); j++) {
// Update sum of current subarray
sum += arr[j] - arr[j - K];
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If ans is true, then return
// index of the current subarray
if (ans != 0) {
return make_pair(ans,
j - K + 1);
}
}
// If the required subarray is
// not possible
return make_pair(-1, 0);
}
// Function to print the subarray whose
// sum is a factorial of any number
void printRes(pair answer,
vector arr, int K)
{
// If no such subarray exists
if (answer.first == -1) {
cout << -1 << endl;
}
// Otherwise
else {
int i = 0;
int j = answer.second;
// Iterate to print subarray
while (i < K) {
cout << arr[j] << " ";
i++;
j++;
}
}
}
// Driver Code
int main()
{
vector arr
= { 23, 45, 2, 4,
6, 9, 3, 32 };
// Given sum K
int K = 5;
// Function Call
pair answer
= sumFactorial(arr, K);
// Print the result
printRes(answer, arr, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
import java.io.*;
class GFG{
// Pair class
public static class Pair
{
int x;
int y;
Pair(int x, int y)
{
this.x = x;
this.y = y;
}
}
// Function to check if a number
// is factorial of a number or not
static int isFactorial(int n)
{
int i = 2;
while (n != 1)
{
// If n is not a factorial
if (n % i != 0)
{
return 0;
}
n /= i;
i++;
}
return i - 1;
}
// Function to return the index of
// the valid subarray
static ArrayList sumFactorial(int arr[],
int K)
{
ArrayList pair = new ArrayList<>();
int i, sum = 0, ans;
// Calaculate the sum of
// first subarray of length K
for(i = 0; i < K; i++)
{
sum += arr[i];
}
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If sum of first K length subarray
// is factorial of a number
if (ans != 0)
{
Pair p = new Pair(ans, 0);
pair.add(p);
return pair;
}
// Find the number formed from the
// subarray which is a factorial
for(int j = i; j < arr.length; j++)
{
// Update sum of current subarray
sum += arr[j] - arr[j - K];
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If ans is true, then return
// index of the current subarray
if (ans != 0)
{
Pair p = new Pair(ans, j - K + 1);
pair.add(p);
return pair;
}
}
// If the required subarray is
// not possible
Pair p = new Pair(-1, 0);
pair.add(p);
return pair;
}
// Function to print the subarray whose
// sum is a factorial of any number
static void printRes(ArrayList answer,
int arr[], int K)
{
// If no such subarray exists
if (answer.get(0).x == -1)
{
// cout << -1 << endl;
System.out.println("-1");
}
// Otherwise
else
{
int i = 0;
int j = answer.get(0).y;
// Iterate to print subarray
while (i < K)
{
System.out.print(arr[j] + " ");
i++;
j++;
}
}
}
// Driver Code
public static void main(String args[])
{
// Given array arr[] and brr[]
int arr[] = { 23, 45, 2, 4,
6, 9, 3, 32 };
int K = 5;
ArrayList answer = new ArrayList<>();
// Function call
answer = sumFactorial(arr,K);
// Print the result
printRes(answer, arr, K);
}
}
// This code is contributed by bikram2001jha Python3
# Python3 program for the above approach
# Function to check if a number
# is factorial of a number or not
def isFactorial(n):
i = 2
while (n != 1):
# If n is not a factorial
if (n % i != 0):
return 0
n = n // i
i += 1
return i - 1
# Function to return the index of
# the valid subarray
def sumFactorial(arr, K):
i, Sum = 0, 0
# Calaculate the sum of
# first subarray of length K
while(i < K):
Sum += arr[i]
i += 1
# Check if sum is a factorial
# of any number or not
ans = isFactorial(Sum)
# If sum of first K length subarray
# is factorial of a number
if (ans != 0):
return (ans, 0)
# Find the number formed from the
# subarray which is a factorial
for j in range(i, len(arr)):
# Update sum of current subarray
Sum = Sum + arr[j] - arr[j - K]
# Check if sum is a factorial
# of any number or not
ans = isFactorial(Sum)
# If ans is true, then return
# index of the current subarray
if (ans != 0):
return (ans, j - K + 1)
# If the required subarray is
# not possible
return (-1, 0)
# Function to print the subarray whose
# sum is a factorial of any number
def printRes(answer, arr, K):
# If no such subarray exists
if (answer[0] == -1):
print(-1)
# Otherwise
else:
i = 0
j = answer[1]
# Iterate to print subarray
while (i < K):
print(arr[j], end = " ")
i += 1
j += 1
# Driver code
arr = [ 23, 45, 2, 4, 6, 9, 3, 32 ]
# Given sum K
K = 5
# Function call
answer = sumFactorial(arr, K)
# Print the result
printRes(answer, arr, K)
# This code is contributed by divyeshrabadiya07C#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Pair class
public class Pair
{
public int x;
public int y;
public Pair(int x, int y)
{
this.x = x;
this.y = y;
}
}
// Function to check if a number
// is factorial of a number or not
static int isFactorial(int n)
{
int i = 2;
while (n != 1)
{
// If n is not
// a factorial
if (n % i != 0)
{
return 0;
}
n /= i;
i++;
}
return i - 1;
}
// Function to return the index of
// the valid subarray
static List sumFactorial(int []arr,
int K)
{
List pair = new List();
int i, sum = 0, ans;
// Calaculate the sum of
// first subarray of length K
for(i = 0; i < K; i++)
{
sum += arr[i];
}
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If sum of first K length subarray
// is factorial of a number
if (ans != 0)
{
Pair p = new Pair(ans, 0);
pair.Add(p);
return pair;
}
// Find the number formed from the
// subarray which is a factorial
for(int j = i; j < arr.Length; j++)
{
// Update sum of current subarray
sum += arr[j] - arr[j - K];
// Check if sum is a factorial
// of any number or not
ans = isFactorial(sum);
// If ans is true, then return
// index of the current subarray
if (ans != 0)
{
Pair p = new Pair(ans, j - K + 1);
pair.Add(p);
return pair;
}
}
// If the required subarray is
// not possible
Pair p1 = new Pair(-1, 0);
pair.Add(p1);
return pair;
}
// Function to print the subarray whose
// sum is a factorial of any number
static void printRes(List answer,
int []arr, int K)
{
// If no such subarray exists
if (answer[0].x == -1)
{
// cout << -1 << endl;
Console.WriteLine("-1");
}
// Otherwise
else
{
int i = 0;
int j = answer[0].y;
// Iterate to print subarray
while (i < K)
{
Console.Write(arr[j] + " ");
i++;
j++;
}
}
}
// Driver Code
public static void Main(String []args)
{
// Given array []arr and brr[]
int []arr = {23, 45, 2, 4,
6, 9, 3, 32};
int K = 5;
List answer = new List();
// Function call
answer = sumFactorial(arr, K);
// Print the result
printRes(answer, arr, K);
}
}
// This code is contributed by shikhasingrajput 输出：
2 4 6 9 3
时间复杂度： O(N)
辅助空间： O(1)