### flaviu2001's blog

By flaviu2001, history, 9 days ago,

1421A — XORwice

Idea and solution: flaviu2001

Hint
Solution

1421B — Putting Bricks in the Wall

Idea: flaviu2001, solution: flaviu2001 and stefdasca

Hint
Solution

1421C — Palindromifier

Idea and solution: flaviu2001

Hint
Solution

1421D — Hexagons

Idea: flaviu2001, solution: flaviu2001 and koala_bear00

Hint
Solution

1421E — Swedish Heroes

Idea and solution: flaviu2001

Hint
Sneaky corner case
Solution
Proof for the pattern

You can also see the video solutions on stefdasca's Youtube channel

• +144

 » 8 days ago, # |   0 Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).
•  » » 7 days ago, # ^ |   0 When will the induction end for the problem E?
 » 8 days ago, # | ← Rev. 4 →   -21 Why is it getting WA on pretest 4? Spoilervoid solve(int test_case){ int i, j; int x, y; cin >> x >> y; vector c(7); for(i=1;i<=6;i++)cin >> c[i]; c[1] = min(c[1] , c[6]+c[2]); c[2] = min(c[2] , c[1]+c[3]); c[3] = min(c[3] , c[2]+c[4]); c[4] = min(c[4] , c[3]+c[5]); c[5] = min(c[5] , c[4]+c[6]); c[6] = min(c[6] , c[5]+c[1]); int ans; if(x <= 0 && y <= x){ int dist = (c[4])*abs(x); ans = dist+c[5]*abs(y-x); }else if(x <= 0 && y >= 0){ int dist = (c[3])*abs(x); ans = dist+c[2]*abs(y); }else if(x <= 0){ int dist = (c[3])*abs(x); ans = dist+c[4]*abs(y); }else if(y >= x){ int dist = (c[1])*abs(x); ans = dist+c[2]*abs(y-x); }else if(y <= 0){ int dist = (c[6])*abs(x); ans = dist+c[5]*abs(y); }else if(y > 0){ int dist = (c[6])*abs(y-x); ans = dist+c[1]*abs(x); }cout << ans << '\n'; return; } UPD: submission
•  » » 8 days ago, # ^ |   +23 Just submit it and wait for anwer...
•  » » 8 days ago, # ^ | ← Rev. 4 →   0 I think the condition of else if(x<=0) is wrong. This condition is reached when x<=0 and x
 » 8 days ago, # |   0 I overkilled B initially by using BFS. Then I realized that it can be done by using just a couple of 'if' conditions.
•  » » 8 days ago, # ^ |   0 Yes,it was a bit tricky. Infact i was also thinking bfs kind of stuff. But later realized that it can be solved just with a greedy aproach.
•  » » 8 days ago, # ^ |   0 I think if we are asked to minimize c then BFS should be use!! Like there will be cases such as S and F has 0 in there adjacent cells but all other cell is having 1 then optimal c will be zero.
•  » » 8 days ago, # ^ |   -73 I tried the bfs approach as well but failed could you pls send me your submission link ? I am interested to see it. Mine is this — don't why it doesn't work tho ? #include using namespace std; int n; int grid[205][205]; bool visited[205][205]; void dfs(int cx, int cy, int obstacle) { if(cx == n-1 && cy == n-1) return; visited[cy][cx] = true; int dx[] = {-1, 0, 0, 1}; int dy[] = {0, 1, -1, 0}; // explore neighbours in all 4 directions for(int i=0; i<4; ++i) { int nx = cx + dx[i]; int ny = cy + dy[i]; if(nx < 0 || nx >= n || ny < 0 || ny >= n) continue; if(grid[ny][nx] == obstacle) { continue; } if(!visited[ny][nx]) { dfs(nx, ny, obstacle); } } } void solve() { char start_; cin >> start_; grid[0][0] = -1; for(int i=0; i> grid[i][j]; } } grid[n-1][n-1] = -1; char end_; cin >> end_; // to tell if we can approach the finish block from top or left side bool top = false; bool left = false; // choose digit 1 dfs(0, 0, 0); if(grid[n-2][n-1] == 1 && visited[n-2][n-1]) { left = true; } if(grid[n-1][n-2] == 1 && visited[n-1][n-2]) { top = true; } // Reset visited array for(int i=0; i> t; while(t--){ cin >> n; solve(); } return 0; } 
•  » » » 8 days ago, # ^ |   0 I tried exactly the same stupid approach :(
•  » » 8 days ago, # ^ |   0 We can't avoid BFS to check if we can reach F right??
•  » » » 7 days ago, # ^ | ← Rev. 2 →   0 No, It can be done in O(1), just think of all the cases.
•  » » » 6 days ago, # ^ |   +1 The thing is you're actually not required to check that (you don't need to optimize for minimum "c")
