I'm trying to code if three numbers are even, odd, or neither (when the numbers have an even AND odd in the input). I'm just stuck since every time it outputs "odd" no matter the numbers. Here's my code so far:
msg = input("Enter a,b,c: ")
nums = msg.split(',')
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if (a, b, c % 2) == 0:
print("even")
else:
print("odd")
my professor mentioned using tuples but I don't know where to incorporate it. Any help is appreciated!
I think your professor meant something like this:
if (a % 2, b % 2, c % 2) == (0, 0, 0):
Your condition in if is wrong, because it basically compares tuple to 0 and is always False. Corrected script:
msg = input("Enter a,b,c: ")
nums = msg.split(',')
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a % 2 == 0 and b % 2 == 0 and c % 2 == 0:
print("even")
else:
print("odd")
Or you can use builtin function all():
if all(number % 2 == 0 for number in [a, b, c]):
print("even")
else:
print("odd")
I'm a complete beginner to programming so forgive me for my naivete.
I wanted to make a program in Python that lets me print a given N number of prime numbers, where N is inputted by the user. I searched a little on "for/while" loops and did some tinkering. I ran a program I saw online and modified it to suit the problem. Here is the code:
i = 1
print("Hi! Let's print the first N prime numbers.")
nPrimes = int(input("Enter your N: "))
counter = 0
while True:
c = 0 #another initialization
for x in range (1, (i + 1)):
a = i % x # "a" is a new variable that got introduced.
if a == 0:
c = c + 1
if c == 2:
print(i, end = " ")
counter = counter + 1
if counter > = nPrimes: #if it reaches the number input, the loop will end.
break
i = i+1
print(": Are your", nPrimes, "prime number/s!")
print()
print("Thanks for trying!")
This should be able to print the amount of prime numbers the user so likes. It is a working code, though I am having difficulty trying to understand it. It seems that the variable c is important in deciding whether or not to print the variable i (which in our case is the supposed prime number during that interval).
We do c + 1 to c every time our variable a has a remainder of 0 in a = i % x. Then, if c reaches 2, the current variable i is printed, and variable c re-initializes itself to 0 once a prime number has been found and printed.
This I can comprehend, but I get confused once the numbers of i get to values 4 and onwards. *How is 4 skipped by the program and not printed when it has 2+ factors in the range that makes its remainder equal to zero? Wouldn't c == 2 for 4 and thus print 4? *And how would the program continue to the next number, 5? (Given that variable N is a large enough input).
Any clarifications would be greatly appreciated. Thank you so much!
From Wikipedia we know:
A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers.
So to find a prime, is to find a natural number, aka an integer, which can only be exactly divided by 1 or itself. This is called Approach of Definition to find primes.
Hence, the following loop traverses through all integers from 1 to i,
and it counts how many times the integer i can be exactly divided by them.
for x in range (1, (i + 1)):
a = i % x # "a" is a new variable that got introduced.
if a == 0:
c = c + 1
And later you judge if the integer i can only be exactly divided by 1 and itself.
If true, you got a prime;
otherwise you just keep on.
if c == 2:
print(i, end = " ")
counter = counter + 1
if counter > = nPrimes: #if it reaches the number input, the loop will end.
break
Meanwhile, you can improve this prime searching algorithm a little bit by changing i = 1 to i = 2 in the beginning and adding an if statement:
# Start from 2 instead of 1
# and end at `i - 1` instead of `i`
for x in range (2, i):
a = i % x # "a" is a new variable that got introduced.
if a == 0:
c = c + 1
# Abandon the loop
# because an integer with factors other than 1 and itself
# is unevitably a composite number, not a prime
if c > 0:
break
if c == 0:
print(i, end = " ")
counter = counter + 1
if counter >= nPrimes: #if it reaches the number input, the loop will end.
break
This twist improves the efficiency of your program because you avoid unnecessary and meaningless amount of work.
To prevent potential infinite loop resulting from while expression,
you should replace while True: with while counter < nPrimes:. And the code turns out to be like this:
#if it reaches the number input, the loop will end.
while counter < nPrimes:
c = 0 #another initialization
# Start from 2 instead of 1
# and end at `i - 1` instead of `i`
for x in range (2, i):
a = i % x # "a" is a new variable that got introduced.
if a == 0:
c = c + 1
# Abandon the loop
# because an integer with factors other than 1 and itself
# is unevitably a composite number, not a prime
if c > 0:
break
if c == 0:
print(i, end = " ")
counter = counter + 1
i = i + 1
If you want to read more about how to improve your program's efficiency in finding primes, read this code in C language. :P
c in this case is used to count the number of numbers that divide evenly into i.
for example, if i = 8: 8 is divisible by 1, 2, 4, and 8. so c = 4 since there are 4 things that divide evenly into it
if i = 5: 5 is divisible by 1 and 5. so c = 2 since there are 2 numbers that divide evenly into it
if i = 4 (where you seem to be confused): 4 is divisible by 1, 2, and 4. so c = 3, not 2.
