原文:
问题–给定两个数字 n 和 k，我们的任务是从 n 中减去一个数字 k，直到数字(n)大于零，一旦 n 变成负或零，我们就开始加 k，直到那个数字变成原来的数字(n)。 注意:不允许使用任何循环。 例:
input : n = 15 k = 5
output : 15 10 5 0 1 5 10 15
input : n = 20 k = 6
output : 20 14 8 2 -4 2 8 14 20
解释–我们可以用递归的思想来做，就是我们一次又一次的调用函数，直到 n 大于零(在每次的函数调用中，我们用 k 减去 n)。一旦数字变成负数或零，我们开始在每个函数调用中添加 k，直到数字变成原始数字。这里我们使用一个函数进行加法和减法，但是为了在加法和减法之间切换，我们使用了一个布尔标志。
c
// c program to print number
// series without using loop
#include
using namespace std;
// function print series
// using recursion
void printnumber(int n, int original, int k, bool flag)
{
// print the number
cout << n << " ";
// change flag if number
// become negative
if (n <= 0)
flag = !flag;
// base condition for
// second_case (adding k)
if (n == original && !flag)
return;
// if flag is true
// we subtract value until
// number is greater then zero
if (flag == true) {
printnumber(n - k, original, k, flag);
return;
}
// second case (addition )
if (!flag) {
printnumber(n k, original, k, flag);
return;
}
}
// driver program
int main()
{
int n = 20, k = 6;
printnumber(n, n, k, true);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print number
// series without using loop
import java.io.*;
import java.util.*;
class gfg
{
public static void printnumber(int n, int original, int k, boolean flag)
{
// print the number
system.out.print(n " ");
// change flag if number
// become negative
if (n <= 0)
flag = !flag;
// base condition for
// second_case (adding k)
if (n == original && !flag)
return;
// if flag is true
// we subtract value until
// number is greater then zero
if (flag == true)
{
printnumber(n - k, original, k, flag);
return;
}
// second case (addition )
if (!flag)
{
printnumber(n k, original, k, flag);
return;
}
}
public static void main (string[] args)
{
int n = 20, k = 6;
printnumber(n, n, k, true);
}
}
// this code is contributed by mohit gupta_omg
python 3
# python program to print number
# series without using loop
def printnumber(n, original, k, flag):
#print the number
print(n, end = " ")
# change flag if number
# become negative
if (n <= 0):
if(flag==0):
flag = 1
else:
flag = 0
# base condition for
# second_case (adding k)
if (n == original and (not(flag))):
return
# if flag is true
# we subtract value until
# number is greater then zero
if (flag == true):
printnumber(n - k, original, k, flag)
return
# second case (addition )
if (not(flag)):
printnumber(n k, original, k, flag);
return
n = 20
k = 6
printnumber(n, n, k, true)
# this code is contributed by mohit gupta_omg
c
// c# program to print number
// series without using loop
using system;
public class gfg {
// function print series
// using recursion
static void printnumber(int n,
int original, int k, bool flag)
{
// print the number
console.write(n " ");
// change flag if number
// become negative
if (n <= 0)
flag = !flag;
// base condition for
// second_case (adding k)
if (n == original && !flag)
return;
// if flag is true
// we subtract value until
// number is greater then zero
if (flag == true)
{
printnumber(n - k, original, k,
flag);
return;
}
// second case (addition )
if (!flag)
{
printnumber(n k, original, k,
flag);
return;
}
}
// driver program
static public void main ()
{
int n = 20, k = 6;
printnumber(n, n, k, true);
}
}
// this code is contributed by vt_m.
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
20 14 8 2 -4 2 8 14 20
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