def sequence(n):
while n != 1:
print n,
if n%2 == 0: # n is even
n = n/2
else: # n is odd
n = n*3+1
sequence(6)
6 3 10 5 16 8 4 2
Why the output doesn't include 1 here?Many many thanx!
try using <= or >= for instance, while n >= 1. That should do what you need :)
You have the while loop set on breaking if n == 1. Try possibly doing n > 0 or n >= 1.
while n != 1:
print n
Once n gets a value of 1, it won't enter the loop, thus not allowing 'n' to be printed.
Related
Here is the link to the question. Essentially, it asks to find the kth number having digit sum as 10. I have tried multiple solutions and also looked upon solutions online. Specifically this one (also shared below). The one with constant time talks about outliers in Arithmetic Progression and uses it to find the nth number having sum as 10. Obviously, the code is incorrect as it fails for test cases when k=1000 etc.
#include <bits/stdc++.h>
using namespace std;
int findNth(int n)
{
int nthElement = 19 + (n - 1) * 9;
int outliersCount = (int)log10(nthElement) - 1;
// find the nth perfect number
nthElement += 9 * outliersCount;
return nthElement;
}
int main()
{
cout << findNth(5) << endl;
return 0;
}
Eventually, I ended up writing combination of Arithmetic Progression + brute force as below
#include <bits/stdc++.h>
using namespace std;
#define ll unsigned long long
int main() {
int n;
cin >> n;
int count = 0;
ll i = 19;
for (; ; i += 9) {
int curr = i;
int localSum = 0;
while (curr) {
localSum += curr%10;
curr /= 10;
}
if (localSum == 10) {
count += 1;
}
if (count == n) {
break;
}
}
cout << i << endl;
return 0;
}
I am wondering, if there is no constant time or better algorithm that does not require me to calculate the sum, but my algorithm always hops in a way that I have number whose digit sum is 10?
Here is a Python solution that you can translate into C++.
cached_count_ds_l = {}
def count_digit_sum_length (s, l):
k = (s, l)
if k not in cached_count_ds_l:
if l < 2:
if s == 0:
return 1
elif l == 1 and s < 10:
return 1
else:
return 0
else:
ans = 0
for i in range(min(10, s+1)):
ans += count_digit_sum_length(s-i, l-1)
cached_count_ds_l[k] = ans
return cached_count_ds_l[k]
def nth_of_sum (s, n):
l = 0
while count_digit_sum_length(s, l) < n:
l += 1
digits = []
while 0 < l:
for i in range(10):
if count_digit_sum_length(s-i, l-1) < n:
n -= count_digit_sum_length(s-i, l-1)
else:
digits.append(str(i))
s -= i
l -= 1
break
return int("".join(digits))
print(nth_of_sum(10, 1000))
The idea is to use dynamic programming to find how many numbers there are of a given maximum length with a given digit sum. And then to use that to cross off whole blocks of numbers on the way to finding the right one.
The main logic goes like this:
0 numbers of length 0 sum to 10
- need longer
0 numbers of length 1 sum to 10
- need longer
9 numbers of length 2 sum to 10
- need longer
63 numbers of length 3 sum to 10
- need longer
282 numbers of length 4 sum to 10
- need longer
996 numbers of length 5 sum to 10
- need longer
2997 numbers of length 6 sum to 10
- answer has length 6
Looking for 1000th number of length 6 that sums to 10
- 996 with a leading 0 sum to 10
- Need the 4th past 99999
- 715 with a leading 1 sum to 10
- Have a leading 1
Looking for 4th number of length 5 that sums to 9
- 495 with a leading 0 sum to 9
- Have a leading 10
Looking for 4th number of length 4 that sums to 9
- 220 with a leading 0 sum to 9
- Have a leading 100
Looking for 4th number of length 3 that sums to 9
- 55 with a leading 0 sum to 9
- Have a leading 1000
Looking for 4th number of length 2 that sums to 9
- 1 with a leading 0 sum to 9
- Need the 3rd past 9
- 1 with a leading 1 sum to 9
- Need the 2nd past 19
- 1 with a leading 2 sum to 9
- Need the 1st past 29
- 1 with a leading 3 sum to 9
- Have a leading 10003
Looking for 1st number of length 1 that sums to 6
- 0 with a leading 0 sum to 6
- Need the 1st past 0
- 0 with a leading 1 sum to 6
- Need the 1st past 1
- 0 with a leading 2 sum to 6
- Need the 1st past 2
- 0 with a leading 3 sum to 6
- Need the 1st past 3
- 0 with a leading 4 sum to 6
- Need the 1st past 4
- 0 with a leading 5 sum to 6
- Need the 1st past 5
- 1 with a leading 6 sum to 6
- Have a leading 100036
And it finishes in a fraction of a second.
