We will hold AtCoder Beginner Contest 169.
- Contest URL: https://atcoder.jp/contests/abc169
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200531T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: kyopro_friends, latte0119, tempura0224, ynymxiaolongbao, yuma000
- Rated range: ~ 1999
The point values will be 100-200-300-400-500-600.
We are looking forward to your participation!
Editorial in english must also be posted as soon as in japanese.
I'll post my solutions after the round ends.
where will You post It ?
As a CF Blog.
Atcoder should pay you for this...
In F, i take input of the interval as pairs and sorted the vector of pairs ,then find the median of that interval. Then do the necessary arithmetic as pointed out by Geothermal in here. This approrach fails in few test cases. What's wrong with my approach with F? Can anyone point out what am i missing? Link of my submission
I believe that for the $$$n$$$ even case, you select the wrong two indices to form the median. Use $$$\frac{n-1}{2}$$$ and $$$\frac{n}{2}$$$, rather than $$$\frac{n}{2}$$$ and $$$\frac{n+1}{2}.$$$
Your suggesstion did help to pass some test cases which were not passing earlier, but still few test cases are failing. Can't think of counter test case which could Hack my code.submission
You don't need to sort vector of pairs. Simply sort individually.
I know from the editorial but the problem is that I don't understand why my solution is wrong?
Your solution is wrong because you have sorted vector of pairs. Let these be pairs. 1 5 2 3 After sorting either they will remain same or 2 3 1 5 There is no possible way answer will be wrong in both the cases because for finding median array needs to be sorted.
I hope it would be a great Contest
Excited that I will learn something new!
Update: Learned when to use double when not to use doubles and how to use doubles!
2nd and the 3rd question seems easy but is really tricky... i dont know on which cases my code is getting wa...can anyone please help
its cheating bro!
same problem lol
Don't ruin the name of India more than already it is.
You don't need to look into name of india.
I'm so sad, it took me less time to D than it took me to do B and I haven't even done C. Everytime, I miss one of B-D. I'm cursed D:
I am missing something in B and getting WA again and again. Any hints please!
I'd love to help but I think that is not allowed (correct me if I'm wrong, I'm new to CP), will definitely help after the contest ends.
Yes please man now
so you have to realize anytime the answer overflows out of limit(of data type) or become > 10^18. also if there is even 1 0, ans = 0. Code:
include <bits/stdc++.h>
using namespace std;
typedef long long ll; typedef unsigned long long ull; typedef unsigned int uint; typedef long double ld;
int main() { ios_base::sync_with_stdio(false); cin.tie(0);
}
I had the same code, only difference was that i used long long for 'ans' instead of long double and i got WA on one of the test cases .Changed to long double and got AC.Why?
Because long long data type could be overflowing from the multiplication of say 10^12 and 10^12.
I solved problem C, and I don't even understand why it works.
Will a CE submission be calculated into the overall penalty?
I submitted two pieces of codes in the wrong language and will my overall penalty increase by 10 min?
No.
How to solve F?
For F, we can notice that ans is simply summation of i from 1 to n of (number of subsets of length i with sum s) * (2 ^ (n — i)).
So now let dp[x][j] = summation of i from to n of (number of subsets of length n with sum x) * (2 ^ (n — i)) for the prefix 1..j
Now, apply normal knapsack to combine answers, but since you're adding length 1 to your previous sums, we will have to divide them 2
dp[x][j] = dp[x][j — 1] + (dp[x — a[j]][j — 1] * modinv(2))
Answer is dp[s][n]
You can even do it in O(n) space without needing the second dimension, like a normal knapsack, moving backward.
Can you please provide the link for your solution madhav_1999
https://atcoder.jp/contests/abc169/submissions/13869671
Thanks you BTW you are my friend now.
:)
Knapsack for All Segments vs Knapsack for All Subsets :D
How to solve E!
If n is odd, get the maximum median (sort array of B's and get its median) and the minimum median (sort array of A's and get it's median). The answer is max median — min median +1.
If n is even, sort the arrays and do: max median = B[n/2] + B[(n/2)-1], min median = A[n/2] + A[(n/2)-1]. This way we avoid working with not integer numbers. So the answer is still max median — min median +1.
