I am trying to get this code to calculate 5 and print numbers by 5's with a while statement of 7, so I want it to loop through, generating a different number 7 times; however, when it gets to a number over 10, I want it to start back over at 0 and ignore the 10.
This is my code:
while z < 7:
firstpickplusfive = int(firstpickplusfive) + 1
counts = counts + 1
if counts == 1:
if firstpickplusfive > 9:
firstpickplusfive = 0
if counts == 5:
print firstpickplusfive
z = int(z) + 1
The code prints the first number, but freezes on printing any others. Why isn't this working?
Your code is not in the loop. Python's code blocks are created with indents:
while z < 7:
firstpickplusfive = int(firstpickplusfive) + 1
counts = counts + 1
if counts == 1:
if firstpickplusfive > 9:
firstpickplusfive = 0
if counts == 5:
print firstpickplusfive
z = int(z) + 1
Is this the result you were trying to achieve:
import random
x = random.randint(1,9)
for i in range(1,8):
print x
x += 5
if x >= 10:
x -= 9
This generates a random number, and adds 5 until it passes 10, then subtracts 9, and it does this seven times. If I understand your question correctly, this is what you were trying to do. Please correct me if I am wrong.
no this is not what I was trying to do here is what I am trying to do.
counts = 0
r = 0
firstpickplusfive = 7
p = firstpickplusfive + 5
while r < 3:
firstpickplusfive = p + counts
if firstpickplusfive > 6:
firstpickplusfive = firstpickplusfive + counts - 10
if p > 9:
p = firstpickplusfive + counts
print firstpickplusfive
counts = counts + 1
r = int(r) + 1
it works alone, but when I add it to the script I am trying to write it doesn't work...if there is a simpler way to do it I would appreciate knowing it.
ie.
number = number + 5 + 0
then
number = number + 5 + 1.....ect which
example 7 + 5 + 0 = 12,
7 + 5 + 1 = 13.........
if the number is equal to 10 then I want it to drop the tens place and keep the 1's place
example 7 + 5 + 0 = 2,
7 + 5 + 1 = 3
Here is an easier method:
for i in range(1,4):
num = 7
num += 5
num +=(i-1)
if num >=10:
num -= 10
print num
i+=1
Try this in your script, does it work?
Related
I'm a beginner and I tried this code to list the sum of all the multiples of 3 or 5 below 100, but it gives a wrong answer and I don't know why.
result = 0
result2 = 0
som = 0
sum2 = 0
below_1000 = True
while below_1000:
result = result+3
if result < 1000:
som += result
else:
below_1000 = False
below_1000 = True
while below_1000:
result2 = result2+5
if result2 < 1000:
sum2 += result2
else:
below_1000 = False
final_result = som+sum2
print(final_result)
Since you first loop over multiples of 3, then again over multiples of 5, you are double-counting a lot of values, specifically values that are multiples of both 3 and 5 (for example 15 or 60).
To write this manually, you can use a for loop over range
total = 0
for i in range(1000):
if i % 3 == 0 or i % 5 == 0:
total += i
>>> total
233168
A more concise way to do this same thing is using a generator expression within the sum function
>>> sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
233168
So I wrote a piece of code in pycharm
to solve this problem:
pick any 5 positive integers that add up to 100
and by addition,subtraction or just using one of the five values
you should be able to make every number up to 100
for example
1,22,2,3,4
for 1 I could give in 1
for 2 i could give in 2
so on
for 21 I could give 22 - 1
for 25 I could give (22 + 2) - 1
li = [1, 1, 1, 1, 1]
lists_of_li_that_pass_T1 = []
while True:
if sum(li) == 100:
list_of_li_that_pass_T1.append(li)
if li[-1] != 100:
li[-1] += 1
else:
li[-1] = 1
if li[-2] != 100:
li[-2] += 1
else:
li[-2] = 1
if li[-3] != 100:
li[-3] += 1
else:
li[-3] = 1
if li[-4] != 100:
li[-4] += 1
else:
li[-4] = 1
if li[-5] != 100:
li[-5] += 1
else:
break
else:
if li[-1] != 100:
li[-1] += 1
else:
li[-1] = 1
if li[-2] != 100:
li[-2] += 1
else:
li[-2] = 1
if li[-3] != 100:
li[-3] += 1
else:
li[-3] = 1
if li[-4] != 100:
li[-4] += 1
else:
li[-4] = 1
if li[-5] != 100:
li[-5] += 1
else:
break
this should give me all the number combinations that add up to 100 out of the total 1*10 ** 10
but its not working please help me fix it so it prints all of the sets of integers
I also can't think of what I would do next to get the perfect sets that solve the problem
After #JohnY comments, I assume that the question is:
Find a set of 5 integers meeting the following requirements:
their sum is 100
any number in the [1, 100] range can be constructed using at most once the elements of the set and only additions and substractions
A brute force way is certainly possible, but proving that any number can be constructed that way would be tedious. But a divide and conquer strategy is possible: to construct all numbers up to n with a set of m numbers u0..., um-1, it is enough to build all numbers up to (n+2)/3 with u0..., um-2 and use um-1 = 2*n/3. Any number in the ((n+2)/3, um-1) range can be written as um-1-x with x in the [1, (n+2)/3] range, and any number in the (um-1, n] range as um-1+y with y in the same low range.
So we can use here u4 = 66 and find a way to build numbers up to 34 with 4 numbers.
Let us iterate: u3 = 24 and build numbers up to 12 with 3 numbers.