•  » » 8 days ago, # ^ |   0 it is just tricky we have to just take care of 4 nodes i.e is the two neighbouring nodes of S and two neighbouring nodes of F and then we just have to make a combination like this 1. all neighbours of S are 1 and all neighbours of F are 0 2. all neighbours of S are 0 and all neighbours of F are 1
•  » » 8 days ago, # ^ |   0 Same..but i could'nt solve it btw
 » 8 days ago, # |   +1 Wow so fast editorial!
 » 8 days ago, # |   +10 C's testcase contains a major hint.
•  » » 8 days ago, # ^ |   +12 In constructive problems, the problem-setter always try to hidden the right construcion approach and even use some misleading approach in sample output, so I prefer to ignore the sample output in such problems.However, this time it did help lol
•  » » 8 days ago, # ^ |   -19 +1
 » 8 days ago, # | ← Rev. 4 →   +16 Another way to go about D is to think of the hexagonal world as a regular 2D grid, where you can move from (0,0) to six directions. [all directions except (-1,1) and (1,-1)]Then it's just checking all the possible routes.NB: The image is not uploading correctly for some reason. If somebody knows why, please let me know. Thanks.
 » 8 days ago, # |   0 Feeling depressed after seeing the solution of C and also that I thinked on it for an Hour , I got the hint part while thinking , but didn't got that we can move n-1 to n+1 .
•  » » 2 days ago, # ^ |   0 I was a little depressed too, it took me 2 days thinking about task C to be able to solve it.A good tip is to just look at the editorial when you think the problem uses some algorithm that you don't know. Otherwise, the best thing to do is not to look at the editorial and fight against task.
 » 8 days ago, # |   0 For A, even OR works (a^(a|b))+(b^(a|b))
•  » » 8 days ago, # ^ |   +12 Just use a^b as the answer.
•  » » » 8 days ago, # ^ |   +3 Can anyone actually prove this?
•  » » » » 8 days ago, # ^ |   0 Greedy approach, you just take xor of bigger number with both the numbers,bigger one vanishes and we get the answer which is basically xor of bigger and smaller one.
•  » » » » 8 days ago, # ^ |   +3 If you're talking about proof of why XOR a^b works, then it's this way. Let's take i'th digit from both numbers (in binary). 1. If both of them are 1, then for x we put there 1 as well. This way we'll make that digit become 0 in our answer. 2. If one of them is 1, then no matter what we put there at x, that place in the answer will be 1. 3. If none of them is 1, then we need to put 0 there at x, that place in the answer will be 0. From these approaches, you can see that if digits in respective places are same, then that digit will be 0 in our answer, otherwise, it will be 1. And this is what XOR does indeed.
•  » » » » » 6 days ago, # ^ |   0 thank you so much!!!!
•  » » » » 7 days ago, # ^ |   0 Because the only one situation that can decrease the answer is that same bit of a and b is 1 So actually x=a&b for example if a=1011 and b=1101 than x=a&b=1001 let's foucus on zeroes which means in that bit a is different with b there are 4 situation which NO.1 2 3 4 a= 00 01 10 11 b= 11 10 01 00 We found that for each case the equation (a xor x)+(b xor x)=11=a xor b for the zero bits of x let think about 1 bits in x wo know x=a&b ,so only if the same bit of a and b is 1 can makes the same bit of x equal 1, and obviously 1 xor 1 equals 0 So for the 1 bits in x, wo also get a equation (a xor x)+(b xor x)=00=a xor b By combining the two formulas, we can get the formula (a xor x)+(b xor x)=a xor b Have fun :>
 » 8 days ago, # |   +6 Another way to solve A is we can find the minimum value if we take same 2 digits like 6^6 is zero. so we can take the minimum value from a and b and put that value in x and solve the equation as it said. Simple as that.
•  » » 8 days ago, # ^ |   0 fantastic!
 » 8 days ago, # | ← Rev. 2 →   0 .
 » 8 days ago, # | ← Rev. 2 →   0 Strong pretests for B https://www.imageupload.net/image/inkedadsiz-li.tW8IJ
 » 8 days ago, # |   +7 Can someone please explain the dp solution in E? Thanks in advance :-)
 » 8 days ago, # |   0 https://paste. ubuntu.com /p/Y 9yB9rvZ89/ What I wanted to say
 » 8 days ago, # |   0
