There were a lot of people in our contest today, we numbered each of them separately, ie the first person 2, the second person 3, etc. In this order, 5, 7, 11, 13,… you understood the sequence.
Your task is to find the number of the N-person.
For example: User enters the value of N = 3.
Desired output: 5
The length of the value entered by the user is 10^9
The code I wrote. this number cannot be deduced
import math
def is_prime(n, div=2):
if div> int(math.sqrt(n)): return True
if n% div == 0:
return False
else:
div+=1
return is_prime(n,div)
a =int(input())
if a == 1:
print('2')
else:
primelist=[]
i=1;
while len(primelist)<999:
if is_prime(i):
primelist.insert(0,i)
i+=1
else: i+=1
print(primelist[-a-1])
Kindly advise.
Related
Im currently using micropython and it does not have the .zfill method.
What Im trying to get is to get the YYMMDDhhmmss of the UTC.
The time that it gives me for example is
t = (2019, 10, 11, 3, 40, 8, 686538, None)
I'm able to access the ones that I need by using t[:6]. Now the problem is with the single digit numbers, the 3 and 8. I was able to get it to show 1910113408, but I need to get 19101034008 I would need to get the zeroes before those 2. I used
t = "".join(map(str,t))
t = t[2:]
So my idea was to iterate over t and then check if the number is less than 10. If it is. I will add zeroes in front of it, replacing the number . And this is what I came up with.
t = (2019, 1, 1, 2, 40, 0)
t = list(t)
for i in t:
if t[i] < 10:
t[i] = 0+t[i]
t[i] = t[i]
print(t)
However, this gives me IndexError: list index out of range
Please help, I'm pretty new to coding/python.
When you use
for i in t:
i is not index, each item.
>>> for i in t:
... print(i)
...
2019
10
11
3
40
8
686538
None
If you want to use index, do like following:
>>> for i, v in enumerate(t):
... print("{} is {}".format(i,v))
...
0 is 2019
1 is 10
2 is 11
3 is 3
4 is 40
5 is 8
6 is 686538
7 is None
another way to create '191011034008'
>>> t = (2019, 10, 11, 3, 40, 8, 686538, None)
>>> "".join(map(lambda x: "%02d" % x, t[:6]))
'20191011034008'
>>> "".join(map(lambda x: "%02d" % x, t[:6]))[2:]
'191011034008'
note that:
%02d add leading zero when argument is lower than 10 otherwise (greater or equal 10) use itself. So year is still 4digit string.
This lambda does not expect that argument is None.
I tested this code at https://micropython.org/unicorn/
edited :
str.format method version:
"".join(map(lambda x: "{:02d}".format(x), t[:6]))[2:]
or
"".join(map(lambda x: "{0:02d}".format(x), t[:6]))[2:]
second example's 0 is parameter index.
You can use parameter index if you want to specify it (ex: position mismatch between format-string and params, want to write same parameter multiple times...and so on) .
>>> print("arg 0: {0}, arg 2: {2}, arg 1: {1}, arg 0 again: {0}".format(1, 11, 111))
arg 0: 1, arg 2: 111, arg 1: 11, arg 0 again: 1
I'd recommend you to use Python's string formatting syntax.
>> t = (2019, 10, 11, 3, 40, 8, 686538, None)
>> r = ("%d%02d%02d%02d%02d%02d" % t[:-2])[2:]
>> print(r)
191011034008
Let's see what's going on here:
%d means "display a number"
%2d means "display a number, at least 2 digits"
%02d means "display a number, at least 2 digits, pad with zeroes"
so we're feeding all the relevant numbers, padding them as needed, and cut the "20" out of "2019".
