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Разбор задач Codeforces Round 571 (Div. 2)
Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 18.11.2024 13:12:56 (k2).
Десктопная версия, переключиться на мобильную.
При поддержке
What about shuffle solutions in F? Just mistake?
Is it possible to add this case to the system tests for F and hopefully a rejudge?
https://pastebin.com/dgWMkY4z
Both the following and possibly many more submissions fail on this case:
https://mirror.codeforces.com/contest/1186/submission/56222396
https://mirror.codeforces.com/contest/1186/submission/56210105
Hey bro, Should it be the answer to your test case?
https://pastebin.com/snkSXV4G
I want a prove for the incorrectness of the greedy solutions in F. Really puzzled...
There is no need for proof of incorrectness, ever. There is only need for counterexamples, and there are many here. You can try the test case I provided above.
One of the problemsetters wrote it: For example, suppose there are 4 groups A, B, C, D, each containing 10 nodes. Suppose all nodes from A are connected to all nodes from B, all nodes from B — to all nodes from C, all nodes from C — to all nodes from D. Then there are 3* 10^2 = 300 edges in total, so we want to leave no more than (300+40)/2 = 170 edges. However, if we choose what to delete randomly, we may just delete all edges between B and C accidentally. Then we won't be able to delete anything else. So we will be left with 200 edges.
Oh I get it now. Thanks a lot for this explaination!
Thanks!
It's easy to see that f(c,d) is even if and only if cntb and cntc have same parity
Someone can help to understand this in problem C? I don't see why that is always true
Let the number of 1s in b that have the same index in c be x. Now total number of mismatched indices are cntb + cntc — 2x. Now this will be even if and only if cntb + cntc is even. Hence their parity should be same.
thanks for your additional explanation.
Shouldn't you add 'y' as the number of '0's that have the same index and subtract 2y also from the value? I know it doesn't make a difference for the solution, but still, to keep things correct, you can add it.
No, actually we are calculating number of places bits differ so bits differ if in first string we have 1 and 0 in second or vice versa. So we have total_bits_that_differ=cnt1-x+cnt2-x.
Can anyone elaborate 3rd solution? Cossack Vous and numbers
Just take every element floor value and mark those point which are integers. Now sum the numbers. Sum can be less than equal to 0 . if 0 then print it, otherwise increment those value which are not marked by 1
Can you give a more formal proof od problem D?
(Problem F) For which case will such a solution fail? (Test case-31)
B is missing :p
B task was explained here
In problem C one of the tags is FFT, how is FFT applied here?
I wanna know as well
You can replace 0's with -1's in both strings, and let c be the convolution of a with b reversed. Let a_i be the substring of a that starts in the ith position and has length equals to b, ma(x,y) the number of matches of the strings x and y, and mi(x,y) the number of mismatches. Now if you check c[i] with len(b)-1<=i<len(a) then c[i] = ma(a_i,b)-mi(a_i,b) (a match adds -1*-1=1 or 1*1=1 to c[i] and a mismatch add -1*1=-1). You also know that ma(a_i,b)+mi(a_i,b)=len(b) thus, from both equations, you have ma(a_i,b) =(len(b)-c[i])/2. Consequently, after calculating the convolution of a and b reversed you iterate over the interval mentioned and add 1 to the answer if (len(b)-c[i])/2 % 2 == 0.
Similar Problem : CodeChef:: String Matching
Problem C submission code. How can i optimize more it is giving TLE on test 7.
You can use prefix sum technique to store the number of ones until ith index in an array/vector and then simply consider all the substrings having length same as the length of string b. Here is my code:code
Or you may not use prefix sums. You can calculate it lazy. code
Can you explain the solution of B task? It is interesting problem
B task was explained here
Thanks a lot
what is the error in this code for problem c??
include<bits/stdc++.h>
using namespace std; int main() { string a,b; cin>>a>>b; int al=a.length(),bl=b.length(),i,j,x,ans=0; for(i=0;i<=al-bl;i++) { int x=i,count=0; for(j=0;j<bl;j++) if(a[j+x]==b[j]) count++; if(count%2==0) ans++; } cout<<ans<<endl; return 0; }
Always remember pastebin is useful and straightly paste it is ugly :^
please can you check now.
https://pastebin.com/xT5Gj9st
In your following loop:(begins at line 8)
The loops are nested, and it will be looped for
(al-bl)*bltimes, with a time complexity of $$$O(n^2)$$$. For $$$n\leq 10^6$$$ mentioned in the statement, it works too slow.thanks.But is there any logical error as it is giving WA for first test case and correct for second testcase.