 » 8 days ago, # |   +3 How didn't this solution 95865345 get RE on pretests?
 » 8 days ago, # | ← Rev. 2 →   0 Edit: nvm
•  » » 8 days ago, # ^ |   0 n is inital strings length not current
•  » » 8 days ago, # ^ |   0 The n in (L, n) is w.r.t the new string, which is of length n+1 (after the first (R n-1)).
 » 8 days ago, # |   -16 y u no post code? maybe v vill learn new technique? or dat is bad??? u have code somewear obviously >:(
 » 8 days ago, # |   +55 After solving A-D : Why did I learn algorithms ? ;ㅅ;
•  » » 8 days ago, # ^ |   -20 Haha, I thought that C may have needed knowledge of the KMP algorithm... :(
•  » » 7 days ago, # ^ | ← Rev. 2 →   +3 Wonsei your solution for problem D looks so simple. Can you please explain your idea and solution in a short? I am not understanding the editorial.
•  » » » 7 days ago, # ^ | ← Rev. 2 →   +3 You can see here that to go from the first dot to the second dot, you can take two choices, c1 or c2+c6. so, we call n1(the cost to the right upper hexagon) minimum of those two values. You do that for all 6 directions. After that, you just go the shortest path to the destination from 0,0 using the new values.
 » 8 days ago, # |   +9 I had a bit different solution to D.it is easy to notice, that we need to move on diagonals, or straight line. And, number of different points we need is quite small. After drawing on piece of paper it is obvious that we need only some cominations of $x, y, x-y, y- x$. So, I added points $(0, x), (0, y), ...$ (ones on staight gorizontal line with $(0; 0)$), then $(x, 0), (y, 0) ...$ — first diagonal, and $(x, x), (y, y) ...$ on second.We have 14 points, its easy to calculater distance between points on same diagonal or line. After doing so I run Floyd-Warshall on this graph, so all solution works in $O(14^3) = O(1)$;) per test case. Code.
•  » » 8 days ago, # ^ |   +3 Did the same without running Floyd-Warshall :)https://github.com/actium/cf/blob/master/1400/20/1421d.cpp
•  » » 8 days ago, # ^ |   0 wow, a nice approach indeed , could you share how did you come up with x-y,y-x etc points, its because of thinking in this manner that i am unable to solve.I mean could you help in visualising how we will reach these points, understanding x,y,0 combinations is easier.
•  » » 8 days ago, # ^ |   0 Great idea! I did it and 9 points is enough.
 » 8 days ago, # | ← Rev. 2 →   +22 How to find the pattern of problem E in 24 minutes like kefaa2? I found it hard for me to solve this kind of problem...
 » 8 days ago, # |   0 Just wanted to confirm something about B: the person cannot move from (1,1) to (2,2) right? Basically diagonally?
•  » » 8 days ago, # ^ |   0 Yes. Only right, left, up, or down..
 » 8 days ago, # | ← Rev. 2 →   0 Why my similar approach isn't working for Cfor example: abcd cbabcd cbabcd cbab cbabcdcbab abc printf("L %d\n",n-1); printf("R %d\n",n-2+1); printf("R %d\n",n-2+ n + 1);
•  » » 8 days ago, # ^ |   +1 Don't know if the solution's correct, but maybe put "3" on the top of the answer
•  » » » 8 days ago, # ^ | ← Rev. 2 →   0 Oh, I had forgot to print 3. got it accepted
•  » » 8 days ago, # ^ |   0 Did you print 3?
 » 8 days ago, # |   0 So fast OoO
 » 8 days ago, # |   -8 On problem B, can anyone please tell me what does this statement really means "Waters can move from a square to any other square adjacent by a side, as long as he is still in the grid". Can Waters move from grid[i][j] to grid[i+1][j+1]?
•  » » 8 days ago, # ^ |   0 Yes waters can move from gird[i][j] to 1 position left, or right, or top, or down, on the condition that that position should be in the grid.
•  » » » 8 days ago, # ^ |   0 In this case, 3 S10 101 01F He cannot reach to grid[n][n],then why we have to invert the squares grid[1][2] and grid[2][1]?
•  » » » » 8 days ago, # ^ |   0 Yeah, that's what I had asked as a question during the contest, the reply was "there can be multiple answers"
•  » » 8 days ago, # ^ |   0 "adjacent by side"grid[i+1][j+1] is NOT adjacent by side to grid[i][j]
