4 one digit numbers and try to get all the possible combos
4511
like 4 + 5 + 1 x 1
Code to get first, 2nd 3rd and 4th numbers
numbers = input("Input 4 numbers separated with , : ")
numlist = numbers.split(",")
print (numlist)
No1 = int(numlist.pop(0))
No2 = int(numlist[0])
No3 = int(numlist[1])
No4 = int(numlist[2])
An exhaustive search:
from random import randrange
from itertools import permutations
def solve(digits):
for xs in permutations(digits):
for ops in permutations('+-*/', 3):
equation = reduce(lambda r, (x, op): '{0} {1} {2}'.format(r, op, float(x)), zip(xs[1:], ops), xs[0])
try:
if eval(equation) == 10:
yield equation.replace('.0','')
except ZeroDivisionError:
pass
digits = [randrange(0,10,1) for x in range(4)]
solutions = list(solve(digits))
if solutions:
print '\n'.join(solutions)
else:
print 'No solution for {0}'.format(digits)
Example output:
2 * 5 / 1 + 0
2 * 5 / 1 - 0
2 * 5 + 0 / 1
2 * 5 - 0 / 1
2 / 1 * 5 + 0
2 / 1 * 5 - 0
5 * 2 / 1 + 0
5 * 2 / 1 - 0
5 * 2 + 0 / 1
5 * 2 - 0 / 1
5 / 1 * 2 + 0
5 / 1 * 2 - 0
0 + 2 * 5 / 1
0 + 2 / 1 * 5
0 + 5 * 2 / 1
0 + 5 / 1 * 2
0 / 1 + 2 * 5
0 / 1 + 5 * 2
or
No solution for [7, 1, 0, 2]
Note: I like the recursive expression generator referred to in the comment above, but I just don't think recursively straight off.
Use a recursive approach:
numbers = input("Input 4 numbers separated with , : ")
numlist = list(numbers)
def f(lst,carry,result):
x = lst.pop(0)
if lst == []:
return carry+x == result or \
carry-x == result or \
carry*x == result or \
carry/x == result
elif carry==None:
carry = x
return f(lst,carry,result)
else:
return f(lst,carry+x,result) or\
f(lst,carry-x,result) or\
f(lst,carry*x,result) or\
f(lst,carry/x,result)
print(f(numlist, None, 10))
Your I/O is wrong (first two lines), and I don't know if this is something that you have already tried.
Related
I have got an a number: like 5
i need to split it into 3 sets like
2
1 4
2
2 3
1
5
Or the number 8:
2
8 4
2
7 5
4
1 2 3 6
I try to
def partition(n):
if n < 5:
return
s = n * (n + 1) // 2
if s % 3 != 0:
return
s //= 3
lst, result = [i for i in range(1, n + 1)], []
for _ in range(2):
subset, s_current = [], s
while s_current > 0:
idx_max = bisect_right(lst, s_current) - 1
subset.append(lst[idx_max])
s_current -= lst[idx_max]
lst.pop(idx_max)
result.append(subset)
result.append(lst)
return result
If it can't make 3 sets, should return -1
But it doesn't work what i want
please, help
Examples:
m n: 2 3
1 * 1 = 1
1 * 2 = 2
1 * 3 = 3
2 * 1 = 2
2 * 2 = 4
2 * 3 = 6
m n: 4 2
1 * 1 = 1
1 * 2 = 2
2 * 1 = 2
2 * 2 = 4
3 * 1 = 3
3 * 2 = 6
4 * 1 = 4
4 * 2 = 8
I have written this code but i said "list assignment index out of range", how can I fix it? thanks
m, n = input('m n: ').split()
x = []
for i in range(0, int(m)):
for j in range(0, int(n)):
x[j] = int(m[i]) * int(n[j])
print(str(i) + ' * ' + str(j) + ' = ',x[j])
m, n = input('m n: ').split()
for i in range(1, int(m)+1):
for j in range(1, int(n)+1):
print(str(i) + ' * ' + str(j) + ' = ', i * j)
I have a programm which draws a picture from numbers in certain way
n = int(input(" Введіть ваше число "))
m = n * 2 - 1
pp = " "
i = 0
while m != 0:
l = []
while m > n:
while i < n:
i += 1
j = n - i
k = i
while j != 0:
l.append(pp)
j -= 1
while k != 0:
l.append(str(k))
k -= 1
m -= 1
a = " "
print(a.join(l))
l = []
i = 0
OUTPUT:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
But now I get a task to draw this picture
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Is there any hint how to reflect it without overwriting the whole code?