I am solving this question on HackerRank.
Problem Statement
You are given a number N, you need to print the
number of positions where digits exactly divides N.
Input format
The first line contains T (number of test cases followed by T lines
each containing N).
Constraints
1 <=T <= 15
0 < N < 1010
Output Format
Number of positions in N where digits in that number
exactly divides the number N.
When I give the input
T = 1
N = 114108089
I get ZeroDivisionError
Even after using the try and exception, error handling technique in Python the answer I get is 0 instead of the correct answer 3.
Code so far:
def find_digits():
try:
ctr = 0
c = int(input())
string_c = str(c)
list1 = []
for ch in string_c:
list1.append(ch)
for i in list1:
print i
b = int(i)
if c % b == 0:
ctr += 1
print ctr
except ZeroDivisionError:
print 0
test_case = int(input()) for s in range(test_case):
find_digits()
Some of the digits in N are zeroes, so b == 0 and when you try:
if c % b == 0:
you get a ZeroDivisionError. Rather than wrap the whole function in try, (which means you will only ever get non-zero output for inputs that contain no zeroes at all), wrap that one section:
try:
if c % b == 0:
ctr += 1
except ZeroDivisionError:
pass
Alternatively, check for non-zero values of b first:
if b and c % b == 0:
If you really want to get clever, the function can be done in one line:
sum(N % i == 0 for i in map(int, str(N)) if i)
You are getting a division by zero error because the digits can be zero.
if c % b == 0:
Here b can be zero.
I've made a program which takes number of test cases as input and for each test case, it needs a number as input. Finally it checks whether the numbers you have entered are fibonacci numbers or not and prints accordingly. I've had no problems running it on my PC.But when i upload it to CodeChef.com(where i saw this quesion), it shows runtime error.
Any help is appreciated and as i'm a noob my code might look lengthy ., any modifications are welcome.Thanks!
Here's my code:
def isperfect(n):
import math
if n < 0:
print("No Solution")
return False
else:
test = int(math.sqrt(n))
return test*test == n
test_cases = int(input())
count = 0
store = []
while count < test_cases:
x = int(input())
store.append(x)
count += 1
for each_item in store:
assert isinstance(each_item, int)
s1 = 5*each_item*each_item-4
s2 = 5*each_item*each_item+4
if(isperfect(s1) == True or isperfect(s2) == True):
print("YES")
else:
print("NO")
This is the most elegant solution i've encountered:
def is_fibonacci(n):
phi = 0.5 + 0.5 * math.sqrt(5.0)
a = phi * n
return n == 0 or abs(round(a) - a) < 1.0 / n
The code is not mine, was posted by #sven-marnach.
The original post:
check-input-that-belong-to-fibonacci-numbers-in-python
The runtime error is apparently due to an exception, but Codechef does not provide any more information. It could be various things including divide by zero, memory exhaustion, assertion failure, ...
Although your program works for many common numbers, it doesn't handle all the inputs that the Codechef constraints allow. In particular, the number to be tested can have up to 1000 digits in it. If you try a large input like a 1000-digit number, you'll find it fails. First it fails because of your assert isinstance(each_item, int); a number of 12 digits or more is not of type int. You can just remove the assertion. The next failure occurs because you are using the floating point sqrt routine, and that requires the integer to be converted into floating point. Python can handle very long integers, but the floating point representation is more limited in its precision, and cannot handle 1000 digit integer conversions. That's harder to fix, but can be done. See this ActiveState recipe for an all-integer solution for square root.
I've figured that this can be done by using Newton- Raphson method, i have replaced the code in the function isperfect() with Newton-Raphson formula code, removed assertion and it worked. Thanks for all your help.
Here's the final code:
def isperfect(n):
x = n
y = (x + n // x) // 2
while y < x:
x = y
y = (x + n // x) // 2
return x*x == n
test_cases = int(input())
count = 0
store = []
while count < test_cases:
x = int(input())
store.append(x)
count += 1
for each_item in store:
s1 = 5*each_item*each_item-4
s2 = 5*each_item*each_item+4
if(isperfect(s1) == True or isperfect(s2) == True):
print("YES")
else:
print("NO")
This is going to be a very efficient way of doing it.