I have been assigned with a task to print prime numbers from a range 2...100. I've managed to get most of the prime numbers but can't figure out how to get rid of 9 and 15, basically multiples of 3 and 5. Please give me your suggestion on how can I fix this.
for n in 2...20 {
if n % 2 == 0 && n < 3{
print(n)
} else if n % 2 == 1 {
print(n)
} else if n % 3 == 0 && n > 6 {
}
}
This what it prints so far:
2
3
5
7
9
11
13
15
17
19
One of effective algorithms to find prime numbers is Sieve of Eratosthenes. It is based on idea that you have sorted array of all numbers in given range and you go from the beginning and you remove all numbers after current number divisible by this number which is prime number. You repeat this until you check last element in the array.
There is my algorithm which should do what I described above:
func primes(upTo rangeEndNumber: Int) -> [Int] {
let firstPrime = 2
guard rangeEndNumber >= firstPrime else {
fatalError("End of range has to be greater than or equal to \(firstPrime)!")
}
var numbers = Array(firstPrime...rangeEndNumber)
// Index of current prime in numbers array, at the beginning it is 0 so number is 2
var currentPrimeIndex = 0
// Check if there is any number left which could be prime
while currentPrimeIndex < numbers.count {
// Number at currentPrimeIndex is next prime
let currentPrime = numbers[currentPrimeIndex]
// Create array with numbers after current prime and remove all that are divisible by this prime
var numbersAfterPrime = numbers.suffix(from: currentPrimeIndex + 1)
numbersAfterPrime.removeAll(where: { $0 % currentPrime == 0 })
// Set numbers as current numbers up to current prime + numbers after prime without numbers divisible by current prime
numbers = numbers.prefix(currentPrimeIndex + 1) + Array(numbersAfterPrime)
// Increase index for current prime
currentPrimeIndex += 1
}
return numbers
}
print(primes(upTo: 100)) // [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
print(primes(upTo: 2)) // [2]
print(primes(upTo: 1)) // Fatal error: End of range has to be greater than or equal to 2!
what is the Prime num : Prime numbers are the positive integers having only two factors, 1 and the integer itself,
//Funtion Call
findPrimeNumberlist(fromNumber: 1, toNumber: 100)
//You can print any range Prime number using this fucntion.
func findPrimeNumberlist(fromNumber:Int, toNumber: Int)
{
for i in fromNumber...toNumber
{
var isPrime = true
if i <= 1 { // number must be positive integer
isPrime = false
}
else if i <= 3 {
isPrime = true
}
else {
for j in 2...i/2 // here i am using loop from 2 to i/2 because it will reduces the iteration.
{
if i%j == 0 { // number must have only 1 factor except 1. so use break: no need to check further
isPrime = false
break
}
}
}
if isPrime {
print(i)
}
}
}
func getPrimeNumbers(rangeOfNum: Int) -> [Int]{
var numArr = [Int]()
var primeNumArr = [Int]()
var currentNum = 0
for i in 0...rangeOfNum{
currentNum = i
var counter = 0
if currentNum > 1{
numArr.append(currentNum)
for j in numArr{
if currentNum % j == 0{
counter += 1
}
}
if counter == 1{
primeNumArr.append(currentNum)
}
}
}
print(primeNumArr)
print(primeNumArr.count)
return primeNumArr
}
Then just call the function with the max limit using this
getPrimeNumbers(rangeOfNum: 100)
What is happening in above code:
The numArr is created to keep track of what numbers have been used
Any number that is prime number is added/appended to primeNumArr
Current number shows the number that is being used at the moment
We start from 0 ... upto our range where we need prime numbers upto (with little modification it can be changed if the range starts from other number beside 0)
Remember, for a number to be Prime it should have 2 divisor means should be only completely divisible by 2 numbers. First is 1 and second is itself. (Completely divisible means having remainder 0)
The counter variable is used to keep count of how many numbers divide the current number being worked on.
Since 1 is only has 1 Divisor itself hence its not a Prime number so we start from number > 1.
First as soon as we get in, we add the current number being checked into the number array to keep track of numbers being used
We run for loop to on number array and check if the Current Number (which in our case will always be New and Greater then previous ones) when divided by numbers in numArr leaves a remainder of 0.