•  » » 7 days ago, # ^ |   0 No as grid[i][j] and grid[i+1][j+1] are not adjacent by a side. They don't have a common side.
 » 8 days ago, # |   0 In A) when they say the first bit, I would have said the first bit was the most left hand one. However, I also would have said that b's first bit is 0. Where am I wrong?
 » 8 days ago, # | ← Rev. 2 →   0 I swear I got the answer for B here, https://codeforces.com/submissions/Lithosphere , does anyone know why I failed pretest 2?
•  » » 8 days ago, # ^ |   0 In problem A,I just did a loop to find the x and for all the pair in the test case 1,I got 0.
•  » » 8 days ago, # ^ |   0 I can't view your solution through this link.
•  » » » 8 days ago, # ^ |   0 https://dpaste.org/BG46couldnt workout how to share my contest just be using codeforces links then
 » 8 days ago, # |   +11 I really hope this hint/solution format of editorials becomes the norm :)
 » 8 days ago, # |   -12 Fuck You Ringo!
 » 8 days ago, # |   0 Explain this solution to me please. what is delta? what is delta_a what is delta_b?? I think a and b are the mvoes along the i and j axis. but i cant find formula for delta_a and delta_b. Thanks 95917963 stefdasca
•  » » 8 days ago, # ^ |   +1 I'm not sure why they are named that way in the code, but I recognize that it looks just like what I did. If you set up the linear equations as a matrix, delta is the determinant of the matrix and delta_a/delta_b are the matrix product of inv(A) * (x, y). When divided by delta (the determinant) they give the solutions to the linear equations (i.e. how many steps in each direction are needed).
•  » » » 8 days ago, # ^ | ← Rev. 2 →   0 Oh thanks so much. I got what you mean. Also derived it in image below for any noobs like mehttps://drive.google.com/file/d/1NIn2mh3Ew3IFtHE_6uyeqPH5XspwexIx/view?usp=sharing
 » 8 days ago, # |   0 I tried every possible case in 2nd question. Can anyone tell me where it went wrong ? https://codeforces.com/contest/1421/submission/95878916
•  » » 8 days ago, # ^ | ← Rev. 3 →   -10 YOUR LAST TESTCASE IS WRONG: IT IS WRITTEN => else if (a[0][1] == '0' && a[1][0] == '1' && a[n - 1][n - 2] == '0' && a[n - 2][n - 1] == '0') { cout << "1\n" << 1 + 1 << " " << 0 + 1; } SHOULD BE: else if (a[0][1] == '0' && a[1][0] == '1' && a[n - 1][n - 2] == '0' && a[n - 2][n - 1] == '0') { cout << "1\n" << 0 + 1 << " " << 1 + 1; } Accepted 95920491your cout line is wrong broo
•  » » » 7 days ago, # ^ |   0 Thank You :) got the mistake, silly one because I wrote such a brute force solution it became difficult for me to debug.
•  » » » » 7 days ago, # ^ |   0 its fine. I also wrote brute force. better for u to use small variables for a[0][1], a[1][0] etc. I used br,bl,tr,tl for bottom right, bottom left, top right, top left etc.
 » 7 days ago, # |   +12 Magical DP for E: 95899682
•  » » 7 days ago, # ^ |   0 Many red coders have used the dp for E. Can you please explain the dp solution? Thanks :-)
•  » » » 7 days ago, # ^ |   +3 Basically, the first dimension describes the number of minus signs placed modulo 3, and the other dimension is a boolean flag describing whether or not we "broke" the plus-minus-plus pattern or not. In the end, the number of minuses has to be equal to $2n+1$ modulo 3 and the flag must be on.
•  » » » » 7 days ago, # ^ |   0 I have read the editorial and I understand the corner case and the proof.But I still can not understand the way of dp transfer.I have seen this solution: https://codeforces.com/contest/1421/submission/95897364, it is more clear in state transfer.It seems that when we put '+' in an even position (or put '-' in an odd position), we can transfer the illegal state to the legal state.Why we can do this?If you have time, can you explain that? Thank you a lot!
•  » » » 7 days ago, # ^ | ← Rev. 2 →   0 the problem is equivalent to : assign + and — to every number , query the max sum,but "+-+-+-"is illegal and (the number of '-' + n) mod 3 =1. then the dp is obvious.
»
7 days ago, # |
Rev. 2   -29