Oh god I was trying to use sweep type algorithm and check for each end point, but your answer is much much easier. Could not implement my approach but will try to refine it
Thanks for your approach but can you prove it!
I am not able to understand why every value between min and max can be attained? Can you explain plz.
From the minimum median, you just need to icrease the median by 1 to go to the next value (in case of n pair, increase first the biggest one and after the smaller one in the middle). You can do this until reach the maximum median.
Thanks to 1358C, idea is very similar. You can observe minimum median will be
median of array xand maximum medium will bemedian of array y.So values in the range
median of array xtomedian of array ywill be covered. Take care when size of array is even.Submission
In the second part, there are essentially two tricks for the problem 1) Sort the array. If the value at index i is 0, then we cannot return -1(no matter what). 2) Multiplying two extremely large numbers even when the data type is long long int will result in negative values, which will result in wrong answers. So, a*b>=c is equivalent to b>=(c/a)(take care of the border cases) Link to my code https://atcoder.jp/contests/abc169/submissions/13810433
Okay, so in the 3rd question, I simply multiplied a and b, and applied the floor function for the product. Link to my code in C++ https://atcoder.jp/contests/abc169/submissions/13814577
Submitting the same code, giving WA
I just wonder how to solve B,C...
Is there anything wrong??? https://atcoder.jp/contests/abc169/submissions/13881928 https://atcoder.jp/contests/abc169/submissions/13872116
In problem C, there can be floating-point error.
But how to avoid??
I used string for storing $$$B$$$. Then calculated $$$100*B$$$. link
thank you!
Use b=b*100+0.01;
instead of b*=100;
thank you!
Adding the extra 0.01 (or even 0.001) works, but how does it work? 055D
If we convert double to long long, the fractional part would be discarded. So in this kind of problem, if we add small real number and convert it, we can see the same effect as rounding.
How will there be a floating point error? And how will b=b*100+0.01 save from it?
In your solution for problem B, if resu is 10^17 and next number is 99 then it might overflow
Thank you !
This is sarcasm..Right???
NO NO NO!! I really don't solve them!
You had a bad day don't worry
For B, just check 0 case. Then before multiplying with a[i], check if 1e18/a[i] < cur_product. If it is, simply print -1. Otherwise you can safely multiply.
https://atcoder.jp/contests/abc169/submissions/13804989 I did the same... But getting WA in 2 TCs..
Typecast your 1e18 into int before dividing to avoid precision loss.
Changing to ans > (int)(1e18)/i should give you AC
Thank you !
i was getting TLE Second Question
I'm guessing you may have tried to multiply all the numbers together in Python or a similar language? Since the intermediate products can have $$$O(N)$$$ digits, multiplying them takes $$$O(N)$$$ time, so this approach takes $$$O(N^2)$$$ time in the worst case.
My solutions to all problems are at https://codeforces.me/blog/entry/78195.
Can someone please tell why this fails for C?
You need to round it, rather than taking its floor. For example
float('2.51') = 2.509999999..., soint(float('2.51')*100) = 250.thank you sir
In C, why is math.trunc(a*b) in python giving WA,but in CPP, this is accepted-
nvm, got it
can anybody find mistake in my D solution.
You are not looking for primes greater than √x.
Try for 4000000028.
Any care to explain why is this wrong for problem C ?
It gives WA for random_12.txt
Edit : I always define int as long long in my template.
I'd guess the issue is double precision. Precision is an issue in general, but in particular, the answer here can be up to $$$10^{16}$$$, which is too large for the number of precision bits in a double.
But here multiplications are in long long, the only place where precision could affect is assigning "b = c*100".
Ah, right; my bad--I misread the solution. The issue is precision in that conversion: I changed b = (c*100) to b = (c*100+0.5) and the solution was accepted.
Thanks for the solution.
Any advice on how to avoid these errors in future ?
You need to understand why this error happened. The reason is that if you read a number as a double (lets say 10) the double can be represented as 9.9999... or maybe 10.000...01 (both of which are 10 indeed woth a very small error intrinsic to the floating point representation)) but in c++ if you convert them to an integer it truncates the value automaticaly, so the result could be either 9 or 10. The solution would be doing the comversion as x_int = round(x_double) or x_int = x_double + 0.5.
Thank you very much. This is what I was looking for.
