### arsijo's blog

By arsijo, 11 months ago, ,

• +119

 » 11 months ago, # |   0 What about shuffle solutions in F? Just mistake?
 » 11 months ago, # | ← Rev. 2 →   +47 Is it possible to add this case to the system tests for F and hopefully a rejudge?https://pastebin.com/dgWMkY4zBoth the following and possibly many more submissions fail on this case:https://codeforces.com/contest/1186/submission/56222396https://codeforces.com/contest/1186/submission/56210105
•  » » 11 months ago, # ^ |   +1 Hey bro, Should it be the answer to your test case?https://pastebin.com/snkSXV4G
 » 11 months ago, # |   +3 I want a prove for the incorrectness of the greedy solutions in F. Really puzzled...
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 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.
•  » » 11 months ago, # ^ |   +1 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.
•  » » » 11 months ago, # ^ |   0 Oh I get it now. Thanks a lot for this explaination!
•  » » » 11 months ago, # ^ |   0 Thanks!
 » 11 months ago, # | ← Rev. 2 →   0 thanks for editorial sir
 » 11 months ago, # |   +3 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
•  » » 11 months ago, # ^ |   +12 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.
•  » » » 11 months ago, # ^ |   +4 thanks for your additional explanation.
•  » » » 11 months ago, # ^ | ← Rev. 2 →   0 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.
•  » » » » 11 months ago, # ^ | ← Rev. 2 →   +3 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.
 » 11 months ago, # |   0 Can anyone elaborate 3rd solution? Cossack Vous and numbers
•  » » 11 months ago, # ^ |   0 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
 » 11 months ago, # |   0 Can you give a more formal proof od problem D?
 » 11 months ago, # |   0 (Problem F) For which case will such a solution fail? (Test case-31)
 » 11 months ago, # |   0 B is missing :p
•  » » 11 months ago, # ^ |   +6 B task was explained here
 » 11 months ago, # |   +1 In problem C one of the tags is FFT, how is FFT applied here?
•  » » 11 months ago, # ^ |   +3 I wanna know as well
•  » » » 11 months ago, # ^ |   +9 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
•  » » 11 months ago, # ^ |   0 Similar Problem : CodeChef:: String Matching
 » 11 months ago, # |   0 Problem C submission code. How can i optimize more it is giving TLE on test 7.
•  » » 11 months ago, # ^ |   0 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
•  » » 11 months ago, # ^ |   0 Or you may not use prefix sums. You can calculate it lazy. code
 » 11 months ago, # |   -10 Can you explain the solution of B task? It is interesting problem
•  » » 11 months ago, # ^ |   +3 B task was explained here
•  » » » 11 months ago, # ^ | ← Rev. 2 →   0 Thanks a lot
 » 11 months ago, # |   0 Can someone explain how to calculate f(x,y) in "E problem ". I didn't get the tutorial above. Or if there is any other approach to this problem?
»
11 months ago, # |
-36

