I am having trouble solving the following question:
Write a program that draws “modular rectangles” like the ones below. The user specifies the width and height of the rectangle, and the entries start at 0 and increase typewriter fashion from left to right and top to bottom, but are all done mod 10. Example: Below are examples of a 3 x 5 rectangular:
The following code is what I have tried to solve the problem:
I know it's bad but I still don't know how to print the 5 till 9 numbers on top of each other.
width = int(input("Enter the width of the rectangle:"))
height = int(input("Enter the height of the rectangle:"))
for x in range(0, width, 1):
for y in range(0, height, 1):
print(y, end = ' ')
print()
Thank you all in advance.
You're on the right track, but you have a few issues. Firstly you need to iterate y before x, since you process the columns for each row, not rows for each column. Secondly, you need to compute how many values you have output, which you can do with (y*width+x). To output a single digit, take that value modulo 10. Finally range(0, width, 1) is just the same as range(width). Putting it all together:
width = 5
height = 3
for y in range(height):
for x in range(width):
print((y*width+x)%10, end=' ')
print()
Output:
0 1 2 3 4
5 6 7 8 9
0 1 2 3 4
Related
I am doing a question that gives me a start coordinate, a end coordinate and the number of times of moving.Every time you can add 1 or minus 1 to x or y coordinate based on previous coordinate and the number of moving limit the time the coordinate can move. At last, I need to identify whether there is a possibility to get to the end coordinate
I decide to use recursion to solve this problem however, it does not end even if I wrote return inside a if else statement. Do you mind to take a look at it.
This is the code
# https://cemc.uwaterloo.ca/contests/computing/2017/stage%201/juniorEF.pdf
# input
start = input()
end = input()
count = int(input())
coo_end = end.split(' ')
x_end = coo_end[0]
y_end = coo_end[1]
end_set = {int(x_end), int(y_end)}
#processing
coo = start.split(' ')
x = int(coo[0])
y = int(coo[1])
change_x = x
change_y = y
sum = x + y+count
set1 = set()
tim = 0
timer = 0
ways = 4** (count-1)
def elit(x, y, tim,timer, ways = ways):
print(tim,timer)
tim = tim +1
co1 = (x, y+1)
co2 = (x+1, y)
co3 = (x, y-1)
co4 = (x-1, y)
if tim == count:
tim =0
set1.add(co1)
set1.add(co2)
set1.add(co3)
set1.add(co4)
print(timer)
timer = timer +1
if timer == ways:
print('hiii')
return co1, co2, co3, co4 #### this is the place there is a problem
elit(co1[0],co1[1],tim,timer)
elit(co2[0],co2[1],tim,timer)
elit(co3[0],co3[1],tim, timer)
elit(co4[0],co4[1],tim, timer)
#print(elit(change_x,change_y,tim)) - none why
elit(change_x,change_y,tim, timer)
#print(list1)
for a in set1:
if end_set != a:
answer = 'N'
continue
else:
answer = "Y"
break
print(answer)
In addition, if you have any suggestions about writing this question, do you mind to tell me since I am not sure I am using the best solution.
one of example is
Sample Input
3 4 (start value)
3 3 (end value)
3 (count)
Output for Sample Input
Y
Explanation
One possibility is to travel from (3, 4) to (4, 4) to (4, 3) to (3, 3).