Being skeptic, if x_double is 4.50 is there any chance I will get 5 instead 4 using your suggested solution ?
Yes, it would be a problem. But you did not quite get how to use what I said. If a number has at most x decimal digits, you may want to multiply it to 10^x and convert to integer, so you do int_x = 10^x*double_x+0.5 (but if double_x has more the x digits this may not work correctly)
Now I got it. Thanks for the help.
Do not assume double arithmetic is correct. Often it is not, the result is only near to the mathematic correct result.
Deal with that or use integers.
I didn't, that's why I converted double to int and then calculated the result.
I am yet to find any solution which always works for doubles.
I disagree with that. You just need to unserstand how big the error can be in the worst case, often the error is smaller than 1, so you can often get the integer part right for sure if the constraints allow that
That is the strategy I would call "deal with it".
Use
long doubleinstead ofdoubleWhy does long double over double changes anything ? considering value of is B < 10, having 2 digits are decimal point.
I used this formula:
answer = floor( a * long double(b) )Yes, that works well. I used the similar thing to get AC later. Thanks for the answer.
Range of input is too big which will have precision loss. Try using long long and long double
Why does long double over double changes anything ? considering value of is B < 10, having 2 digits are decimal point.
In the fourth question, I simply used the Sieve of Eratosthenes to get the prime numbers less than 1e6, after that I iterate from 2 to the largest prime less than 1e6, and simply code out for the number of ways the number can be broken down! Just be careful when we multiply two numbers of digits 5 and 6! Link to my code- https://atcoder.jp/contests/abc169/submissions/13855795
Why this fails for problem B: https://atcoder.jp/contests/abc169/submissions/13804989
How to solve c? I multiplied a*b and took floor(a*b). What's wrong? submission
See, Double can store large values but on the cost of significant digits,in Long you will get more significant digits before decimal point but then you can not store digits after the decimal points
But I also tried (long)(b*100)*a/100. What's wrong with this?
problem B solution : https://atcoder.jp/contests/abc169/submissions/13824612
Please someone help me out with Problem C ....... My Submission
Precision issue. Do long long x = round(b*100) and AC
Whats wrong wit C
import java.math.BigDecimal; import java.util.Scanner;
public class Main {
public static void main(String args[]) { Scanner sc=new Scanner(System.in); long a=sc.nextLong(); long b=(long)(sc.nextDouble()*100); long ans=(a*b)/100; System.out.println(ans); }My Simple and short Solution for C (Used Bigdecimal Library in JAVA)
You can do it this way but still whats wrong in my code.
I usually get one hit on both Codeforces and Atcoder, but this time seems horrible xD
Dude you are really patient!
Whats wrong in D
import java.util.Scanner;
public class Main {
public static void main(String args[]) { Scanner sc=new Scanner(System.in); long n=sc.nextLong(); long copyn=n; long x=(long)Math.sqrt(n); long copyx=x; long ans=0; for(int i=2;i<=x;i++) { int count=1; while(n%((Math.pow(i,count)))==0) { n= (long) (n/Math.pow(i,count)); count++; ans++; } while(n%(Math.pow(i,count))==0) { n= (long) (n/Math.pow(i,count)); } } if(copyn==1) { System.out.println(0); } else if(n>x) { System.out.println(1+ans); } else { System.out.println(ans); } }}
What is the problem with this solution of C?Can someone help?
include<bits/stdc++.h>
using namespace std; typedef long long ll;
define mod 1000000007
int main() { ll a; float b1; cin>>a>>b1;
b1=b1*100; ll b=b1; ll s=a*b; s=s/100; printf("%lld\n",s); }
Maybe precision issues? Try doing b1 = round(b1*100)
Well, if the input is 1.890000 then it becomes 189.000000 and then 189 by typecast. There shouldn't be any problem.
You dont understand how it works. The number may be interpreted as 188.99999999... which is the same as 189.00000 but when you cast to int in c++ the value is truncated, it sort of an undefined behaviour, the value may be casted to 188 or 189, use the round to avoid that.
Yeah, I got it. That was the problem actually. Thanks a lot.
https://atcoder.jp/contests/abc169/submissions/13900202 [WA] https://atcoder.jp/contests/abc169/submissions/13900438 [AC]
Both are identical solutions for problem B of AtCoder Beginner Contest 169. Does anyone knows why this happen?
