Editorial of Round #489(Div.2)

7 years ago, # ^ |

← Rev. 13 →

If at some point you have X dresses, at the end of that month you may have 2X or 2X-1 dresses. If you draw that as a binary tree, and write down all possible numbers of dresses after K months, it's easy to see that they will represent a sequence of 2^k consequent numbers. More generally it will look like this: 2^k*x,2^k*x-1,2^k*x-2,..., 2^k*x-(2^k-1) Those are all possibilities for numbers of dresses. Summing them up you get, 2^k*x*2^k-(1 + 2 + ... + (2^k-1)) = 2^k*x*2^k-(2^k-1)*2^(k-1) Remember we need to double that value since the last month numbers of dresses only doubles, the wardrobe doesn't eat any. So the sum is 2^(k+1)*x*2^k-(2^k-1)*2^k. Now divide that with 2^k to get the arithmetic mean, you get 2^(k+1)*x-2^k + 1. It's all about writing it down.

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prithviking98

7 years ago, # ^ |

this problem was actually easier than B xD

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Single_Ready_To_Mingle

7 years ago, # ^ |

True that

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7 years ago, # ^ |

I agree. If there were not some tricky test cases in C (like if x is a multiple of 10^9+7) actually people would do it more than B.

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PetitAPetit

7 years ago, # ^ |

← Rev. 2 →

Can you explain why you divided on 2^k at the end?

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7 years ago, # ^ |

← Rev. 3 →

The expected value of number of dresses is what we are looking for. Expected value is equivalent to arithmetic mean which we calculate by dividing sum of the numbers by how many numbers there are. ((a1+a2+...+an)/n). Therefore, having 2^k possibilities for number of dresses and their sum already calculated, we calculate the arithmetic mean like this: sum/2^k (sum = 2^(k+1)*x*2^k-2^(k-1)*2^k).

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sharmajatin741

7 years ago, # |

anyone please explain solution of problem D I am not getting it through editorials

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7 years ago, # ^ |

← Rev. 2 →

First, we'll analyze this problem in an O(n^2) solution:

Initialize 2 arrays PrefSum[n] and PrefProd[n] in which PrefSum[i] is the sum of first i number in the array, and PrefProd[i] is the product of first i number in the array. After that, we count the subsegment by for loops.

Looking at this solution, we can easily notice that k*SumOfSegment cannot exceed 10^5*10^8*10^5=10^18, so that ProductOfSegment cannot exceed 10^18<2^60, which means a subsegment can never have 60 or more numbers that greater than 1.

With this fact, we can reduce the complexity of this solution to only O(n*60).

The trick here is you have to find the way to skip the 1's in the array using pre-calculated arrays, without leaving out any possible answer. When skipping each group of consecutive 1's, we can prove that there are only at most 1 possible answer between this group. That way, we can keep each loop in O(60) complexity.

Here's my code (C++14): https://ideone.com/JLNwsZ

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Scandium

7 years ago, # ^ |

Hi, the explanation you provided is very helpful. Why the PrefSum[n] and PrefProd[n]are storing the values for first i numbers? Shouldn't they be 2D arrays storing values for (i,j) implying sum and product from ith element to jth element if we want to consider all subsegment?

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http://codeforces.net/contest/992/submission/39416834 :(

7 years ago, # ^ |

You could get sum of the subsegment from i to j by PrefSum[j]-PrefSum[i-1]. Same mechanism works for PrefProd. The prefix sum is very useful in reducing the memory and time complexity of your solution.

To initial the prefix sum array, you start with pref[1]:=A[1] then for each i:=2 to n: pref[i]:=pref[i-1]+A[i]

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aakashjaiswal

7 years ago, # ^ |

i did the same thing but i dont know why i am getting wa

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7 years ago, # ^ |

Your failed test consists of n 1's. I think the problem is in the way you skip the 1's and add the subsegments to the answer. Try to debug on these kind of test!

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SayMyNameSweetHeart

7 years ago, # ^ |

Wouldn't PrefProd[i] overflow? How can we keep on multiplying array entries and storing in PrefProf[i]? Thanks.

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7 years ago, # ^ |

That's just my very early O(n^2) approach for this problem so that I could show you how I reduce my complexity in this problem. I know that PrefProd[i] could overflow (it should be :)) ) but that doesn't really matter in my real solution.

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-_-_-

7 years ago, # |

Actually A can be solved in O(N) where N = 2e5 Just store all negative numbers as

1e5 + abs(a[i])

where a[i] < 0.

Space is increased here but still better than N*log(N) on an average

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hell_hacker

7 years ago, # |

← Rev. 2 →

Why does this idea not work for D?

For every array index which is 1 I store the number of 1s after that so that I can directly jump to the next index. (for example, 1 1 3 1 1 1 4 I get 2 1 — 3 2 1 -)

Now a normal nested for loop and if array[i] == 1 then if sum * k was less than prod and after adding no. of 1s following that 1 if sum * k >= prod, answer is incremented.

And normal check for prod == k*sum for elements other than 1.