the detailed question can be seen in this file https://cemc.uwaterloo.ca/contests/computing/2017/stage%201/juniorEF.pdf
It is question 3. Thank you
thank you guys
the function is returning properly however by the time you reach the recursive depth to return anything you have called so many instances of the function that it seems like its in an infinite loop
when you call elite the first time the function calls itself four more times, in the example you have given timer is only incremented every 3 cycles and the function only return once timer hits 16 thus the function will need to run 48 times before returning anything and each time the function will be called 4 more times, this exponential growth means for this example the function will be called 19807040628566084398385987584 times, which depending on your machine may well take until the heat death of the universe
i thought i should add that i think you have somewhat over complicated the question, on a grid to get from one point to another the only options are the minimum distance or that same minimum with a diversion that must always be a multiple of 2 in length, so if t the movement is at least the minimum distance or any multiple of 2 over the result should be 'Y', the minimum distance will just be the difference between the coordinates on each axis this can be found by add in the difference between the x and y coordinates
abs(int(start[0]) - int(end[0])) + abs(int(start[1]) -int(end[1]))
the whole function therefore can just be:
def elit():
start = input('start: ').split(' ')
end = input('end: ').split(' ')
count = int(input('count: '))
distance = abs(int(start[0]) - int(end[0])) + abs(int(start[1]) -int(end[1]))
if (count - distance) % 2 == 0:
print('Y')
else:
print('N')
input:
3 4
3 3
3
output:
Y
input:
10 4
10 2
5
output:
N
I'm trying to create a row of numbers in python turtle, similar to something like this, with spaces between each number:
0 1 2 3 4 5 6 8 9
This is my code so far in a loop:
import turtle
pen = turtle.Turtle()
numbers = ["0","1","2","3","4","5","6","7","8","9"]
coordinates = 100,100
for i in range(10):
pen.penup()
pen.goto(coordinates)
pen.pendown()
pen.write(numbers[i])
coordinates = coordinates
Whenever I run this, it writes the numbers 0-9, but in the same spot at the coordinates 100,100.
I want each number to be 50 apart, so after the number 0 is at 100,100 the number 1 should be at 100,150. So, at the very end, what would I put after the "coordinates word" so that each number would be 50 apart?
After you write each number, the coordinate needs to be updated as well.
In your last line coordinates = coordinates, you could try
coordinates = (coordinates[0], coordinates[1] + 50)
So that the y value is updated by 50 with each iteration.
I just finished a challenge on Dcoder ("Love for Mathematics") using Python. I failed two test-cases, but got one right. I used somewhat of a lower level of Python for the same as I haven't explored more yet, so I'm sorry if it looks a bit too basic.The Challenge reads:
Students of Dcoder school love Mathematics. They love to read a variety of Mathematics books. To make sure they remain happy, their Mathematics teacher decided to get more books for them.
A student would become happy if there are at least X Mathematics books in the class and not more than Y books because they know "All work and no play makes Jack a dull boy".The teacher wants to buy a minimum number of books to make the maximum number of students happy.
The Input
The first line of input contains an integer N indicating the number of students in the class. This is followed up by N lines where every line contains two integers X and Y respectively.
#Sample Input
5
3 6
1 6
7 11
2 15
5 8
The Output
Output two space-separated integers that denote the minimum number of mathematics books required and the maximum number of happy students.
Explanation: The teacher could buy 5 books and keep student 1, 2, 4 and 5 happy.
#Sample Output
5 4
Constraints:
1 <= N <= 10000
1 <= X, Y <= 10^9
My code:
n = int(input())
l = []
mi = []
ma = []
for i in range(n):
x, y = input().split()
mi.append(int(x))
ma.append(int(y))
if i == 0:
h=ma[0]
else:
if ma[i]>h:
h=ma[i]
for i in range(h):
c = 0
for j in range(len(mi)):
if ma[j]>=i and mi[j]<=i:
c+=1
l.append(c)
great = max(l)
for i in range(1,len(l)+1):
if l[i]==great:
print(i,l[i])
break
My Approach:
I first assigned the two minimum and maximum variables to two different lists - one containing the minimum values, and the other, the maximum. Then I created a loop that processes all numbers from 0 to the maximum possible value of the list containing maximum values and increasing the count for each no. by 1 every time it lies within the favorable range of students.
In this specific case, I got that count list to be (for the above given input):
[1,2,3,3,4,4,3,3,2 ...] and so on. So I could finalize that 4 would be the maximum no. of students and that the first index of 4 in the list would be the minimum no. of textbooks required.
