给定一个由N 个整数组成的数组arr[]和一个整数K，任务是找到一个长度为K的子数组，其元素之和等于任意数的阶乘。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.
朴素方法：解决问题的最简单方法是计算所有长度为K 的子数组的总和，并检查这些总和中的任何一个是否是任何数字的阶乘。如果发现任何子数组都为真，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：为了优化上述方法，其思想是使用滑动窗口技术来计算长度为 K 的所有子数组的总和，然后检查总和是否为阶乘。以下是步骤：
- 计算前K 个数组元素的总和并将总和存储在一个变量中，例如sum 。
- 然后遍历剩余的数组并不断更新sum ，通过从前一个子数组中减去第一个元素并添加当前数组元素来得到当前大小为K的子数组的和。
- 要检查总和是否是一个数的阶乘，请将总和除以 2、3 等，直到不能再除以。如果数字减少到 1，则总和是数字的阶乘。
- 如果上述步骤中的和是一个数的阶乘，则存储该子数组的开始和结束索引以打印子数组。
- 完成上述步骤后，如果没有找到这样的子数组，则打印“-1” 。否则，打印存储了开始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;
// Calculate 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;
// Calculate 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
# Calculate 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;
// Calculate 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)
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