problem:
Enter 10 integers and create a program that outputs the number of multiples of 3 and the number of multiples of 5 respectively.
ex)input:
10 15 36 99 100 19 46 88 87 13
ex)print:
Multiples of 3 : 4
Multiples of 5 : 3
my code:
t=0
f=0
a=list(input().split())
for i in range(11):
if int(a[i])%3==0: #index
t+=1
else:
int(a[i]) % 5==0
f += 1
print('Multiples of 3 :',t)
print('Multiples of 5 :',f)
error:
if int(a[i])%3==0:
IndexError: list index out of range
enter code here
what's wrong?
I don't know
plz help me I an so angry
Try this python code snippet
t=0
f=0
n = int(input("Enter number of elements : "))
for i in range(0, n):
ele = int(input())
if (ele % 3) == 0:
t = t + 1
elif (ele % 5) == 0:
f = f + 1
else:
print("element not counted")
print('Multiples of 3 :',t)
print('Multiples of 5 :',f)
I tried this code for list of 3 values of 3, 5 and 10 respectively
Related
i want have comparison among input numbers .general shape of input numbers is like this :
n
x1 x2
x3 x4
x5 x6
x7 x8
n indicates how many input line we have and on each line we have 2 number with space ,just 2 number,
the first part of line is price and the second is quality
for example like this :
4
1 5
7 9
5 6
20 30
in line two '1' is price and '5' is quality
if i can find more quality with lower price i print 'yes' if not print 'no'
i write this but i can not change list to dictionary
x = int(input())
count =0
y= []
while count!=x:
y.append(input())
count+=1
z= []
for i in y:
z.append(i.split())
def dictionary(x):
d={x[0]:x[1]}
return d
for h in z:
a=dictionary(h)
for example i want
2
1 10
7 3
'yes'
or another example :
4
1 5
7 9
5 6
20 30
'no'
i hope someone answer this
n = int(input())
q,p = map(int, input().split())
for _ in range(n-1):
q1,p1 = map(int, input().split())
if q1 > q and p1 < p:
print("yes")
break
else:
print("no")
This works. The code is really simple.
I am new to coding and I am finding this site really helpful. So I have been trying to solve this problem and I am getting erroneous results, so I would be really grateful if you could help me out here.
The Problem: Find the sum of all the multiples of 3 or 5 below 1000. (For example, if we list all the positive integers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9, which sum is 23.)
My code:
count = 0
count1 = 0
for x in range(1000):
if x % 5 == 0:
count = count + x
if x % 3 == 0:
count1 = count1 + x
print count1 + count
What am I doing wrong?
Thanks in advance!
You want an elif in your code so you don't count the same x twice but a simpler way is to use an or with a single count variable:
count = 0
for x in range(1000):
if x % 5 == 0 or x % 3 == 0:
count += x
Which can be done using sum:
print(sum(x for x in range(3, 1000) if not x % 5 or not x % 3))
For completeness, a working version using your own code:
count = 0
count1 = 0
for x in range(1000):
if x % 5 == 0:
count += x
elif x % 3 == 0:
count1 += x
print count1 + count
ifs are always evaluated; so, for instance, when x is 15 it is evenly divisible by 5 and 3 so you count 15 twice, an elif is only evaluated if the previous if/elif evaluates to False so using elif only one occurrence of x will be added to the total.
Below 10 there is no number being multiple of both 5 and 3. But below 1000 there are several numbers divided exactly by 3 and 5 also (15, 45 ...).
So you need:
count=0
for x in range(1000):
if x % 5 == 0 or x % 3 == 0:
count=count + x
print count
i am trying to loop though a series of numbers and assign a constant index to them with restarting a loop at a certain number.
input numbers
0
3
6
9
12
15
18
21
0
3
6
9
12
15
18
21
......
the expected output should be
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
my code looks like this. I works but never stops.
How could i iterate over the input data?
count = 1
for line in in_file:
should_restart = True
while should_restart:
should_restart = False
for i in xrange(0,22,3):
print count
if i == 21:
should_restart = True
count+=1
break
Thanks!
Supposing you have an in_file like this (that's just an example yours is different but I wanted to do it with your given input):
in_file = "0 3 6 9 12 15 18 21 0 3 6 9 12 15 18 21"
in_file = in_file.split(" ")
you setup an end-condition for your counter and a starting count:
end_condition = 21
count = 1
then you iterate through your list:
for line in in_file:
line = int(line) # This is only needed if you have string-data
within the loop print the current count:
print(count, end=' ')
and later increment the counter if you reach the final value:
if line == end_condition:
count += 1
Problem originally is in this link. I wrote a Python code but I got 64 points (total points is 100) and this indicates that my code has some missing points. I passed 11 of 16 test cases but 5 test cases have problematic for me. Could you say where my code has some missing points and how can I fix it?
import math
m = int(raw_input())
liste = []
y_liste = []
md = 0
ad = 0
sum = 0
sum2 = 0
for k in range(m):
temp = str(raw_input())
liste.append(temp)
liste[k] = liste[k].split(" ")
liste[k] = [int(i) for i in liste[k]]
for k in range(m):
md += liste[k][k]
ad += liste[k][m-k-1]
if md == ad:
print 0
else:
for k in range(m):
for l in range(m):
sum2 += liste[l][k]
sum += liste[k][l]
if sum2 != md and -(k+1) is not y_liste:
y_liste.append(-(k+1))
if sum != md and (k+1) is not y_liste:
y_liste.append(k+1)
sum2 = 0
sum = 0
if md != ad:
y_liste.append(0)
print len(y_liste)
y_liste.sort()
for i in y_liste:
print i
Problem Statement
Magic Square
Johnny designed a magic square (square of numbers with the same sum for all rows, columns and diagonals i.e. both the main diagonal - meaning the diagonal that leads from the top-left corner towards bottom-right corner - and the antidiagonal - meaning the diagonal that leads from top-right corner towards bottom-left corner). Write a program to test it.
