一个人只握手一次的握手次数
一个聚会有 N 个人。找出握手的总数，这样一个人只能握手一次。
示例:
Input : 5
Output : 10
Input : 9
Output : 36
我们可以看到这个问题的递归性质。
// n-th person has (n-1) choices and after
// n-th person chooses a person, problem
// recurs for n-1.
handshake(n) = (n-1) + handshake(n-1)
// Base case
handshake(0) = 0
下面是上面递归公式的实现。
C++
// Recursive C++ program to count total
// number of handshakes when a person
// can shake hand with only one.
#include <bits/stdc++.h>
using namespace std;
// Function to find all possible handshakes
int handshake(int n)
{
// When n becomes 0 that means all the
// persons have done handshake with other
if (n == 0)
return 0;
else
return (n - 1) + handshake(n - 1);
}
// Driver code
int main()
{
int n = 9;
cout << " " << handshake(n);
return 0;
}
// This code is contributed by shivanisinghss2110
C
// Recursive C program to count total
// number of handshakes when a person
// can shake hand with only one.
#include <stdio.h>
// function to find all possible handshakes
int handshake(int n)
{
// when n becomes 0 that means all the
// persons have done handshake with other
if (n == 0)
return 0;
else
return (n - 1) + handshake(n - 1);
}
int main()
{
int n = 9;
printf("%d", handshake(n));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Recursive Java program to
// count total number of
// handshakes when a person
// can shake hand with only one.
import java.io.*;
class GFG
{
// function to find all
// possible handshakes
static int handshake(int n)
{
// when n becomes 0 that
// means all the persons
// have done handshake
// with other
if (n == 0)
return 0;
else
return (n - 1) + handshake(n - 1);
}
// Driver Code
public static void main (String[] args)
{
int n = 9;
System.out.print(handshake(n));
}
}
// This code is contributed
// by chandan_jnu
Python 3
# Recursive Python program
# to count total number of
# handshakes when a person
# can shake hand with only one.
# function to find all
# possible handshakes
def handshake(n):
# when n becomes 0 that means
# all the persons have done
# handshake with other
if (n == 0):
return 0
else:
return (n - 1) + handshake(n - 1)
# Driver Code
n = 9
print(handshake(n))
# This code is contributed
# by Shivi_Aggarwal
C
// Recursive C# program to
// count total number of
// handshakes when a person
// can shake hand with only one.
using System;
class GFG
{
// function to find all
// possible handshakes
static int handshake(int n)
{
// when n becomes 0 that
// means all the persons
// have done handshake
// with other
if (n == 0)
return 0;
else
return (n - 1) + handshake(n - 1);
}
// Driver Code
public static void Main (String []args)
{
int n = 9;
Console.WriteLine(handshake(n));
}
}
// This code is contributed
// by Arnab Kundu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Recursive PHP program to
// count total number of
// handshakes when a person
// can shake hand with only one.
// function to find all
// possible handshakes
function handshake($n)
{
// when n becomes 0 that means
// all the persons have done
// handshake with other
if ($n == 0)
return 0;
else
return ($n - 1) + handshake($n - 1);
}
// Driver Code
$n = 9;
echo(handshake($n));
// This code is contributed
// by Shivi_Aggarwal
?>
java 描述语言
<script>
// Recursive JavaScript program to
// count total number of
// handshakes when a person
// can shake hand with only one.
// function to find all
// possible handshakes
function handshake(n) {
// when n becomes 0 that
// means all the persons
// have done handshake
// with other
if (n === 0)
return 0;
else
return n - 1 + handshake(n - 1);
}
// Driver Code
var n = 9;
document.write(handshake(n));
</script>
Output:
36
我们可以通过扩展递归得出一个直接公式。
handshake(n) = (n-1) + handshake(n-1)
= (n-1) + (n-2) + handshake(n-2)
= (n-1) + (n-2) + .... 1 + 0
= n * (n - 1)/2
C++
// Recursive CPP program to count total
// number of handshakes when a person
// can shake hand with only one.
#include <stdio.h>
// function to find all possible handshakes
int handshake(int n)
{
return n * (n - 1)/2;
}
int main()
{
int n = 9;
printf("%d", handshake(n));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Recursive Java program to
// count total number of
// handshakes when a person
// can shake hand with only one.
class GFG
{
// function to find all
// possible handshakes
static int handshake(int n)
{
return n * (n - 1) / 2;
}
// Driver code
public static void main(String args[])
{
int n = 9;
System.out.println(handshake(n));
}
}
// This code is contributed
// by Arnab Kundu
Python 3
# Recursive Python program
# to count total number of
# handshakes when a person
# can shake hand with only one.
# function to find all
# possible handshakes
def handshake(n):
return int(n * (n - 1) / 2)
# Driver Code
n = 9
print(handshake(n))
# This code is contributed
# by Shivi_Aggarwal
C
// Recursive C# program to
// count total number of
// handshakes when a person
// can shake hand with only one.
using System;
class GFG
{
// function to find all
// possible handshakes
static int handshake(int n)
{
return n * (n - 1) / 2;
}
// Driver code
static public void Main ()
{
int n = 9;
Console.WriteLine(handshake(n));
}
}
// This code is contributed by Sachin
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Recursive PHP program to
// count total number of
// handshakes when a person
// can shake hand with only one.
// function to find all
// possible handshakes
function handshake($n)
{
return $n * ($n - 1) / 2;
}
// Driver Code
$n = 9;
echo(handshake($n));
// This code is contributed
// by Shivi_Aggarwal
?>
java 描述语言
<script>
// Recursive Javascript program to
// count total number of
// handshakes when a person
// can shake hand with only one.
// Function to find all
// possible handshakes
function handshake(n)
{
return n * parseInt((n - 1) / 2, 10);
}
// Driver code
let n = 9;
document.write(handshake(n));
// This code is contributed by rameshtravel07
</script>
Output:
36
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