In [65]:
import scipy.optimize as so
from numpy import *
In [66]:
def fib(n):
s5=sqrt(5.)
return sqrt(0.2)*(((1+s5)/2.)**n-((1-s5)/2.)**n)
def apx_fib(n):
s5=sqrt(5.)
return sqrt(0.2)*(0.5*(1+s5))**n
def solve_fib(f):
_f=lambda x, y: (apx_fib(x)-y)**2
return so.fmin_slsqp(_f,0., args=(f,),iprint=0)
def test_fib(fibn):
if fibn<1:
print 'No, it is not'
else:
n=int(round(solve_fib(fibn)))
if int(round(fib(n)))==int(fibn):
print 'Yes, it is. (%d)'%n
else:
print 'No, it is not'
In [67]:
asarray(fib(arange(1,20)), dtype='int64')
Out[67]:
array([ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
144, 233, 377, 610, 987, 1597, 2584, 4181])
In [68]:
test_fib(34)
Yes, it is. (9)
In [69]:
test_fib(4181)
Yes, it is. (19)
In [70]:
test_fib(4444)
No, it is not
Fairly simple and efficient way of doing it
def isInFib(n):
if n == 0: return False
elif n == 1: return True
else:
A = 1
B = 1
FLIP = True
while(True):
new = A + B
if new > n: return False
elif new == n: return True
else:
if(FLIP):
A = new
FLIP = not FLIP
else:
B = new
FLIP = not FLIP
Explanation of Algorithm
I first check if my input is equal to 0 or 1, and return appropriate.
If my input is greater than 1, we go into the else, infinite loop.
A and B represent the last two previous numbers in the sequence. We add A and B to get new, the current Fibonacci number.
We check if new is equal to our input number, if that's true we return true, we are done and complete the function.
If it's greater, that means our number is not in the Fibonacci sequence, since we have surpassed it.
if it's less, we need to keep going! and this is where I think it get's confusing to explain. I want to set up A or B as my current fibonacci sequence number (new), but I have make sure that I keep switching between them, since I don't want one to get left behind. Fibonacci sequence takes the previous 2 numbers and adds them together. I want A and B to be my last two previous sums.
I'll use an example
1,1,2,3,5,8,13
A and B are initially 1. So new is equal to 2 first. I then check if I'm over my input number or equal to it. But if my input number is less. I want to keep going.
I want A to equal new (value = 2) then, before we get to our next iteration of the while loop. So that new will equal 2 + 1 as A + B on the next iteration.
But, then THE next iteration of the loop, I want to set B as 3, and I want to leave A being equal to 2. So I have to keep switching between putting the current fibonacci number I'm at, in A or B. And that's why I have the flip logic! It just keeps switching between True and False.
Try this function:
def isfib(number):
num1 = 1
num2 = 1
while True:
if num2 <= number:
if num2 == number:
return True
else:
tempnum = num2
num2 += num1
num1 = tempnum
else:
return False
Here's how it works:
Set num1 and num2 to 0.
If num2 isn't less than or equal to the number return False.
If num2 is equal to the number return True.
Set add num1 to num2, set num1 to num2's original value and go back to step 2.
num=int(input("Enter a number : "))
n1=0
n2=1
i=1
lst=[]
lst.append(n1)
lst.append(n2)
while i<num-1:
n3=n1+n2
n1=n2
n2=n3
lst.append(n3)
i=i+1
for i in lst:
if (num == i):
print("Fibonacci number")
break
else:
print("Not fibonacci")
def isfib(number):
num1 = 0
num2 = 1
while True:
if num2 <= number:
if num2 == number:
return True
else:
tempnum = num2
num2 += num1
num1 = tempnum
else:
return False
n=int(input())
number=n
fibonacci=isfib(number)
if (fibonacci):
print("true")
elif n==0:
print("true")
else:
print("false")
I am trying to write a program that first of all checks if you can get from number a to number b by only swapping 2 digits. My below code is designed to start at 53150220288 and then check if it is possible to reach number 537163806382 by only changing 2 digits. So basically i am trying to only change 2 digits of the number 537163806382 so that the new numbers are multiples of 99. Then output them to a text file. I am writing this program to do it for me but it is actually for a maths competition.
f = open('Blank.txt', 'w')
a = 53150220288
b = 537163806382
b = str(b)
c = 0
while a <= 1000000000:
for i in range(len(b)):
if b[i] == a[i]:
c = c
else:
c = c + 1
if c == 2:
a = str(a)
print(a, file=f)
else:
c = 0
a = int(a)
a = a + 99
f.close()
Problem with the above code is that it outputs absolutely nothing! I don't know why?
Your program never enters the while loop:
a = 53150220288
while a <= 1000000000:
The obvious reason is that your program doesn't enter the loop.
so do this:
a = 53150220288
while a <= 1000000000: