I am a student, new to python. I am trying to code a program that will tell if a user input number is fibonacci or not.
num=int(input("Enter the number you want to check\n"))
temp=1
k=0
a=0
summ=0
while summ<=num:
summ=temp+k
temp=summ
k=temp
if summ==num:
a=a+1
print("Yes. {} is a fibonnaci number".format(num))
break
if a==0:
print("No. {} is NOT a fibonacci number".format(num))
#The program is working for only numbers upto 2.
You don't need quite so many variables. A Fibonacci integer sequence can be generated with the single assignment
a, b = b, a + b
# Short for
# temp = a
# a = b
# b = temp + b
Repeated applications sets a to the numbers in the pattern in sequence. The choice of initial values for a and b determine which exact sequence you get. The Fibonacci numbers are generated when a = 0 and b = 1 are used.
a = 0
b = 1
a, b = 1, 0 + 1 # 0, 1
a, b = 1, 1 + 1 # 1, 2
a, b = 2, 1 + 2 # 2, 3
a, b = 3, 2 + 3 # 3, 5
# etc
(As another example, the Lucas numbers are generated if you start with a = 2 and b = 1.)
All you need to do is return True if n == a at each step, iterating until n > a. If n wasn't one of the a values generated by the loop, you'll return False.
n = int(input(...))
a = 0
b = 1
while a <= n:
if n == a:
return True
a, b = b, a + b
return False
I'd suggest something like this:
fib_terms = [0, 1] # first two fibonacci terms
user_input= int(input('Enter the number you want to check\n'))
# Add new fibonacci terms until the user_input is reached
while fib_terms[-1] <= user_input:
fib_terms.append(fib_terms[-1] + fib_terms[-2])
if user_input in fib_terms:
print(f'Yes. {user_input} is a fibonacci number.')
else:
print(f'No. {user_input} is NOT a fibonacci number.')
Check this:
def fibonacci_list(limit):
u_n_1 = 1
u_n_2 = 1
u_n = 1
out_list = [u_n]
while u_n <= limit:
out_list.append(u_n)
u_n = u_n_1 + u_n_2
u_n_2 = u_n_1
u_n_1 = u_n
return out_list
def is_fibonacci_number(n):
if n in fibonacci_list(n):
return True
return False
def main():
n = int(input('Enter a number:'))
if is_fibonacci_number(n):
print(str(n) + ' is a fibonacci number ')
else:
print(str(n) + ' is not a fibonacci number ')
if __name__ == '__main__':
main()
bool checkfibonacci(int n)
{
int a = 0;
int b = 1;
if (n == a || n == b) return true;
int c = a + b;
while(c <= n)
{
if(c == n) return true;
a = b;
b = c;
c = a + b;
}
return false;
}
Something like this will do it.
welcome to the python community.
In Your code temp means the previous number and k means the previous number of temp. The problem is that you change the temp before you assign the next value to k. To solve that you just have to sawp the lines 8 and 9. In the new order you first assign the value of temp to k then summ to temp. That's it!
I suggest you to do it by a list in this way:
fib[i] = fib[i-1] + fib[i-2]
It will improve the clarity of your code.
Related
I want to print the digits of an integer in order, and I don't want to convert to string. I was trying to find the number of digits in given integer input and then divide the number to 10 ** (count-1), so for example 1234 becomes 1.234 and I can print out the "1". now when I want to move to second digit it gets confusing. Here is my code so far:
Note: Please consider that I can only work with integers and I am not allowed to use string and string operations.
def get_number():
while True:
try:
a = int(input("Enter a number: "))
return a
except:
print("\nInvalid input. Try again. ")
def digit_counter(a):
count=0
while a > 0:
count = count+1
a = a // 10
return count
def digit_printer(a, count):
while a != 0:
print (a // (10 ** (count-1)))
a = a // 10
a = get_number()
count = digit_counter(a)
digit_printer(a, count)
I want the output for an integer like 1234 as below:
1
2
3
4
Using modulo to collect the digits in reversed order and then print them out:
n = 1234
digits = []
while n > 0:
digits.append(n%10)
n //=10
for i in reversed(digits):
print(i)
Recursive tricks:
def rec(n):
if n > 0:
rec(n//10)
print(n%10)
rec(1234)
Finding the largest power of 10 needed, then going back down:
n = 1234
d = 1
while d * 10 <= n:
d *= 10
while d:
print(n // d % 10)
d //= 10
This question already has answers here:
What is a debugger and how can it help me diagnose problems?
(2 answers)
Closed 2 years ago.
Consider:
Enter image description here
Input: 20
17
999997
Output: 2^2 * 5
17
757 * 1321
My code:
a = int(input())
# Find the factors first
for i in range(2, a+1):
s = 0
b = a
d = 0
# See if it is a prime number
if a%i == 0:
for x in range(1, i+1):
if a%x == 0:
d = d + x
if (d-1)/i == 1:
d = 0
print(i)
else:
s = 0
b = a
d = 0
continue
d = 0
# I will see how many prime numbers
while(b>0):
if (b/i)%1 == 0:
s = s + 1
b = b/i
else:
b = 0
if b == 1:
b = 0
print(s)
I will find the factors first, and then see if it is a prime number. If so, I will see how many prime numbers it is
if i input 12, it outputs 2 2
Enter link description here
I believe you need the output of the following.
import math
a = int(input())
while (a % 2 == 0):
print(2)
a = int(a/2)
while (a % 3 == 0):
print(3)
a = int(a/3)
for i in range(5, math.ceil(math.sqrt(a)), 6):
while (a % i == 0):
print(i)
a = int(a / i)
while (a % (i + 2) == 0):
print(i + 2)
a = int(a / (i + 2))
if (a > 3):
print(a)
This will give you the prime factors for a given number. As I can understand, it is what you are looking for.
a = int(input("Enter a number:"))
for i in range(2, a + 1):
if a % i != 0:
continue
# SETTING THE DEFAULT VALUES AT THE BEGINNING OF EVERY ITERATION OF THE LOOP
s = 0
b = a
d = 0
for x in range(1, i + 1):
if b % x == 0:
d = d + x
if (d - 1) / i == 1:
d = 0
print(i)
else:
# s = 0 # NO LONGER NEEDED, AS WE RESET THEM AT THE BEGINNING OF THE LOOP
# b = a
# d = 0
continue
while b > 0:
if (b / i) % 1 == 0:
s = s + 1
b = b / i
else:
b = 0
if b == 1:
b = 0
print(s)
a /= i**s # THIS LINE IS IMPORTANT
You were close. You forgot to set the default values at the beginning of every iteration of the loop, so they sometimes didn't have the right values ; and you should set a to a different value by dividing it by the factor you found (i**s, so i to the power of s).
As has been mentioned, your code also follows an odd coding style. I suggest you stop putting newlines between each statement, and start separating operators with spaces (example: range(3+5) is bad, range(3 + 5) is more readable)
You are using too many loops here and that's why you are getting too much confused. Here is the code which serve the same purpose (if I understand your problem correctly)
a = int(input("Enter a number: "))
i = 2
factors = []
while i <= a:
if (a%i) == 0:
factors.append(i)
a = a/i
else:
i = i + 1
print(factors)
here I am returning a list, if you want you can change the type accordingly.
Here are the inputs/outputs:
Enter a number: 17
[17]
Enter a number: 100
[2, 2, 5, 5]
Enter a number: 12
[2, 2, 3]
I made this code about a number and it's power. It will ask a number and it's power and show the output like a horizontal list.. Like
Number = 2
Power = 3.... then output will be like=
1
2
4
Number and power can be +/-.
But I want to sum those numbers like Sum = 7 after it shows
1
2
4
I have no idea how to do it after the output. I am new to programming maybe that's why can't figure out this problem.
Here is the code in Python :
A =float(input("Number:"))
B =float(input("Power:"))
print("Result of Powers:")
i = 0
while i < B:
print(A**i)
i = i + 1
while i >= B:
print(A**i)
i = i - 1
You could simplify this with numpy as follows
import numpy as np
A =float(input("Number:"))
B =int(input("Power:"))
print("Result of Powers:")
power = np.arange(B)
power_result = A ** power
sum_result = np.sum(power_result)
print(power_result)
print(sum_result)
I made B into an int, since I guess it makes sense. Have a look into the numpy documentation to see, what individual functions do.
You can create another variable to store the sum
and to print values on the same line use end=" " argument in the print function
a = float(input("Number:"))
b = int(input("Power:"))
sum = 0.0
i = 0
while b < 0:
ans = a**i
i = i - 1
print(ans, end=" ")
sum = sum + ans
b += 1
while b >= 0:
ans = a**i
i = i + 1
print(ans, end=" ")
sum = sum + ans
b -= 1
print("\nSum = " + str(sum))
I'm not sure what you want to achieve with the second loop. This works:
A =float(input("Number:"))
B =float(input("Power:"))
print("Result of Powers:")
i = 0
n_sum = 0
while i < B:
n_sum += A**i
print(A**i)
i = i + 1
while i >= B:
n_sum += A**i
print(A**i)
i = i - 1
print(n_sum)
Prompt: Write a program that adds all the digits in an integer. If the resulting sum is more than one digit, keep repeating until the sum is one digit. For example, the number 2345 has the sum 2+3+4+5 = 14 which is not a single digit so repeat with 1+4 = 5 which is a single digit.
This is the code I have so far. It works out for the first part, but I can't figure out how to make it repeat until the sum is a single digit. I'm pretty sure I'm supposed to nest the code I already have with another while statement
n = int(input("Input an integer:"))
sum_int=0
while float(n)/10 >= .1:
r= n%10
sum_int += r
n= n//10
if float(n)/10 > .1: print(r, end= " + ")
else: print(r,"=",sum_int)
this is a sample output of the code
Input an integer: 98765678912398
8 + 9 + 3 + 2 + 1 + 9 + 8 + 7 + 6 + 5 + 6 + 7 + 8 + 9 = 88
8 + 8 = 16
1 + 6 = 7
This should work, no division involved.
n = int(input("Input an integer:"))
while n > 9:
n = sum(map(int, str(n)))
print(n)
It basically converts the integer to a string, then sums over the digits using a list comprehension and continues until the number is no greater than 9.