Task
Write a program that will check if the given square is magic (i.e. has the same sum for all rows, columns and diagonals).
Input
First line: N , the size of the square (1 <= N <= 600).
Next N lines: The square, N space separated integers pre line, representing the entries per each row of the square.
Output
First line: M , the number of lines that do not sum up to the sum of the main diagonal (i.e. the one that contains the first element of the square). If the Square is magic, the program should output 0.
Next M lines: A sorted (in incremental order ) list of the lines that do not sum up to the sum of the main diagonal. The rows are numbered 1,2,…,N; the columns are numbered -1,-2,…,-N; and the antidiagonal is numbered zero.
Note: There is a newline character at the end of the last line of the output.
Sample Input 1
3
8 1 6
3 5 7
4 9 2
Sample Output 1
0
Sample Input 2
4
16 3 2 13
5 10 11 8
6 9 7 12
4 15 14 1
Sample Output 2
3
-2
-1
0
Explanation of Sample Output 2
The input square looks as follows: http://i.stack.imgur.com/JyMgc.png
(Sorry for link but I cannot add image due to reputation)
The square has 4 rows (labeled from 1 to 4 in orange) and 4 columns (labeled from -1 to -4 in green) as depicted in the image above. The main diagonal and antidiagonal of the square are highlighted in red and blue respectively.
The main diagonal has sum = 16 + 10 + 7 +1 = 34.
The antidiagonal has sum = 13 + 11 + 9 + 4 = 37. This is different to the sum of the main diagonal so value 0 corresponding to the antidiagonal should be reported.
Row 1 has sum = 16 + 3 + 2 + 13 = 34.
Row 2 has sum = 5 + 10 + 11 + 8 = 34.
Row 3 has sum = 6 + 9 + 7 + 12 = 34.
Row 4 has sum = 4 + 15 + 14 + 1 = 34.
Column -1 has sum = 16 + 5 + 6 + 4 = 31. This is different to the sum of the main diagonal so value -1 should be reported.
Column -2 has sum = 3 + 10 + 9 + 15 = 37. This is different to the sum of the main diagonal so value -2 should be reported.
Column -3 has sum = 2 + 11 + 7 + 14 = 34.
Column -4 has sum = 13 + 8 + 12 + 1 = 34.
Based on the above, there are 3 lines that do not sum up to the sum of the elements of the main diagonal. Since they should be sorted in incremental order, the output should be:
3
-2
-1
0
Your explanation doesn't discuss this clause which is a potential source of error:
if md == ad:
print 0
else:
It says that if the main diagonal and antidiagonal add up to the same value, print just a 0 (no bad lines) indicating the magic square is valid (distinct from reporting a 0 in the list of bad lines). Consider this valid magic square:
9 6 3 16
4 15 10 5
14 1 8 11
7 12 13 2
If I swap 13 and 11, the diagonals still equal each other but the square is invalid. So the above code doesn't appear to be correct. In the else clause for the above if statement, you test:
if md != ad:
y_liste.append(0)
a fact you already know to be true from the previous/outer test so your code seems to be out of agreement with itself.
i'm writing a program that takes a triangular number, and finds a collection of up to three other triangular numbers whose sum is equal to the first one. All the code i have currently is below. If you run the program and type 10 for example for the input it will print out a list of the first 10 triangular numbers along with combinations of smaller triangular numbers whose sum is the 10th one.
what i'm trying to do, is to prevent the same answer from coming up in a different order if you use the 10 example again, the output is:
6 + 21 + 28 = 55
6 + 28 + 21 = 55
10 + 45 = 55
21 + 6 + 28 = 55
21 + 28 + 6 = 55
28 + 6 + 21 = 55
28 + 21 + 6 = 55
45 + 10 = 55
but obviously 6 + 21 + 28 = 55 is the same as 6 + 28 + 21 = 55, also 10 + 45 = 55 and 45 + 10 = 55 and so on... they're just in a different order. So does anyone know a simple way i could check to make sure the group of values isn't repeating in any order? If you still have questions about what i'm trying to do, just post a comment and ill try and explain this better.
Code:
def f(x):
newX = 0
for i in range(0,x):
newX = newX + i
return newX
def main():
lst = []
inpt = input("What number triangular number would you like to test: ")
for i in range(2,inpt+2):
x = f(i)
lst.append(x)
find(x, lst)
#x is the triangular number, lst is list of all of them up to it
def find(x, lst):
a = 0
b = 0
c = 0
length = len(lst)
print(lst)
last = lst[length-1]
for j in lst:
a = j
for k in lst:
b = k
if(a+b == x):
if(a != b):
print(str(a) + " + " + str(b) + " = " + str(x))
for i in lst:
c = i
if(a+b+c == x):
if(a != b and a != c and b != c):
print(str(a) + " + " + str(b) + " + " + str(c) + " = " + str(x))
print
main()
a hint
>>> [1, 2, 3] == [1, 3, 2]
False
>>> sorted([1, 3, 2])
[1, 2, 3]
>>> [1, 2, 3] == sorted([1, 3, 2])
True
Canonicalization is the process whereby there is one-and-only-one ordering for a set of elements. This makes comparisons between different orderings of the the same elements become equal.