# define ll long long int

int main() {

string s;
cin>>s;
ll n= s.size();
ll i,ans=0;
for(i=0;i<(s.size()/2);i++)
{
if(s[i] == s[n-i-1])
ans++;
}
if(ans == n/2)
{
cout<<"0"<<endl;
}
else
{
cout<<"1"<<endl;
cout<<"L"<<" "<<n<<endl;
}

}

•  » » 7 days ago, # ^ |   0 the question clearly says only up to n-1,where n is the length of the string and u are using n.
 » 7 days ago, # |   +6 Can someone explain what's the intuition behind the pattern in Problem E. DP aside, it seems very hard to just observe that pattern from the problem. The tutorial proves that it's right but how can we come up with something like that in the contest.
•  » » 7 days ago, # ^ | ← Rev. 2 →   +8 You can brute force it and look at the set of sequences of signs you get
 » 7 days ago, # | ← Rev. 3 →   0 UPD: Found the bug!
 » 7 days ago, # |   0 Just exact same procedure I had followed for problem c as editorial
 » 7 days ago, # |   0 Can anyone tell me the detailed solution of Problem D. I got the part that we need to consider only two edges but how to calculate the distance after that?
•  » » 7 days ago, # ^ | ← Rev. 2 →   +1 find two or one direction to get to the desired hexagon. Next, try to reduce the cost of transition by adjacent hexagons (c [i] = min (c [i], c [i + 1] + c [i — 1]);). Then you find the number of steps and find the answer in long long. I found the distance just by brute force, but I'm sure that it can be better)).
•  » » » 6 days ago, # ^ | ← Rev. 2 →   0 Update: got it.
•  » » 7 days ago, # ^ |   0
 » 7 days ago, # |   0 I didn't think about quadrants and shortest paths in problem D
 » 7 days ago, # |   +1 Can someone please explain a solution which involves linear algebra/matrices on problem D?
 » 7 days ago, # |   +34 More like IfElseForces.
•  » » 6 days ago, # ^ |   0 I think so. You can even solve A-D problem without any algorithm.
 » 7 days ago, # | ← Rev. 2 →   0 An interesting adaption based on B: For a given graph, what is the minimum number of invertions we need to make so that Pink Floyd can successfully go through (1,1) to (n,n)?For example: 4 S000 0000 0001 001F Then the answer should be 1. For if we invert the cell (3,4), Pink Floyd can go from (1,1) to (n,n). And it can be proved that 1 is the most optimal answer. So how to solve this?
•  » » 7 days ago, # ^ |   0 An optimal set of inversions must either only change 0-cells to 1-cells and create a path consisting of 1-cells, or only change 1-cells to 0-cells and create a path consisting of 0-cells. So, you can run two instances of either 0-1 BFS or Dijkstra's algorithm, one to find the cost of making a path of 0-cells, and one to find the cost of making a path of 1-cells. Then the answer will be the minimum of these two values.
•  » » » 7 days ago, # ^ | ← Rev. 2 →   0 Thx a lot !
