递归练习题|第 7 集
原文:https://www . geesforgeks . org/practice-questions-for-recursion-set-7/
问题 1 预测以下程序的输出。下面的 fun()一般做什么？
C++
#include <iostream>
using namespace std;
int fun(int n, int* fp)
{
int t, f;
if (n <= 2) {
*fp = 1;
return 1;
}
t = fun(n - 1, fp);
f = t + *fp;
*fp = t;
return f;
}
int main()
{
int x = 15;
cout << fun(5, &x) << endl;
return 0;
}
C
#include <stdio.h>
int fun ( int n, int *fp )
{
int t, f;
if ( n <= 2 )
{
*fp = 1;
return 1;
}
t = fun ( n-1, fp );
f = t + *fp;
*fp = t;
return f;
}
int main()
{
int x = 15;
printf("%d\n",fun(5, &x));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
import java.io.*;
class GFG {
static int fp = 15;
static int fun ( int n)
{
int t, f;
if ( n <= 2 )
{
fp = 1;
return 1;
}
t = fun ( n - 1);
f = t + fp;
fp = t;
return f;
}
public static void main (String[] args)
{
System.out.println(fun(5));
}
}
// This code is contributed by shubhamsingh10
Python 3
fp = 15
def fun ( n ):
global fp
if ( n <= 2 ):
fp = 1
return 1
t = fun ( n - 1 )
f = t + fp
fp = t
return f
# Driver code
print(fun(5))
# This code is contributed by shubhamsingh10
C
using System;
class GFG{
static int fp = 15;
static int fun ( int n)
{
int t, f;
if ( n <= 2 )
{
fp = 1;
return 1;
}
t = fun ( n - 1 );
f = t + fp;
fp = t;
return f;
}
static public void Main ()
{
Console.Write(fun(5));
}
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
//Javascript Implementation
var fp = 15;
function fun( n )
{
var t, f;
if ( n <= 2 )
{
fp = 1;
return 1;
}
t = fun ( n - 1 );
f = t + fp;
fp = t;
return f;
}
// Driver Code
document.write(fun(5));
// This code is contributed by shubhamsingh10
</script>
Output
5
程序计算第 n 个斐波那契数。语句 t = fun ( n-1，fp)给出第(n-1)个斐波那契数，fp 用于存储第(n-2)个斐波那契数。fp 的初始值(在上面的程序中是 15)无关紧要。下面的递归树显示了从 1 到 10 的所有步骤，用于 fun(5，&x)的执行。
(1) fun(5, fp)
/ \
(2) fun(4, fp) (10) t = 5, f = 8, *fp = 5
/ \
(3) fun(3, fp) (9) t = 3, f = 5, *fp = 3
/ \
(4) fun(2, fp) (8) t = 2, f = 3, *fp = 2
/ \
(5) fun(1, fp) (7) t = 1, f = 2, *fp = 1
/
(6) *fp = 1
问题 2: 预测以下程序的输出。
C++
#include <iostream>
using namespace std;
void fun(int n)
{
if(n > 0)
{
fun(n - 1);
cout << n <<" ";
fun(n - 1);
}
}
int main()
{
fun(4);
return 0;
}
// This code is contributed by shubhamsingh10
C
#include <stdio.h>
void fun(int n)
{
if(n > 0)
{
fun(n-1);
printf("%d ", n);
fun(n-1);
}
}
int main()
{
fun(4);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
import java.util.*;
class GFG{
static void fun(int n)
{
if(n > 0)
{
fun(n - 1);
System.out.print(n+" ");
fun(n - 1);
}
}
public static void main(String[] args)
{
fun(4);
}
}
// This code is contributed by Shubhamsingh10
Python 3
def fun(n):
if(n > 0):
fun(n - 1)
print(n,end=" ")
fun(n - 1)
# driver code
fun(4)
# This code is contributed by shubhamsingh10
C
using System;
class GFG{
static void fun(int n)
{
if(n > 0)
{
fun(n - 1);
Console.Write(n+" ");
fun(n - 1);
}
}
static public void Main ()
{
fun(4);
}
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
function fun(n)
{
if(n > 0)
fun(n - 1);
document.write(n+" ")
fun(n - 1);
}
// driver code
fun(4)
// This code is contributed by bobby.
</script>
Output
1 2 1 3 1 2 1 4 1 2 1 3 1 2 1
fun(4)
/
fun(3), print(4), fun(3) [fun(3) prints 1 2 1 3 1 2 1]
/
fun(2), print(3), fun(2) [fun(2) prints 1 2 1]
/
fun(1), print(2), fun(1) [fun(1) prints 1]
/
fun(0), print(1), fun(0) [fun(0) does nothing]
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