Maybe, someone playing with the testcases :)
Or maybe first solution is sent to Problem A not Problem B?
I think I got a beautiful solution for F: the answer is just the coefficient of x^s in polynominal (2 + x^a1)(2 + x^a2)...(2 + x^an) here '^' is for power. we can just ignore higher order coefficient than x^s.
https://atcoder.jp/contests/abc169/submissions/13847100
please ignore unremoved code for problem D, E.
In B:
For the statement:
if(inp[i] <= 1e18/ans)I got WA. But when i wrote :if(inp[i] <= 1000000000000000000/ans)I got AC. Is there any differences between 1e18 and 1000000000000000000?? Otherwise what is the reason??You should cast 1e18 to long long otherwise the result of the division becomes a float.
Thanks man. That means 1e18 format is double type.
Yup!
In editorial there is a typo in problem D English version.
Wrong:$$$O(N)$$$
Correct:$$$O(sqrt(N))$$$
Can AnyBody please tell how can i reduce the submission time of my solution to the problem F?
https://atcoder.jp/contests/abc169/submissions/13920879
I have seen people getting AC in less than 100ms while mine is giving AC in 1000 ms.
thanks in advance. :).
In the editorial of problem F it says:
dp[i][j] = When the choices for the first i options is already determined, the number of combination such that the sum of ak for each k that the first option was chosen is equal to jWhat does it even mean?
Can anyone help me debug my solution? I get TLE on 6-7 test cases, no idea why. Thank you!
i * i can overflow since n goes upto $$$10^{12}$$$
Use long long instead.
T_T I don't know how I missed that. Thanks!
what is my wrong with my code in problem-c? https://atcoder.jp/contests/abc169/submissions/13926562
For problem D, Maybe this sounds stupid, But can someone please explain why is it optimal to compute for each prime and its powers first, then similarly with next prime and its powers, and so on instead of going from 2 onwards ?
same thing I was also wondering .
Can anyone provide any proof for problem-E for both odd and even case.
If we increase one value by $$$1$$$ and leave the rest as it is, then median can stay the same or increase by $$$1$$$ for odd $$$n$$$ and stay the same or increase by \frac{1}{2}
Unable to parse markup [type=CF_MATHJAX]
is even. Now let's set $$$X_i = A_i$$$ and increase all of them one by one till we reach $$$X_i = B_i$$$. At first median equals median of $$$A_i$$$, and after all operations equals median of $$$B_i$$$ and at each step it will either stay the same or increase by $$$1$$$ in case of odd $$$n$$$ and by $$$\frac 12$$$ for even $$$n$$$ thus will reach all integers for odd $$$n$$$ and halfintegers for even $$$n$$$ from range $$$[\text{median}(A_1, \ldots, A_n), \text{median}(B_1, \ldots, B_n)]$$$.To prove we can't reach any other value we only need to see, that median of an integer sequence is an integer for odd $$$n$$$ and a half integer for even $$$n$$$ and see that we can't reach a value outside that segment, but this is straightforward conclusion from the fact, that if all values increase or stay the same, then the median can only increase or stay the same as well.
If anyone is still stuck at D here is a Solution
I am Facing problems in the C . No matter what I submit , I am shown wrong answer in 6/22 cases . I am not understanding where is the error or fault exists in that code .. If anyone knows , please give me the solution . My code is given below ::::
include<stdio.h>
int main() { long long int a,pro1; double b,pro=1.0; scanf("%lld %lf",&a,&b); b=b*100; pro=(a*b); pro1=(long long int)(pro/100); printf("%lld",pro1); return 0; }
For big values of $$$a$$$ the variable $$$pro$$$ has simply not enough bits to store the exact result of $$$a*b$$$. In this case the double type truncates bits, which results in pro1 beeing some other value than the mathematecally expected one.
Can someone help me where I'm going wrong in problem D, I'm getting hand_22 as the only test case as WA. Here is my code https://atcoder.jp/contests/abc169/submissions/14102528 Thanks
ull i=1;You have this just before the for loop, but it should be inside it, so that the while loop always starts at 1. Since all the primefactors in mp are independent of each other.Thanks a lot for reading my code and identifying my mistake.