Simplified Code

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madhav_thakker

7 years ago, # |

Can anyone explain the problem E to me? I have trouble understanding the query part? Particularly, We find the leftmost element which may be king-shaman yet. We can realise that it's the leftmost element, which doesn't lie in the good prefix (as there isn't king-shaman according the definition), which have a value at least . It can be done using segment tree with maximums, with descent. The left-most element that may be king is (i+1) ? And how is this O(logN) ? Thanks

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7 years ago, # |

can somebody tell me, in B why we are running the for loop from i = 1 to i^2 < (y/x)

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shouryap

7 years ago, # ^ |

We are looking for all divisors of $\text{[math]}$ Of course, if j is a divisor of p, then so is $\text{[math]}$ , and one of i or $\text{[math]}$ is less than or equal to $\text{[math]}$ . So it suffices to iterate over the divisors of p which are less than or equal to its square root; the ones that are greater than or equal to the square root of p will automatically "show up" as $\text{[math]}$ for some divisor $\text{[math]}$ of p.

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7 years ago, # ^ |

nice explanation , thanks

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ameya

7 years ago, # |

If somebody could tell me why my solution for D gives the wrong answer on Test 133, I would be very grateful.
I keep the positions of the nonzero elements in a vector, traverse this vector, while also keeping a track of the number of ones to the left and the right of the current subsegment we're considering.

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ameya

7 years ago, # ^ |

Got it — overflow on finding the product was the problem. AC now!

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raja1999

7 years ago, # |

+11

In Problem E Solution 1,

We can realise that it's the leftmost element, which doesn't lie in the good prefix (as there isn't king-shaman according the definition), which have a value at least p[i]. It can be done using segment tree with maximums, with descent.

How can we find this index in log(N)?

I could only think of (log(n))^2 approach.

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tfg

7 years ago, # ^ |

Do a maximum query segment tree. When you visit the segments that cover the range, if the maximum element is what you need you go down else just return n. Now, you are in a range that has the answer you need, you just need to decide wether to go left/right. If the maximum element to the left if what you need, go left otherwise go right.

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oos1111

6 years ago, # ^ |

I'm still no clear. Could you please explain a little bit more?
Here is a snippet of code from solution1:

int down(int i, int l, int r, LL S){
    if (r-l==1) return l;
    int mid = (l+r)/2;
    if ((LL) rmq_max[2*i+1] >= S) return down(2*i+1, l, mid, S);
    return down(2*i+2, mid, r, S);
}
int get(int i, int l, int r, int l1, int r1, LL S){
    if (l1 >= r1) return n;
    if (l==l1 && r==r1){
        if ((LL) rmq_max[i] < S) return n;
        return down(i, l, r, S);
    }
    int mid = (l+r)/2;
    int res = get(2*i+1, l, mid, l1, min(r1, mid), S);
    if (res != n) return res;
    return get(2*i+2, mid, r, max(l1, mid), r1, S);
}
LL get_sum(int i, int l, int r, int l1, int r1){
    if (l1 >= r1) return 0;
    if (l==l1 && r==r1) return rmq_sum[i];
    int mid = (l+r)/2;
    return get_sum(2*i+1, l, mid, l1, min(r1, mid)) + get_sum(2*i+2, mid, r, max(l1, mid), r1);
}

The get method is to find the leftmost element which is larger than value S. And if we ignore the down method called in it, I think the time complexity of it is the same as get_sum since in the worst case the res is n every time and it will call itself twice every time. And now we take down into account. The down method is log(N) itself. So the time complexity of the method should be log(N)^2. How can we tell that the complexity of get is log(N)?

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tfg

6 years ago, # ^ |

get() visits the "cover" of the range, so the same segments that something like get_sum() would visit. Once it gets to a range that it calls down, it will get an index and will go back returning it (if there's no answer there, it won't call down). So the first part (visiting the ranges that cover the query range) is O(log) and the second part (down) is O(log) too.

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oos1111

6 years ago, # ^ |

Oh I got it. down would be called once at most(ignoring recursive call). Thank you very much!

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7 years ago, # |

In problem B , test 7 we have the input 1 1000000000 158260522 158260522, whose answer is 1. How can we have a number b/w 1 and 1000000000 whose hcf is greater than the number itself, shouldn`t the answer be 0.

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shouryap

7 years ago, # ^ |

1 and 1000000000 are the lower and upper bounds of the two numbers. The HCF and LCM both have to be 158260522.

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7 years ago, # ^ |

oh, i miscalculated the number of zeroes

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Kerim.K

7 years ago, # |

+21

Thanks for such a good contest. I like problems, they are interesting to me and especially I like 3rd solution of E problem.

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Osama_Alkhodairy

7 years ago, # |

Any sqrt decomposition solution for E passed?

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Heisenbug

6 years ago, # ^ |

I made one just now.

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EsmaelSY

7 years ago, # |

← Rev. 2 →

In problem B, Is There any efficient way to solve it if the "LCM" condition removed, I mean:

Given three numbers l r x , count number of pairs (a,b) such that l <= a,b <= r and GCD(a,b) = x.

where : (1 ≤ l ≤ r ≤ 10^9, 1 ≤ x ≤ 10^9).

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rusan

7 years ago, # |

Why complexity of problem A is nlogn we are just looping through the n elements then why not O(n) where did logn came from??

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