Read the statement, $$$f(b,c)$$$ returns the number of positions where the two strings are different, not the same.
another solution for C:
let the changes be equal to total position such that b[i]!=b[i+1].
now let cnt stores the total mismatch between first substring of a and b.
do cnt=cnt%2. if cnt is even: ans++
To calculate mismatch between just next substring of a and b: {
1.Now position where b[i]!=b[i+1] there will be +1( if previous match become mismatch ) or -1( if previous mismatch become a match ) in value of cnt and in either case cnt value change from odd to even or even to odd.
to keep it simple lets just +1 for those positions because we only concern about odd/even of the cnt.
so overall total +1 in cnt will be by + changes
2.take care about new mismatch and previous mismatch
(new) +1 the cnt if mismatch between last char of b and last char of current substring
(old) +1 the cnt if mismatch between first char of b and first char of current substring
if cnt is even: ans++
}
Link to my submission
how problem C can be solve using fft ?
Let $$$n=|a|,m=|b|$$$. Suppose the string is 0-indexed. Let $$$f[i]=a[i],g[i]=b[m-i]$$$. You can find that
$$$f(a[i\dots i+m-1],b[0\dots m-1])=\sum\limits_{j=0}^{m-1}(a[i+j]-b[j])^2$$$
$$$=\sum\limits_{j=0}^{m-1}(f[i+j]-g[m-j])^2$$$
$$$=\sum\limits_{j=0}^{m-1}(f[i+j]^2+g[m-j]^2-2f[i+j]g[m-j])$$$
$$$=\sum\limits_{j=0}^{m-1}f[i+j]^2+\sum\limits_{j=0}^{m-1}g[m-j]^2-2\sum\limits_{j=0}^{m-1}f[i+j]g[m-j]$$$
The first two parts can be solved with prefix sums. The last part, just do FFT with $$$f$$$ and $$$g$$$(let the product be $$$h$$$), then it is $$$2h[i+m]$$$.
The last part is always even, so we dont need to do fft
Thanks for pointing it out! (Another way to explain why $$$f$$$ is even if and only if $$$cnt_a$$$ and $$$cnt_b$$$ have the same parity? lol)
Could you please see my submission for problem D, I am getting wrong on the 77th test case. (https://mirror.codeforces.com/contest/1186/submission/56261077) UPD: Got it, it was integer overflow problem
same bro even i am getting wrong answer on D my sol is 56262408 56262408
Use long long for sum variable
The variable in which you are storing sum of floor or ceil
can someone explain me in simple way in problem C why having same parity gives even no of change?
f(b, c) == even no of changes
let number of 1 in string b = x1, number of 1 in string c = x2, number of common 1 in both string = x,
so, number of changes = x1 + x2 — 2*x, which should be even. /** 2*x is always even **/, x1 + x2 should be even , it implies x1 and x2 should be both even or both odd, hence should have same parity
there can be two types of mismatch in the problem
let one string be c and other be b;
the two types can be if there c[i]=0 and b[i] = 1;
other case can be if c[i]=1 and b[i] = 0
so the number of mismatch of first type is numc-x and second type is numb-x .
where numc and numb are number of 1 in c and b respectively.
total mismatch = numc+numb-2*x
to make this even numc and numb must have same parity
In problem E, while doing this multiplication, ((fieldx−1)⋅(fieldy−1)−1)⋅n⋅m / 2 it can lead to overflow. Can someone suggest how to handle the overflow? I think it requires some sort of trick.
My solution is here: https://mirror.codeforces.com/contest/1186/submission/56266238
It is failing on test 5 due to overflow (at least that's what I think).
Maximum answer can be 10^18 so that's not the problem(overflow). But you calculated your dp wrong. Your code with correctly calculated dp. Code
Thank you very much!