For the output you are expecting, a very simple way to do it is with this code :
n = 5
for i in range(n):
print(' '.join([str(x+1) for x in range(i+1)]))
Output :
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
You can try this. Its simple.
for i in range(1,n+1):
for j in range(1, i+1):
print(j, end="")
print()
So I'm given the following diagram:
And I'm being asked to find the area of each polygon given n. The area is basically the sum of all blue squares since each square has an area of 1. So when n = 1, the area is one. When n = 2, the area is 5. Because of the relationship between each polygon, I know that I could knock this down using set theory.
n Area(n)
1 1
2 A(n-1) + (4 * (n-1)) = 5
3 A(n-1) + (4 * (n-1)) = 13
4 A(n-1) + (4 * (n-1)) = 25
5 A(n-1) + (4 * (n-1)) = 41
However, I didn't have as much luck trying to represent this in code:
def shapeArea(n):
prev_output = 0
output = 0
if n == 1:
output = 1
elif n > 1:
for i in range(n):
prev_output = n-1 + (4 * (n-1))
output = prev_output + (4 * (n-1))
return output
For example: For n = 2, I'm getting an output of 9 instead of 5.
You were close :-)
Here are the small fix-ups:
def shapeArea(n):
output = 1
for i in range(1, n):
output += 4 * i
return output
Running this:
for n in range(1, 6):
print(n, shapeArea(n))
Gives this output:
1 1
2 5
3 13
4 25
5 41
of course with Gauss's theorem the code would look like this:
def ShapeArea(n):
return 2*(n**2) - (2*n)+1
In a recursive solution, it is much simple from your logic.
def shape_area(n):
if n == 1:
return 1
return shape_area(n-1) + 4*(n-1)
You can use Gauss' trick to make a closed-form formula:
2*Area(n) =
1 + 4 + 8 + 12 + ... + 4(n-1)
+
1 + 4(n-1) + 4(n-2) + 4(n-3) + ... + 4
--------------------------------------------------
2 + 4n + 4n + 4n + ... + 4n
= (n-1)4n + 2
= 4n^2 - 4n + 2
So Area(n) = 2n2 - 2n + 1
def shape_area(n):
return sum([1] + [4*i for i in range(1, n)])
The Problem Statement:
Sanjay is addicted to alcohol. Every night he drinks 4 bottles of vodka. He is going to his home. At first, he takes a step forward (which is 5m) but beacuse he is drunk, after his each step in forward direction, his body gets imbalanced and he takes a step backward (which is 3m).
Each step takes 1 min to complete. The distance from the bar to home is n metres. Calculate the time taken by him to reach his home.
Input Format:
single line containing one integer n.
Constraints:
0 <= n < 10^18
Output Format
single integer describing the time taken by him to reach home.
from math import *
n = int(input())
x = 0
m = 0
n = n % 1000000007
n = n % 1000000007
while x < n:
x += 5
m += 1
if x >= n:
break
x -= 3
m += 1
print(m)
But the time limit is exceeding in the last test case i.e. for n = 10^18 like numbers
Sample Input 0
11
Sample Output 0
7
The time taken is simply n/2 * 2
He advances 2 meters each "cycle" 5 forward 3 back
So we see how many "cycles" go into n (n / 2m) this will result
In the number of "cycles" taken to reach his house
Then we simply multiply by the amount of time taken per cycle (2 minutes)
to get the total time taken (t = n/2 * 2).
Try reducing the problem. Let time_taken(dist) be the function that tells us how long it takes to get home. Then the following hold:
time_taken(1) == 1
time_taken(2) == 1
time_taken(3) == 1
time_taken(4) == 1
time_taken(5) == 1
time_taken(6) == 1 * 2 + time_taken(4) (since 5-3 = 2)
== 1 * 2 + 1
time_taken(7) == 1 * 2 + time_taken(5)
== 1 * 2 + 1
time_taken(11) == 1 * 2 + time_taken(9)
== 2 * 2 + time_taken(7)
== 3 * 2 + time_taken(5)
== 3 * 2 + 1
time_taken(26) == 1 * 2 + time_taken(24)
== 2 * 2 + time_taken(22)
== ...
== 11 * 2 + time_taken(4)
== 11 * 2 + 1
if n > 5:
time_taken(n) == 1 * 2 + time_taken(n - 2)
== 2 * 2 + time_taken(n - 4)
== ...
== (formula here) * 2 + time_taken(4 or 5)