Incidentally the million'th is 20111220000010, the billionth is 10111000000002000000010000002100, and the trillionth is 10000000100000100000100000000000001000000000000100000000010110001000.
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
Just starting with Python and doing some challenges, this one on Collatz numbers.
I am stuck at the start however, where the range that I am passing to the collatz method is not iterating over the given range.
What am I missing here?
def collatz(number):
for i in number:
if i % 2:
return i // 2
else:
return 3 * (i + 1)
try:
print(collatz(range(0,10)))
except ZeroDivisionError:
print("Zero Division")
except TypeError:
print "Type Error"
for i in range(0,10):
print(collatz(i))
Short answer: the 'return' statement is causing your function to exit on the first iteration, thus preventing further iterations from occurring.
Long answer:
Here's a breakdown of how your code is running.
collatz(number) receives a list [0,1,...10]
A loop is created for [0,1,...10] starting from 0
1st, start with i = 0
0 % 2 evals to 0, which is False, which means we skip to the else statement
We return 3 * (0 + 1), which evaluates to 3.
We finish the call to collatz(number)
The correct approach is answered in some other folks' posts.
algorithm
If the number is even, divide it by two.
If the number is odd, triple it and add one.
code
def f(n):
if n % 2 == 0:
return n // 2
else:
return 3 * n + 1
def a(i, n):
if i == 0:
return n
else:
return f(a(i - 1, n))
def collatz(n):
i = 0
c = 0
out = []
while c != 1:
c = a(i, n)
out.append(c)
i += 1
return out
print(collatz(6))
output
[6, 3, 10, 5, 16, 8, 4, 2, 1]
consider that
0 -- is the first
1 -- is the second
2 -- is the third
.....
9 -- is the 10th
11 -- is the 11th
what is an efficient algorithm to find the nth palindromic number?
I'm assuming that 0110 is not a palindrome, as it is 110.
I could spend a lot of words on describing, but this table should be enough:
#Digits #Pal. Notes
0 1 "0" only
1 9 x with x = 1..9
2 9 xx with x = 1..9
3 90 xyx with xy = 10..99 (in other words: x = 1..9, y = 0..9)
4 90 xyyx with xy = 10..99
5 900 xyzyx with xyz = 100..999
6 900 and so on...
The (nonzero) palindromes with even number of digits start at p(11) = 11, p(110) = 1001, p(1100) = 100'001,.... They are constructed by taking the index n - 10^L, where L=floor(log10(n)), and append the reversal of this number: p(1101) = 101|101, p(1102) = 102|201, ..., p(1999) = 999|999, etc. This case must be considered for indices n >= 1.1*10^L but n < 2*10^L.
When n >= 2*10^L, we get the palindromes with odd number of digits, which start with p(2) = 1, p(20) = 101, p(200) = 10001 etc., and can be constructed the same way, using again n - 10^L with L=floor(log10(n)), and appending the reversal of that number, now without its last digit: p(21) = 11|1, p(22) = 12|1, ..., p(99) = 89|8, ....
When n < 1.1*10^L, subtract 1 from L to be in the correct setting with n >= 2*10^L for the case of an odd number of digits.