One more step u2 = 8 and build numbers up to 4 with 2 numbers.
Ok: u0 = 1 and u1 = 3 give immediately:
1 = u0
2 = 3 - 1 = u1 - u0
3 = u1
4 = 3 + 1 = u1 + u0
Done.
Mathematical disgression:
In fact u0 = 1 and u1 = 3 can build all numbers up to 4, so we can use u2 = 9 to build all numbers up to 9+4 = 13. We can prove easily that the sequence ui = 3i verifies sum(ui for i in [0, m-1]) = 1 + 3 + ... + 3m-1 = (3m - 1)/(3 - 1) = (um - 1) / 2.
So we could use u0=1, u1=3, u2=9, u3=27 to build all numbers up to 40, and finally set u4 = 60.
In fact, u0 and u1 can only be 1 and 3 and u2 can be 8 or 9. Then if u2 == 8, u3 can be in the [22, 25] range, and if u2 == 9, u3 can be in the [21, 27] range. The high limit is given by the 3i sequence, and the low limit is given by the requirement to build numbers up to 12 with 3 numbers, and up to 34 with 4 ones.
No code was used, but I think that way much quicker and less error prone. It is now possible to use Python to show that all numbers up to 100 can be constructed from one of those sets using the divide and conquer strategy.
Why in python a times table program doesn't work like this?
n = int(input("Type a number: "))
count = 10
while count < 0:
print(f"{count} x {n} = {n * count}")
count = count - 1
But this works correctly:
n = int(input("Type a number: "))
for i in range(1, 11):
print(f"{i} x {n} = {n * i}")
The result is (or should be), for example:
1 x 5 = 5
2 x 5 = 10
3 x 5 = 15
4 x 5 = 20
5 x 5 = 25
6 x 5 = 30
7 x 5 = 35
8 x 5 = 40
9 x 5 = 45
10 x 5 = 50
As a beginner I need to understand... and is it possible to use while in that situation?
This is python 3.x
You want to start count from 1 so that your loop increments rather than decrements. Your while loop conditional check is also incorrect as count is never less than 0. Try this:
n = int(input("Type a number: "))
count = 1
while count <= 10:
print(f"{count} x {n} = {n * count}")
count = count + 1
You need to set the condition in the while to be true to go through the loop. Thus we will say do while count is greater than or equal to zero.
n = int(input("Type a number: "))
count = 10
while count >= 0:
print(f"{count} x {n} = {n * count}")
count = count - 1
For some reason, when I try to set a while function, it doesn't do what I want.
For example:
import random
x = 0
while x <= 10:
print random.randint(1, 100)
x += x + 1
This only runs four times. But if I change it to while x <= 1000: it runs ten times. Any idea why?
EDIT
I am sorry, But think this is a legitimate question. Please stop downvoting me! Next time I will do more research into the syntax for problems as simple as this. Sorry.
To increment x by 1 you should use
x += 1
or
x = x + 1
You have combined them into x += x + 1, which adds x+1 to x each time.
Your problem is x += x + 1 is essentially adding x + (x + 1). Simply remove the initial +.
You are adding x + 1 to x. It should be x+=1, that makes you finish sooner:
Start: x = 0
After first iteration: x + x + 1 = 0 + 0 + 1 = 1
After second iteration: x + x + 1 = 1 + 1 + 1 = 3
After third iteration: x + x + 1 = 3 + 3 + 1 = 7
After fourth iteration: x + x + 1 = 7 + 7 + 1 = 15, which is greater that 10 and stops.
You are multiplying x by 2 and adding 1, so it's not an error just is the answer to another question.
I have a task to make a program that will sum the first 100 Fibonacci numbers. I checked my output in Python, and my output in QBasic 64 and they aren't same. I checked with different inputs also.
Input: 10
Output: 89
-----------
Input: 100
Output: 573147844013817084101
Is it correct ?
Here is my code:
n = int(input())
print()
p = 0
d = 1
z = p + d
print(str(p) + ' + ' + str(d) + ' = ' + str(z))
for i in range(n - 2):
p = d
d = z
z = p + d
print(str(p) + ' + ' + str(d) + ' = ' + str(z))
print('Sum:', z)
EDIT: Code edited again, check it now. I just found on Wikipedia.. It depends from what number you start the loop. So if I use (0, 1, 1, 2, 3, 5, 8, 13, 21, and 34) as first 10 Fibonacci numbers, the sum is going to be 88, not 89.
The sums of the first ten and 100 fibonacchi number would be 88 and 573147844013817084100, respectively:
>>> cache = {}
>>> def fib(n):
if n == 0: return 0
if n == 1: return 1
if not n in cache:
cache[n] = fib(n - 1) + fib(n - 2)
return cache[n]
>>> sum([fib(i) for i in range(10)])
88
>>> sum([fib(i) for i in range(100)])
573147844013817084100
In your loop you are already starting the iteration at the 3rd position, since you set. So set your range to (n -2).
0: 1
1 : 1
2 : 1
3 : 2
4 : 3
5 : 5
To get the sum of the Fibonacci numbers, using zero as the first in the series, you need to do this:
def run_it(n):
N2 = 0
N1 = 0
N = 0
z = N
for i in range(n):
print(N,z)
N2 = N1
N1 = N
if N is 0: N = 1
else: N = N1 + N2
z = z + N
run_it(int(input('Number: ')))
To calculate the sum using one as the start of the series, change the initial value of N from zero to one.