As $$$x,y \leqslant 10^9$$$ and $$$n,m \leqslant 1000$$$, we have:
$$$Field_x=\lfloor \dfrac{x-1}{n} \rfloor \leqslant 10^6$$$ and so is $$$Field_y$$$
Thus, $$$((Field_x-1)\times (Field_y-1)-1)\times n\times m \leq 10^{18}$$$, which is in range of
long long.Use
long longfor calculation.Where is author's source code for F?
In problem C, It's easy to see that f(c,d) is even if and only if cntb and cntc have same parity. In other words if cntc≡cntd(mod2) then f(c,d) is even.
what is the meaning of parity here? I have looked up all the comments but cant found any clear explanation.
A and B have same parity if
A = B(mod 2)
just to be sure, you mean A = B%2?
no A % 2 = B % 2
For problem F, you don't need to "find Euler cycle and do the following steps for each component independently."
Because adding fictive vertex makes the graph connected.
It doesn't if a component has an euler cycle by itself(meaning all degrees are even).
I think I have a good idea of understanding Problem C.
Find out how many different positions of two strings A and B are equivalent to $$$\sum{a_i\bigoplus b_i}$$$
Because it only judges parity, it is equivalent to $$$XOR(a_i)\bigoplus XOR(b_i)$$$
Can you explain a little bit better your solution?
The task of the problem is to find out how many different positions of two strings A and B(A is the substring of string a,B is the string b)
As I said above,it is equivalent to $$$XOR(a_i)\bigoplus XOR(b_i)$$$
Because of two adjacent substrings of string a is just a character difference,so we can solve it in $$$O(n)$$$
In problem C, I don't understand the explanation for the problem. Please explain the pseudo code for the solution.
There is my Python code. It looks wery similar with pseudo code.
I don't understand solution problem D, anybody explain better?,i was seeing some solutions and most of them have this solution
https://mirror.codeforces.com/contest/1186/submission/56278052
Let a[] be the given array of doubles whose sum is 0
We need to find b[] such that b[i] = ceil(a[i]) or floor(a[i]) such that sum of all elements in b[] is 0.
Fact: ceil(x) — floor(x) = 1 for any real number with a non zero decimal part ceil(x) = floor(x) for any real number with zero decimal part
So let's initialize b[] such that b[i] = ceil(a[i]). Let's call the sum of this array as s. We need to make s = 0, and we can do so only by subtracting 1 from a set of 's' b[i]s (whose corresponding a[i] has a non zero decimal part).
Can some one explain me last part of editorial E .I mean how author arrived at if bitcnt(a)+bitcnt(b) is an odd number, then the matrix is inversed. Also how do we determine if a field is inverted or not if its coordinates (x,y) are odd,odd or odd,even ?
One important note — (In tutorial of Problem E)in last statement "if bitcnt(a)+bitcnt(b) is an odd number, then the matrix is inversed" , indexing is considered from zero i.e suppose we are referring to first row and second column then that is (0,1). readers might get confused since in problem and in most of the part of tutorial of problem E indexing is considered from 1. Also bit count refers to number of 1's in binary representation of number and not least number of bits required to represent the number in binary.
For Problem $$$E$$$, $$$bitcnt(a)+bitcnt(b)$$$ tells you whether the matrix at position $$$(a,b)$$$ is inverted or not due to the following reason:
Imagine the whole infinite a grid as a big quadrant, which consists of $$$4$$$ quadrants, each of which consists of another $$$4$$$ quadrants, and so on. Then if you started looping on bits from the most to the least significant position (let the current bit in the loop be the $$$i_{th}$$$ bit), then checked the $$$i_{th}$$$ bit in both $$$a$$$ and $$$b$$$. If bits are $$$(0,0)$$$ you go to the top left quadrant, $$$(0,1)$$$ to the top right, $$$(1,0)$$$ to the bottom left, $$$(1,1)$$$ to the bottom right. Then you repeat the process for the quadrant you went to using the $$$(i-1)_{th}$$$ bit and so on. This shows then when you eventually reach your matrix, it will be inverted only if the parity of $$$bitcnt(a)+bitcnt(b)$$$ is odd.
Where is the author's code for problem F???
E problem ， if bitcnt(a)+bitcnt(b) is an odd number, then the matrix is inversed. Who can prove it ？