In this problem you just need to loop through the integers

*i*from 1 to*n*determine, is*i*lucky, if yes, then try, if*n*mod*i*= 0, then answer is "YES". If there are no such*i*, answer is "NO".Notice, that answer is either one digit or -1. So, if there are no 4 and 7 digits, then answer is -1. Else, if there is more or equel number of digits 4, then answer is 4, else answer is 7.

Let generate all lucky number between 1 and 10

^{10}. Consider all segment [1;*L*0], [*L*0 + 1;*L*1], [*L*1 + 1;*L*2] ... Then the result equals to product of intersection of segments [1;*L*0] and [1;*n*] by*L*0 plus size of intersection of [*L*0 + 1;*L*1] and [1;*n*] multiplied by*L*1 and so on.Notice, that if there exits such

*i*that*i*mod 2 = 0 and*d*_{i}= 4 and*d*_{i + 1}= 7 and*d*_{i - 1}= 4 then after that operation there will be a loop. So, let we simple do all operation from left ro right, and, when we will win our loop, just return a rusult (which variates by*k*mod 2 (*k*is one that leaves after operation from left side).If

*n*≤ 14 let find out, is*k*more than*n*! or not? If yes, then return -1. Then, notice, that since*k*≤ 10^{9}, then at most 13 elements from right (suffix) will change. So, element from left of this part (prefix) will not change (then we can just find a number of lucky numbers on than range). To find result for the rest of the permutation (suffix) we need to find*k*-th permutation of number from 1 to*t*(*t*is maximun integer, such that*t*! ≤*k*). After we find that permutation, we can just loop through that permutation and count result.Lei calculate arrays

*L*and*R*. Let for all lucky*i*,*L*[*i*] = number of operation needed to move every segment, which's right end is to left from*i*to*i*.*R*[*i*] = number of operation needed to move every segment which left end is to right to*i*. How to calculate such array? Just use a sorting. Let arrange array*A*of pairs of integers, first - number, second equals to 0, if that number end of some segment, 1 if that number is lucky. Then, iterate from left to right, counting number of segment end we counted and*s*_{i},*s*_{i}=*s*_{i - 1}+ (*A*_{i}-*A*_{i - 1}) **c*_{i}. The same way you can use to*R*. Now, to find the answer we must find (using method of two pointers of binary search) pair of indexes*x*and*y*, such that*i*≤*j*,*L*_{j}+*R*_{i}≤*k*,*Lucky*_{j}-*Lucky*_{i}+ 1 ≤ min_size_of_input_segment. Also, is this problem you should arrange function*Mul*(*a*,*b*), which return*min*(*a***b*,*INF*). Since simple multiplication overflows, then you can use multipling modulo 2^{64}or double or min-long-arithmetic.In this problem you can use many different algorithms, here is one of them. Obviously, number of different lucky number is 30, because

*A*_{i}always is ≤ 10000. Let*D*_{i}- difference between minimum lucky number which is greater than or equal to*A*_{i}and*A*_{i}. Now, we need to have 5 (but you can use less number) of operation on*D*: Subtract(l, r, d) - subtract number*d*from all*A*_{i}(*l*≤*i*≤*r*), Minimum(l, r) - minumum number of all*D*_{i}(*l*≤*i*≤*r*), Count(l, r) - how many times that minimum occared in that interval, Left(l, r) = leftmost occarence of that minimum, Set(i, d) - assign*d*to*D*_{i}(*D*_{i}=*d*). Now, we can do our operations. If our operation is "count", then we need to find minimum number*d*(*d*= Minimum(l, r)), if it is equal to 0, then answer is Count(l, r, 0), otherwise, answer is 0. If out operation is "add", then we need to Subtract(l, r, d), but now some*D*_{i}might be less than 0. So, while, Minimum(l, r) ≤ 0, let j = Left(l, r), assign*D*_{j}new value (use Set(j,*D*_{j}')), which can be calculated with complaxity*O*(1).Complexity of this algorithm is

*O*(*m***log*(*n*) +*n***log*(*n*) **C*), where*C*is equal to 30.
dfrom allD_{i}"D_{j}new value" - if you update those where Minimum(l, r) = 0, then the Count queries will fail right?Lucky Array (Div1 E) ,

How to implementSegment_Tree?

Who can explain the solution in detail ?

## include<stdio.h>

int main(){ int n ; scanf("%d" , &n); printf((n%4 && n%7 && n%47 && n%74 && n%477)?"NO":"YES"); return 0; }

why did it work?

Ideally it should not have worked but due to weak test cases, his solution worked.

In

Problem 122A, A. Lucky Division, what is the meaning of evenly divided, because this has been accepting the answers for 49 as YES. I don't understand why it should be a YES?Because 49 is divided by 7

Yeah, I do understand that, but what is the meaning of term

evenly dividedin the problem statement. I didn't understand the context of this word because 7 divides 49, 7 times which isn't even."Evenly" means "exactly" here

Thanks for the Help, Mate. Noobie to this world. ^_^

evenly divided means completely divisible And 49 is completely divisible by 7 (which is a lucky number), so 49 is almost lucky number. That is why the answer is YES.

Lucky Array 121E — 72 testcases and still passes O(n*m*log(10^4)) :0 Code

It's actually faster than $$$O(n\cdot m\cdot log(n))$$$ because you only update the BIT when a number becomes or "unbecomes" lucky, this can only happen at most 60 times per number, so it's actually $$$O(n(m+60 \log n))$$$.

Still more than 4 sec. :D

It's a perfectly valid solution, the heaviest part $$$O(nm)$$$ has a small constant because it's very simple and cache-friendly. I have just written a blog post about this, I believe people will find it useful.

I have implemented the same code and it passed all test cases, and worst time was around 2 secs.

Submission

In Problem (Div 2A), How 799 is YES? Here every digit is not 4 or 7 and also the number is not evenly divided by 4 or 7.

799 = 47 * 17; 47 is lucky number.

Can someone explain the logic behind 2C ? The editorial is very short and unclear.

103256425 why I get wrong answer on test 21 can anybody help?

IN problem Div1(A) or Div2(c) how to create those lucky numbers from 1 to 10^10?

For each length of number L, the are only 2^L possible lucky numbers (since there are 2 possibilities for each digit: 4 or 7).

Generate instead all possible bitmasks of length L (for L = 1 to 10), and replace 0s with 4s, 1s with 7s. Note that you need to include bitmasks with leading zeros.

Python implementation: https://paste2.org/YfMYvaXe

thank you..@jimm89

what wrong in this program??

<?php function division($y){ for($x=1; $x<=1000; $x++){ if($y%7==0 || $y%4==0){ echo"YES"; break; } else{ echo"NO"; break; } } } division(); ?>