给定n个朋友,每个人可以保持单身,也可以与其他朋友配对。每个朋友只能配对一次。找出使朋友保持单身或配对的方式的总数。
例子 :
Input : n = 3
Output : 4
Explanation:
{1}, {2}, {3} : all single
{1}, {2, 3} : 2 and 3 paired but 1 is single.
{1, 2}, {3} : 1 and 2 are paired but 3 is single.
{1, 3}, {2} : 1 and 3 are paired but 2 is single.
Note that {1, 2} and {2, 1} are considered same.
f(n) = ways n people can remain single
or pair up.
For n-th person there are two choices:
1) n-th person remains single, we recur
for f(n - 1)
2) n-th person pairs up with any of the
remaining n - 1 persons. We get (n - 1) * f(n - 2)
Therefore we can recursively write f(n) as:
f(n) = f(n - 1) + (n - 1) * f(n - 2)
由于上述递归公式具有重叠的子问题,因此我们可以使用动态编程来解决。
C++
// C++ program for solution of
// friends pairing problem
#include
using namespace std;
// Returns count of ways n people
// can remain single or paired up.
int countFriendsPairings(int n)
{
int dp[n + 1];
// Filling dp[] in bottom-up manner using
// recursive formula explained above.
for (int i = 0; i <= n; i++) {
if (i <= 2)
dp[i] = i;
else
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
}
return dp[n];
}
// Driver code
int main()
{
int n = 4;
cout << countFriendsPairings(n) << endl;
return 0;
}
Java
// Java program for solution of
// friends pairing problem
import java.io.*;
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int dp[] = new int[n + 1];
// Filling dp[] in bottom-up manner using
// recursive formula explained above.
for (int i = 0; i <= n; i++) {
if (i <= 2)
dp[i] = i;
else
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
}
return dp[n];
}
// Driver code
public static void main(String[] args)
{
int n = 4;
System.out.println(countFriendsPairings(n));
}
}
// This code is contributed by vt_m
Python3
# Python program solution of
# friends pairing problem
# Returns count of ways
# n people can remain
# single or paired up.
def countFriendsPairings(n):
dp = [0 for i in range(n + 1)]
# Filling dp[] in bottom-up manner using
# recursive formula explained above.
for i in range(n + 1):
if(i <= 2):
dp[i] = i
else:
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]
return dp[n]
# Driver code
n = 4
print(countFriendsPairings(n))
# This code is contributed
# by Soumen Ghosh.
C#
// C# program solution for
// friends pairing problem
using System;
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int[] dp = new int[n + 1];
// Filling dp[] in bottom-up manner using
// recursive formula explained above.
for (int i = 0; i <= n; i++) {
if (i <= 2)
dp[i] = i;
else
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
}
return dp[n];
}
// Driver code
public static void Main()
{
int n = 4;
Console.Write(countFriendsPairings(n));
}
}
// This code is contributed by nitin mittal.
PHP
Javascript
C++
// C++ program for solution of friends
// pairing problem Using Recursion
#include
using namespace std;
int dp[1000];
// Returns count of ways n people
// can remain single or paired up.
int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
int main()
{
memset(dp, -1, sizeof(dp));
int n = 4;
cout << countFriendsPairings(n) << endl;
// this code is contributed by Kushdeep Mittal
}
Java
// Java program for solution of friends
// pairing problem Using Recursion
class GFG {
static int[] dp = new int[1000];
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 1000; i++)
dp[i] = -1;
int n = 4;
System.out.println(countFriendsPairings(n));
}
}
// This code is contributed by Ita_c.
Python3
# Python3 program for solution of friends
# pairing problem Using Recursion
dp = [-1] * 1000
# Returns count of ways n people
# can remain single or paired up.
def countFriendsPairings(n):
global dp
if(dp[n] != -1):
return dp[n]
if(n > 2):
dp[n] = (countFriendsPairings(n - 1) +
(n - 1) * countFriendsPairings(n - 2))
return dp[n]
else:
dp[n] = n
return dp[n]
# Driver Code
n = 4
print(countFriendsPairings(n))
# This code contributed by PrinciRaj1992
C#
// C# program for solution of friends
// pairing problem Using Recursion
using System;
class GFG {
static int[] dp = new int[1000];
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
static void Main()
{
for (int i = 0; i < 1000; i++)
dp[i] = -1;
int n = 4;
Console.Write(countFriendsPairings(n));
}
}
// This code is contributed by DrRoot_
PHP
2)
{
$dp[$n] = countFriendsPairings($n - 1) + ($n - 1) *
countFriendsPairings($n - 2);
return $dp[$n];
}
else
{
$dp[$n] = $n;
return $dp[$n];
}
}
// Driver Code
$n = 4;
echo countFriendsPairings($n)
// This code is contributed by Ryuga
?>
Javascript
C++
#include
using namespace std;
// Returns count of ways n people
// can remain single or paired up.
int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
int main()
{
int n = 4;
cout << countFriendsPairings(n);
return 0;
}
// This code is contributed by mits
Java
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
public static void main(String[] args)
{
int n = 4;
System.out.println(countFriendsPairings(n));
}
}
// This code is contributed by Ravi Kasha.