You could utilize recursion.
Try this:
def sum_of_digits(n):
s = 0
while n:
s += n % 10
n //= 10
if s > 9:
return sum_of_digits(s)
return s
n = int(input("Enter an integer: "))
print(sum_of_digits(n))
you can try this solution,
if n=98 then your output will be 8
def repitative_sum(n):
j=2
while j!=1:
n=str(n)
j=len(n)
n=list(map(int,n))
n=sum(n)
print(n)
You don't need to convert your integer to a float here; just use the divmod() function in a loop:
def sum_digits(n):
newnum = 0
while n:
n, digit = divmod(n, 10)
newnum += digit
return newnum
By making it a function you can more easily use it to repeatedly apply this to a number until it is smaller than 10:
n = int(input("Input an integer:"))
while n > 9:
n = sum_digits(n)
print(n)
def add_digits(num):
return (num - 1) % 9 + 1 if num > 0 else 0
A simple, elegant solution.
I'm not sure if it's anti-practice in Python because I know nothing about the language, but here is my solution.
n = int(input("Input an integer:"))
def sum_int(num):
numArr = map(int,str(num))
number = sum(numArr)
if number < 10:
print(number)
else:
sum_int(number)
sum_int(n)
Again I am unsure about the recursion within a function in Python, but hey, it works :)
If you like recursion, and you must:
>>> def sum_digits_rec(integ):
if integ <= 9:
return integ
res = sum(divmod(integ, 10))
return sum_digits(res)
>>> print(sum_digits_rec(98765678912398))
7
def digit_sum(num):
if num < 10:
return num
last_digit = num % 10
num = num / 10
return digit_sum(last_digit + digit_sum(num))
input_num = int(input("Enter an integer: "))
print("result : ",digit_sum(input_num))
This may help you..!
Try with strings comprehension:
new = input("insert your number: ")
new = new.replace(" ","")
new =sum([int(i) for i in new])
if new not in range (10):
new = str(new)
new = sum ([int(i) for i in new])
print (new)
Note that the answers which convert int to str are relying on the python conversion logic to do internally the divmod calculations that we can do explicitly as follows, without introducing the non-numeric string type into an intrinsically numerical calculation:
def boilItDown(n):
while n >= 10:
t = 0
while n:
d, m = divmod(n, 10)
n, t = d, t + m
n = t
return n
n = 98765678912398
print(boilItDown(n))
Output:
7
I have attempted this problem like this :
a = input("Enter number : ")
s = 3
w = 1
while a>0:
digit=a%10
if n%2 == 0:
p = p*digit
else:
s = s+digit
a=a/10
n=n+1
print "The sum is",s
it works perfectly for even no of digits but for odd no of digits like for 234 it shows the sum as 6 and product 3
No explicit loop:
import operator
from functools import reduce # in Python 3 reduce is part of functools
a = input("Enter number : ")
lst = [int(digit) for digit in a]
add = sum(lst[1::2])
mul = reduce(operator.mul, lst[::2],1)
print("add =",add,"mul =",mul,"result =",add+mul)
Producing:
Enter number : 234
add = 3 mul = 8 result = 11
You have to start with n = 0 for this to work
a = int(input("Enter number"))
s = 0
p = 1
n = 0
while a>0:
digit=a%10
if n%2 == 0:
p *= digit
else:
s += digit
a /= 10
n += 1
print "The sum is",s
print "Product is",p
Easy mistake to make about the numbering. The first item of any string, list or array is always index 0. For example, be careful in future to take away 1 from the value returned from len(list) if you are iterating through a list's items with a for loop e.g.
for x in range(len(list)-1):
#your code using list[x]
Here's the mathematical version:
n = input('Enter a number: ')
digits = []
while n > 0:
digits.append(n%10)
n //= 10
s = 0
p = 1
i = 0
for digit in reversed(digits):
if i%2 == 0:
s += digit
else:
p *= digit
i += 1
print 'Sum of even digits:', s
print 'Product of odd digits:', p
print 'Answer:', s+p
I have tried to make this as simple as possible for you.
Here's a function that does the same thing:
def foo(n):
s = 0
p = 1
for i, digit in enumerate(str(n)):
if i%2 == 0:
s += digit
else:
p *= digit
return s+p
def foo(num):
lst = [int(digit) for digit in str(num)]
mul, add = 1, 0
for idx, val in enumerate(lst):
if not idx % 2:
mul *= val
else:
add += val
return add, mul
And using it:
>>> foo(1234)
(6, 3)
>>> foo(234)
(3, 8)
This function will take an integer or a string representation of an integer and split it into a list of ints. It will then use enumerate to iterate over the list and preform the required operations. It returns a 2 element tuple.