But only 1 test-case worked and two failed. I would really appreciate it if anyone could help me out here.
Thank You.
This problem is alike minimum platform problem.
In that, you need to sort the min and max maths books array in ascending order respectively. Try to understand the problem from the above link (platform problem) then this will be a piece of cake.
Here is your solution:
n = int(input())
min_books = []
max_books = []
for i in range(n):
x, y = input().split()
min_books.append(int(x))
max_books.append(int(y))
min_books.sort()
max_books.sort()
happy_st_result = 1
happy_st = 1
books_needed = min_books[0]
i = 1
j = 0
while (i < n and j < n):
if (min_books[i] <= max_books[j]):
happy_st+= 1
i+= 1
elif (min_books[i] > max_books[j]):
happy_st-= 1
j+= 1
if happy_st > happy_st_result:
happy_st_result = happy_st
books_needed = min_books[i-1]
print(books_needed, happy_st_result)
Try this, and let me know if you need any clarification.
#Vinay Gupta's logic and explanation is correct. If you think on those lines, the answer should become immediately clear to you.
I have implemented the same logic in my code below, except using fewer lines and cool in-built python functions.
# python 3.7.1
import itertools
d = {}
for _ in range(int(input())):
x, y = map(int, input().strip().split())
d.setdefault(x, [0, 0])[0] += 1
d.setdefault(y, [0, 0])[1] += 1
a = list(sorted(d.items(), key=lambda x: x[0]))
vals = list(itertools.accumulate(list(map(lambda x: x[1][0] - x[1][1], a))))
print(a[vals.index(max(vals))][0], max(vals))
The above answer got accepted in Dcoder too.
I am building a multiplication table in python using while loops. The output is strange though as the numbers line up. The table is fine in terms of completion but it looks ugly. How can i straighten the last three columns? I would post a pic of the output but i am a new user and it will not let me.
width = int(input("Please enter the width of the table:"))
def print_times_table(width):
row = 0
col = 0
width += 1
spaces = ' '
while row < width:
col = 0
while col < width:
print(row*col, spaces, end="")
col += 1
print("\n", end='')
row +=1
print_times_table(width)
output: http://i.stack.imgur.com/C9AzE.jpg
First, you don't need a picture; you should be able to show the output as text just by indenting four spaces (or using the {} icon):
Please enter the width of the table:4
0 0 0 0 0
0 1 2 3 4
0 2 4 6 8
0 3 6 9 12
0 4 8 12 16
The problem is that you're assuming each number will be the same width. This works up to 3x3, because everything is one character wide, but for 5x5, some numbers are one character, some are two (and of course it gets even worse at 10x10).
The easy way to fix this is to force each cell to be the same width.
First, you have to calculate the biggest size you'll need. But that's easy: it's the size of width*width.
Next, you have to know how to force the numbers to use up that many characters. Python has a few ways to do this. I'll show how to do it with old-fashioned field specifiers, but you should look into how to convert this to new-style format strings. (If this is homework, someone teaching you Python 3 will probably grade you down for using old-style fieldspecs. If it's just for your own self-education, it's worth figuring out how to do it both ways.) Or, alternatively, you should look at how to convert it to use a variable-width format spec ('%*d') instead of building a static '%4d' spec. (nneonneo's answer should give a clue to that.)
width = int(input("Please enter the width of the table:"))
def print_times_table(width):
row = 0
col = 0
fieldspec = '%' + str(len(str(width * width))) + 'd'
width += 1
while row < width:
col = 0
while col < width:
print(fieldspec % (row*col,), ' ', end="")
col += 1
print("\n", end='')
row +=1
print_times_table(width)
Now you get:
Please enter the width of the table:4
0 0 0 0 0
0 1 2 3 4
0 2 4 6 8
0 3 6 9 12
0 4 8 12 16
Use a variable-width field specifier:
print('%*d' % (8, row*col), end='')
This automatically adds enough padding to fill the number out to (here) 8 spaces. You can pass that spacing parameter in as an argument, too.