If Remainder is 0, we add 1 to the counter.
Since we are already ignoring 1, the max number of counter for a prime number should be 1 which means only divisible by itself (only because we are ignoring it being divisible by 1)
Hence if counter is equal to 1, it confirms that the number is prime and we add it to the primeNumArr
And that's it. This will give you all prime numbers within your range.
PS: This code is written on current version of swift
Optimised with less number of loops
Considered below conditions
Even Number can not be prime number expect 2 so started top loop form 3 adding 2
Any prime number can not multiplier of even number expect 2 so started inner loop form 3 adding 2
Maximum multiplier of any number if half that number
var primeNumbers:[Int] = [2]
for index in stride(from: 3, to: 100, by: 2) {
var count = 0
for indexJ in stride(from: 3, to: index/2, by: 2) {
if index % indexJ == 0 {
count += 1
}
if count == 1 {
break
}
}
if count == 0 {
primeNumbers.append(index)
}
}
print("primeNumbers ===", primeNumbers)
I finally figured it out lol, It might be not pretty but it works haha, Thanks for everyone's answer. I'll post what I came up with if maybe it will help anyone else.
for n in 2...100 {
if n % 2 == 0 && n < 3{
print(n)
} else if n % 3 == 0 && n > 6 {
} else if n % 5 == 0 && n > 5 {
} else if n % 7 == 0 && n > 7{
} else if n % 2 == 1 {
print(n)
}
}
Is there a more elegant way to write this? I dont want to use a for loop
if i==1 || i==6 || i==11 || i==16 || i==21 || i==26 || i==31 || i==36
function
end
basically i is the index of a vector, after each fifth element of this vector (starting with the first ) a specific function is applied. i starts with 1 and it increments after the if statement and just if it equals these values of the if condition the if statement is valid
EDITTED FOR MATLAB CODE OF MODULO
output = mod(input, 5); //this will output 1 if it is 1, 5, 11, 16
//input is your 1, 5, 11, 16 etc
//output is the result of modulo. else it is 0, 2, 3, 4
if(output == 1)
[previous answer]
i forgot how to write this in matlab but with your values, put it this way.
if(number%5==1)
any input 1 or 6 or 11 or any else that can add 5 to it, you'll end up in 1. else it will return false
I've got a list of some integers, e.g. [1, 2, 3, 4, 5, 10]
And I've another integer (N). For example, N = 19.
I want to check if my integer can be represented as a sum of any amount of numbers in my list:
19 = 10 + 5 + 4
or
19 = 10 + 4 + 3 + 2
Every number from the list can be used only once. N can raise up to 2 thousand or more. Size of the list can reach 200 integers.
Is there a good way to solve this problem?
4 years and a half later, this question is answered by Jonathan.
I want to post two implementations (bruteforce and Jonathan's) in Python and their performance comparison.
def check_sum_bruteforce(numbers, n):
# This bruteforce approach can be improved (for some cases) by
# returning True as soon as the needed sum is found;
sums = []
for number in numbers:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
return n in sums
def check_sum_optimized(numbers, n):
sums1, sums2 = [], []
numbers1 = numbers[:len(numbers) // 2]
numbers2 = numbers[len(numbers) // 2:]
for sums, numbers_ in ((sums1, numbers1), (sums2, numbers2)):
for number in numbers_:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
for sum_ in sums1:
if n - sum_ in sums2:
return True
return False
assert check_sum_bruteforce([1, 2, 3, 4, 5, 10], 19)
assert check_sum_optimized([1, 2, 3, 4, 5, 10], 19)
import timeit
print(
"Bruteforce approach (10000 times):",
timeit.timeit(
'check_sum_bruteforce([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
print(
"Optimized approach by Jonathan (10000 times):",
timeit.timeit(
'check_sum_optimized([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
Output (the float numbers are seconds):
Bruteforce approach (10000 times): 1.830944365834205
Optimized approach by Jonathan (10000 times): 0.34162875449254027
The brute force approach requires generating 2^(array_size)-1 subsets to be summed and compared against target N.