Python3
# Returns count of ways n people
# can remain single or paired up.
def countFriendsPairings(n):
a, b, c = 1, 2, 0;
if (n <= 2):
return n;
for i in range(3, n + 1):
c = b + (i - 1) * a;
a = b;
b = c;
return c;
# Driver code
n = 4;
print(countFriendsPairings(n));
# This code contributed by Rajput-Ji
C#
using System;
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
public static void Main(String[] args)
{
int n = 4;
Console.WriteLine(countFriendsPairings(n));
}
}
// This code has been contributed by 29AjayKumar
PHP
输出 :
10
时间复杂度: O(n)
辅助空间: O(n)
另一种方法:(使用递归)
C++
// C++ program for solution of friends
// pairing problem Using Recursion
#include
using namespace std;
int dp[1000];
// Returns count of ways n people
// can remain single or paired up.
int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
int main()
{
memset(dp, -1, sizeof(dp));
int n = 4;
cout << countFriendsPairings(n) << endl;
// this code is contributed by Kushdeep Mittal
}
Java
// Java program for solution of friends
// pairing problem Using Recursion
class GFG {
static int[] dp = new int[1000];
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 1000; i++)
dp[i] = -1;
int n = 4;
System.out.println(countFriendsPairings(n));
}
}
// This code is contributed by Ita_c.
Python3
# Python3 program for solution of friends
# pairing problem Using Recursion
dp = [-1] * 1000
# Returns count of ways n people
# can remain single or paired up.
def countFriendsPairings(n):
global dp
if(dp[n] != -1):
return dp[n]
if(n > 2):
dp[n] = (countFriendsPairings(n - 1) +
(n - 1) * countFriendsPairings(n - 2))
return dp[n]
else:
dp[n] = n
return dp[n]
# Driver Code
n = 4
print(countFriendsPairings(n))
# This code contributed by PrinciRaj1992
C#
// C# program for solution of friends
// pairing problem Using Recursion
using System;
class GFG {
static int[] dp = new int[1000];
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
if (dp[n] != -1)
return dp[n];
if (n > 2)
return dp[n] = countFriendsPairings(n - 1) + (n - 1) * countFriendsPairings(n - 2);
else
return dp[n] = n;
}
// Driver code
static void Main()
{
for (int i = 0; i < 1000; i++)
dp[i] = -1;
int n = 4;
Console.Write(countFriendsPairings(n));
}
}
// This code is contributed by DrRoot_
的PHP
2)
{
$dp[$n] = countFriendsPairings($n - 1) + ($n - 1) *
countFriendsPairings($n - 2);
return $dp[$n];
}
else
{
$dp[$n] = $n;
return $dp[$n];
}
}
// Driver Code
$n = 4;
echo countFriendsPairings($n)
// This code is contributed by Ryuga
?>
Java脚本
输出 :
10
时间复杂度: O(n)
辅助空间: O(n)
由于上述公式类似于斐波那契数,因此我们可以使用迭代解决方案来优化空间。
C++
#include
using namespace std;
// Returns count of ways n people
// can remain single or paired up.
int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
int main()
{
int n = 4;
cout << countFriendsPairings(n);
return 0;
}
// This code is contributed by mits
Java
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
public static void main(String[] args)
{
int n = 4;
System.out.println(countFriendsPairings(n));
}
}
// This code is contributed by Ravi Kasha.
Python3
# Returns count of ways n people
# can remain single or paired up.
def countFriendsPairings(n):
a, b, c = 1, 2, 0;
if (n <= 2):
return n;
for i in range(3, n + 1):
c = b + (i - 1) * a;
a = b;
b = c;
return c;
# Driver code
n = 4;
print(countFriendsPairings(n));
# This code contributed by Rajput-Ji
C#
using System;
class GFG {
// Returns count of ways n people
// can remain single or paired up.
static int countFriendsPairings(int n)
{
int a = 1, b = 2, c = 0;
if (n <= 2) {
return n;
}
for (int i = 3; i <= n; i++) {
c = b + (i - 1) * a;
a = b;
b = c;
}
return c;
}
// Driver code
public static void Main(String[] args)
{
int n = 4;
Console.WriteLine(countFriendsPairings(n));
}
}
// This code has been contributed by 29AjayKumar
的PHP
输出 :
10
时间复杂度: O(n)
辅助空间: O(1)