Is ther any other way to get sum 1 to 100 with recursion?-CodePudding

I'm studing recursive function and i faced question of

"Print sum of 1 to n with no 'for' or 'while' "

ex ) n = 10

answer = 55

n = 100

answer = 5050

so i coded

import sys
sys.setrecursionlimit(1000000)

sum = 0
def count(n):
    global sum
    sum  = n
    if n!=0:
        count(n-1)
count(n = int(input()))
print(sum)

I know it's not good way to get right answer, but there was a solution

n=int(input())

def f(x) :
    if x==1 :
        return 1
    else :
        return ((x 1)//2)*((x 1)//2) f(x//2)*2

print(f(n))

and it works super well , but i really don't know how can human think that logic and i have no idea how it works.

Can you guys explain how does it works? Even if i'm looking that formula but i don't know why he(or she) used like that And i wonder there is another solution too (I think it's reall important to me)

I'm really noob of python and code so i need you guys help, thank you for watching this

CodePudding user response：

Here is a recursive solution.

def rsum(n):
    if n == 1: # BASE CASE
        return 1
    else: # RECURSIVE CASE
        return n   rsum(n-1)

You can also use range and sum to do so.

n = 100
sum_1_to_n = sum(range(n 1))

CodePudding user response：

There is 2 ways to find the answer

1. Recursion

def sum(n):
    if n == 1:
        return 1
    if n <= 0:
        return 0
    else:
        return n   sum(n-1)
        
   
print(sum(100))

This is a simple recursion code, when you try to apply the reccurent function F_n = n F_(n-1) to find the answer

2. Formula

Let S = 1 2 3 ... n

Then lets do something like this

S = 1 2 3 ... n

S = n (n - 1) (n - 2) ... 1

Let's combine them and we get

2S = (n 1) (n 1) ... (n 1) - n times

From that you get

S = ((n 1) * n) / 2

So for n = 100 you get

S = 101 * 100 / 2 = 5050

So in python you will get something like

sum = lambda n: ( (n   1) * n) / 2
        
   
print(sum(100))

CodePudding user response：

you can try this:

def f(n):
if n == 1:
    return 1
return n   f(n - 1)

print(f(10))

this function basically goes from n to 1 and each time it adds the current n, in the end, it returns the sum of n n - 1 ... 1