The run time can be dramatically improved by simply splitting the problem in two. Store, in sets, all of the possible sums for one half of the array and the other half separately. It can now be determined by checking for every number n in one set if the complementN-n exists in the other set.
This optimization brings the complexity down to approximately: 2^(array_size/2)-1+2^(array_size/2)-1=2^(array_size/2 + 1)-2
Half of the original.
Here is a c++ implementation using this idea.
#include <bits/stdc++.h>
using namespace std;
bool sum_search(vector<int> myarray, int N) {
//values for splitting the array in two
int right=myarray.size()-1,middle=(myarray.size()-1)/2;
set<int> all_possible_sums1,all_possible_sums2;
//iterate over the first half of the array
for(int i=0;i<middle;i++) {
//buffer set that will hold new possible sums
set<int> buffer_set;
//every value currently in the set is used to make new possible sums
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums1.insert(myarray[i]);
//transfer buffer into the main set
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums1.insert(*set_iterator);
}
//iterator over the second half of the array
for(int i=middle;i<right+1;i++) {
set<int> buffer_set;
for(set<int>::iterator set_iterator=all_possible_sums2.begin();set_iterator!=all_possible_sums2.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums2.insert(myarray[i]);
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums2.insert(*set_iterator);
}
//for every element in the first set, check if the the second set has the complemenent to make N
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
if(all_possible_sums2.find(N-*set_iterator)!=all_possible_sums2.end())
return true;
return false;
}
Ugly and brute force approach:
a = [1, 2, 3, 4, 5, 10]
b = []
a.size.times do |c|
b << a.combination(c).select{|d| d.reduce(&:+) == 19 }
end
puts b.flatten(1).inspect
Didn't find any similar question about this.
This is a final round Facebook question:
You are given a ring of boxes. Each box has a non-negative number on it, can be duplicate.
Write a function/algorithm that will tell you the order at which you select the boxes, that will give you the max sum.
The catch is, if you select a box, it is taken off the ring, and so are the two boxes next to it (to the right and the left of the one you selected).
so if I have a ring of
{10 3 8 12}
If I select 12, 8 and 10 will be destroyed and you are left with 3.
The max will be selecting 8 first then 10, or 10 first then 8.
I tried re-assign the boxes their value by take its own value and then subtracts the two next to is as the cost.
So the old ring is {10 3 8 12}
the new ring is {-5, -15, -7, -6}, and I will pick the highest.
However, this definitely doesn't work if you have { 10, 19, 10, 0}, you should take the two 10s, but the algorithm will take the 19 and 0.
Help please?
It is most likely dynamic programming, but I don't know how.
The ring can be any size.
Here's some python that solves the problem:
def sublist(i,l):
if i == 0:
return l[2:-1]
elif i == len(l)-1:
return l[1:-2]
else:
return l[0:i-1] + l[i+2:]
def val(l):
if len(l) <= 3:
return max(l)
else:
return max([v+val(m) for v,m in [(l[u],sublist(u,l)) for u in range(len(l))]])
def print_indices(l):
print("Total:",val(l))
while l:
vals = [v+val(m) for v,m in [(l[u],sublist(u,l)) for u in range(len(l)) if sublist(u,l)]]
if vals:
i = vals.index(max(vals))
else:
i = l.index(max(l))
print('choice:',l[i],'index:',i,'new list:',sublist(i,l))
l = sublist(i,l)
print_indices([10,3,8,12])
print_indices([10,19,10,0])
Output:
Total: 18
choice: 10 index: 0 new list: [8]
choice: 8 index: 0 new list: []
Total: 20
choice: 10 index: 0 new list: [10]
choice: 10 index: 0 new list: []
No doubt it could be optimized a bit. The key bit is val(), which calculates the total value of a given ring. The rest is just bookkeeping.