To use a variable width using the new-style formatting syntax:
print('{:{width}}'.format(row*col, width=8), end='')
why not just use your variable in combonation with .ljust? like so:
width = int(13)
def table(width):
row = 1
col = 1
width += 1
spaces = ' '
while row < width:
col = 1
while col < width:
print(str(row*col).rjust(5, ' '), end="")
col += 1
print("\n", end='')
row +=1
table(width)
it works the exact same(you can use whatever numbers you wish, doesn't matter) and just adjust accordingly depending on how large your numbers are getting...
I have the following file I'm trying to manipulate.
1 2 -3 5 10 8.2
5 8 5 4 0 6
4 3 2 3 -2 15
-3 4 0 2 4 2.33
2 1 1 1 2.5 0
0 2 6 0 8 5
The file just contains numbers.
I'm trying to write a program to subtract the rows from each other and print the results to a file. My program is below and, dtest.txt is the name of the input file. The name of the program is make_distance.py.
from math import *
posnfile = open("dtest.txt","r")
posn = posnfile.readlines()
posnfile.close()
for i in range (len(posn)-1):
for j in range (0,1):
if (j == 0):
Xp = float(posn[i].split()[0])
Yp = float(posn[i].split()[1])
Zp = float(posn[i].split()[2])
Xc = float(posn[i+1].split()[0])
Yc = float(posn[i+1].split()[1])
Zc = float(posn[i+1].split()[2])
else:
Xp = float(posn[i].split()[3*j+1])
Yp = float(posn[i].split()[3*j+2])
Zp = float(posn[i].split()[3*j+3])
Xc = float(posn[i+1].split()[3*j+1])
Yc = float(posn[i+1].split()[3*j+2])
Zc = float(posn[i+1].split()[3*j+3])
Px = fabs(Xc-Xp)
Py = fabs(Yc-Yp)
Pz = fabs(Zc-Zp)
print Px,Py,Pz
The program is calculating the values correctly but, when I try to call the program to write the output file,
mpipython make_distance.py > distance.dat
The output file (distance.dat) only contains 3 columns when it should contain 6. How do I tell the program to shift what columns to print to for each step j=0,1,....
For j = 0, the program should output to the first 3 columns, for j = 1 the program should output to the second 3 columns (3,4,5) and so on and so forth.
Finally the len function gives the number of rows in the input file but, what function gives the number of columns in the file?
Thanks.
Append a , to the end of your print statement and it will not print a newline, and then when you exit the for loop add an additional print to move to the next row:
for j in range (0,1):
...
print Px,Py,Pz,
print
Assuming all rows have the same number of columns, you can get the number of columns by using len(row.split()).
Also, you can definitely shorten your code quite a bit, I'm not sure what the purpose of j is, but the following should be equivalent to what you're doing now:
for j in range (0,1):
Xp, Yp, Zp = map(float, posn[i].split()[3*j:3*j+3])
Xc, Yc, Zc = map(float, posn[i+1].split()[3*j:3*j+3])
...
You don't need to:
use numpy
read the whole file in at once
know how many columns
use awkward comma at end of print statement
use list subscripting
use math.fabs()
explicitly close your file
Try this (untested):
with open("dtest.txt", "r") as posnfile:
previous = None
for line in posnfile:
current = [float(x) for x in line.split()]
if previous:
delta = [abs(c - p) for c, p in zip(current, previous)]
print ' '.join(str(d) for d in delta)
previous = current
just in case your dtest.txt grows larger and you don't want to redirect your output but rather write to distance.dat, especially, if you want to use numpy. Thank #John for pointing out my mistake in the old code ;-)
import numpy as np
pos = np.genfromtxt("dtest.txt")
dis = np.array([np.abs(pos[j+1] - pos[j]) for j in xrange(len(pos)-1)])
np.savetxt("distance.dat",dis)