 » 7 days ago, # |   0 I am surprised that nobody I've seen did binary search for D... (The official solution seems easier, though :p)
 » 7 days ago, # | ← Rev. 2 →   0 I'm sorry if this is duplicated question. Why problems aren't rated after contests?
•  » » 7 days ago, # ^ |   0 I can take time to set problem ratings after end of round.
 » 7 days ago, # |   +1 I got the idea for B using 16 if-else immediately. But later spent 1 hr thinking how to implement it better. Finally ended up with an 8 if-else implementation.. Should have gone with the 16 if-else implementation itself...
 » 7 days ago, # | ← Rev. 2 →   +9 How to prove the conclusion of E((n+m)%3=1) via induction？
•  » » 4 days ago, # ^ | ← Rev. 2 →   0 This is clearly true for the base cases when $n = 1, 2, 3$. Now, consider a general $n$. Suppose for a moment that you have already performed the sequence of operations $n - 2$ times, and are therefore left with just two remaining elements: $X_1$ and $X_2$. Note that $X_1$ is really just the sum (with possible -1s in front) of some elements of $a$. Let $n_1$ be the number of elements in that sum, and similarly define $n_2$. (Note: $n$ = $n_1$ + $n_2$). Let $m_1$ and $m_2$ be the number of elements with -1 in front in $X_1$ and $X_2$ respectively. By our induction hypothesis, we have that $m_1 + n_1 \equiv 1 \text{ (mod 3)}$ $m_2 + n_2 \equiv 1 \text{ (mod 3)}$Let $m$ be the number of terms with -1 in front in our final sum after doing the operation to $X_1$ and $X_2$. Then $m = n_1 - m_1 + n_2 - m_2$.Finally: $n + m \equiv n_1 + n_2 + n_1 - m_1 + n_2 - m_2 \equiv 2n_1 + 2n_2 - m_1 - m_2 \equiv 2n_1 + 2n_2 - (1 - n_1) - (1 - n_2) \equiv 1 \text{ (mod 3)}$
 » 6 days ago, # |   0 It's very nice to see this form of Tutorial.Thanks a lot!
 » 6 days ago, # |   0 Can we solve problem D with using DFS??.
 » 6 days ago, # |   0 mikemirzayanov any updates when will rating get updated for previous rounds problems ?
 » 6 days ago, # |   0 Can someone explain why ternary search will work for D? I mean ternary search like this:We do ternary search on distance moved along $y=x$ using C1 or C4. Then for remaining we use the Manhattan distance and use the required amount of other direction vectors. Why does this function always form a V shaped graph?
 » 6 days ago, # | ← Rev. 2 →   0 How to do it for multiple bits? Single bit proof is trivial for this : We'll leave it up to you to prove that (a ⊕ (a & b)) + (b ⊕ (a & b)) = a ⊕ b, where ⊕ is the bitwise XOR.Upd: I got the proof. Thanks.
 » 2 days ago, # |   0 If you print: 3 L 2 R 2 R 2n-1 `At C task, you'll get ac too, it's good to think about that.
 » 14 hours ago, # |   0 In problem B, I don't know why in test case 2, the answer c = 2. I thought it should be 0 because after Waters goes into cell (1,2) or (2,1) he has no any cells to go? Blame me if I'm wrong, please just tell me why?