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; }

•  » » 11 months ago, # ^ |   0 Always remember pastebin is useful and straightly paste it is ugly :^
•  » » » 11 months ago, # ^ |   0 please can you check now.https://pastebin.com/xT5Gj9st
•  » » » » 11 months ago, # ^ |   0 In your following loop:(begins at line 8) for(i=0;i<=al-bl;i++) { int x=i,count=0; for(j=0;j
•  » » » » » 11 months ago, # ^ |   0 thanks.But is there any logical error as it is giving WA for first test case and correct for second testcase.
•  » » » » » » 11 months ago, # ^ |   0 Read the statement, $f(b,c)$ returns the number of positions where the two strings are different, not the same.
 » 11 months ago, # | ← Rev. 2 →   +8 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: { Better you draw 1st sample test case in notebook.(think now as we are shifting b to right.)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 + changes2.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 substringif cnt is even: ans++}Link to my submission
 » 11 months ago, # |   0 how problem C can be solve using fft ?
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 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 $2h[i+m]$.
•  » » » 11 months ago, # ^ |   +8 The last part is always even, so we dont need to do fft
•  » » » » 11 months ago, # ^ |   0 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)
•  » » » » » 11 months ago, # ^ | ← Rev. 2 →   0 Could you please see my submission for problem D, I am getting wrong on the 77th test case. (https://codeforces.com/contest/1186/submission/56261077) UPD: Got it, it was integer overflow problem
•  » » » » » » 11 months ago, # ^ | ← Rev. 2 →   0 same bro even i am getting wrong answer on D my sol is 56262408 56262408
•  » » » » » » » 11 months ago, # ^ |   0 Use long long for sum variableThe variable in which you are storing sum of floor or ceil
 » 11 months ago, # |   0 can someone explain me in simple way in problem C why having same parity gives even no of change?
•  » » 11 months ago, # ^ | ← Rev. 3 →   0 f(b, c) == even no of changeslet 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
•  » » » 11 months ago, # ^ |   0 Hello @practise_ , can you please share in-depth resources regarding parity and related topics?
•  » » » » 11 months ago, # ^ |   0 parity simply means having same parity means both are even or both are odd
•  » » » 11 months ago, # ^ |   0 thnks a lot
•  » » 11 months ago, # ^ |   0 Can You please explain me what is the meaning of "parity" here?
•  » » » 11 months ago, # ^ |   0 parity means either even or odd
•  » » » » 11 months ago, # ^ | ← Rev. 2 →   0 so the numbers 2,4 are having same parity and 3,7 are having same parity. please correct me if i am wrong..
•  » » » » » 11 months ago, # ^ |   0 yes u r right 3 6 have diifrnt parirty
•  » » » » » » 11 months ago, # ^ |   0 thank you
•  » » » » 11 months ago, # ^ |   0 Thanks
•  » » 11 months ago, # ^ |   0 there can be two types of mismatch in the problemlet 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] = 0so 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*xto make this even numc and numb must have same parity
 » 11 months ago, # |   0 Can you post Code solution of problem D for better understanding arsijo
•  » » 11 months ago, # ^ |   0 https://codeforces.com/submissions/slowbutdetermined917 see my code it is simple bro !
 » 11 months ago, # |   0 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://codeforces.com/contest/1186/submission/56266238It is failing on test 5 due to overflow (at least that's what I think).
•  » » 11 months ago, # ^ |   0 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
•  » » » 11 months ago, # ^ |   0 Thank you very much!
•  » » 11 months ago, # ^ | ← Rev. 5 →   0 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 long for calculation.
 » 11 months ago, # |   +14 Where is author's source code for F?
 » 11 months ago, # |   0 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.
•  » » 11 months ago, # ^ |   +13 A and B have same parity ifA = B(mod 2)
•  » » » 11 months ago, # ^ |   0 just to be sure, you mean A = B%2?
•  » » » » 11 months ago, # ^ |   +13 no A % 2 = B % 2
•  » » » » » 11 months ago, # ^ |   0 Ohh Ok I got it. I have figured out a way to solve even but how is the participant supposed to know that if cntc≡cntd(mod2) is true than only the string will be even.
 » 11 months ago, # |   0 Can anybody kindly tell me what's wrong with this code for problem C? — https://codeforces.com/contest/1186/submission/56220737Also, is there any way to see all the entries in a large input in a testcase?
 » 11 months ago, # |   0 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.
•  » » 11 months ago, # ^ |   0 It doesn't if a component has an euler cycle by itself(meaning all degrees are even).
 » 11 months ago, # |   0 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 $XOR(a_i)\bigoplus XOR(b_i)$
•  » » 11 months ago, # ^ |   0 Can you explain a little bit better your solution?
•  » » » 11 months ago, # ^ |   0 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 $O(n)$
 » 11 months ago, # |   0 In problem C, I don't understand the explanation for the problem. Please explain the pseudo code for the solution.
•  » » 11 months ago, # ^ |   0 There is my Python code. It looks wery similar with pseudo code.
 » 11 months ago, # |   0 I don't understand solution problem D, anybody explain better?,i was seeing some solutions and most of them have this solutionhttps://codeforces.com/contest/1186/submission/56278052
•  » » 11 months ago, # ^ |   0 Let a[] be the given array of doubles whose sum is 0We 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 partSo 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).
•  » » » 11 months ago, # ^ |   0 Thanks for explanation. I got it now. I have one more thing to ask if you dont mind. I am usually a python user and now I am shifting to c++ for simply the reason that it is way faster. But I am having difficulties finding an IDE for it. Could yo explain which IDE to select? I use debugging a lot
 » 11 months ago, # | ← Rev. 3 →   0 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 ?
 » 11 months ago, # |   +5 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.
 » 11 months ago, # |   0 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.
 » 10 months ago, # |   0 Where is the author's code for problem F???
 » 10 months ago, # |   0 Can anyone explain me this code for problem C clcik here
 » 7 months ago, # |   0 E problem ， if bitcnt(a)+bitcnt(b) is an odd number, then the matrix is inversed. Who can prove it ？
 » 7 weeks ago, # |   0 Here is a clear explanation for Problem C1. 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.Answer: Let, C = 1001, B= 1100. Hamming distance between them is, f(C,B) = CnC — CnB + 2 * X, where CnC and CnB are the count of 1's in C and B respectively. And X is the number of 1's that has same index in both C and B. So how the formula works? have a look at the example: C[0]=B[0]=1 so we have 1 such X. so f(C,B) = 2 + 2 — 2*1 (why 2*X as we have to subtract from both string). Now,1.we know adding or subtracting an even number from a number doesn't change the parity of a number, so we can discard -2*x part!. 2. we know, even+even=even and odd+odd=even, which means they have to have same parity to become even, this CnC and CnB have to satisfy cntc≡cntd(mod2).