This yields the simple algorithm:
p(n) = { L = logint(n,10);
P = 10^(L - [1 < n < 1.1*10^L]); /* avoid exponent -1 for n=1 */
n -= P;
RETURN( n * 10^L + reverse( n \ 10^[n >= P] ))
}
where [...] is 1 if ... is true, 0 else, and \ is integer division.
(The expression n \ 10^[...] is equivalent to: if ... then n\10 else n.)
(I added the condition n > 1 in the exponent to avoid P = 10^(-1) for n=0. If you use integer types, you don't need this. Another choice it to put max(...,0) as exponent in P, or use if n=1 then return(0) right at the start. Also notice that you don't need L after assigning P, so you could use the same variable for both.)
How is this code working for multiple of 5
bool isMultipleof5(int n)
{
/* If n is a multiple of 5 then we make sure that last
digit of n is 0 */
if ( (n&1) == 1 )
n <<= 1;
float x = n;
x = ( (int)(x*0.1) )*10;
/* If last digit of n is 0 then n will be equal to (int)x */
if ( (int)x == n )
return true;
return false;
}
It first makes n divisable by 2.
Next, it checks if it is divisable by 10 by multiplying with 0.1 and again with 10. The idea that if it is divisable by 10, you will get back to the original, and only then.
So, if the modifies n is divisable by 10 - it is certainly divisable by 5 as well, and since modified n is always divisable by 2, if it is divisable by 5 it will be divisable by 10, and the algorithm works.
NOTE: This is very unsuggested and especially might break with large values due to floating point precision issues. using the % operator should be prefered: return (n % 5) == 0
This is how the code works with some examples.
if ( (n&1) == 1 ) //Checks if the number is odd
n <<= 1; //Multiplies the number by 2 if odd
x = ( (int)(x * 0.1) //Divides the number 10 then truncates any decimal places
* 10 ) //Multiplies it back by 10
if ( (int)x == n ) //If the floating point value equals the (semi) original value its divisible by 5
return true;
return false; //Other wise false
Example:
15 & 1 == 1 //15 is odd
15 <<= 1; //n is now 30
30 / 10 = 3;
3 * 10 = 30; //x is now 30
30 == 30 //15 is a multiple of 5
17 & 1 == 1 //17 is odd
17 <<= 1; //n is now 34
34 / 10 = 3.4;
((int)3.4 = 3) * 10 = 30; //x is now 30
30 != 34 //17 is not a multiple of 5.
As others said though just simply use the mod operator %.
This is how it works:
Double the number. Now anything ending in 5 will be divisible 10 (and also divisible by 5). n <<= 1; (the check for oddness is unnecessary (n&1) == 1)
Divide it by 10, and cast away the fractional part. (int)(x*0.1)
Multiply it by 10, so now we have the same number as in step 1 only if the number in step 1 was already divisible by 10.
The use of floating point to divide by 10 makes this algorithm dangerous and probably incorrect for large values.
Try this
bool isMultipleof5(int n)
{
return (n%5) == 0;
}
A simpler way would be
bool isMultipleof5(int n)
{
return 0 == ( n % 5 ) ;
}
#define IS_MULTIPLE_OF_5(n) (((n)%5) ? 0 : 1)
I'd agree that (n % 5) == 0 would be an ideal solution, but that wasn't really the question.
This code works because it first checks if the input is odd. If it is, it multiplies by two. Since all odd multiples of 5 end with a 5, multiplying by 2 gives a number that ends with 0.
Then it checks if the last digit is 0. This can only happen if it started as a 0 (i.e. was even, we didn't change it) or if it was odd and ended in a 5 (we multiplied by 2). So, if it ends in 0 then the input must have been divisible by 5.
I'd add that this is also an awkward way to check the value of the last digit. I'd suggest n % 10 == 0 instead, but like others mentioned... you could have just used n % 5 == 0 in the first place ;).