### JaySharma1048576's blog

By JaySharma1048576, 4 months ago,

I am sorry for the weak tests in B, for C being a little standardish and for misjudging the relative difficulties of E and F. Nevertheless, I hope you enjoyed the round. Here is the editorial. Do provide your feedback on each problem so that I can improve upon them the next time.

Hint 1
Hint 2
Tutorial
Solution
Feedback

Hint 1
Tutorial
Solution
Feedback

Hint 1
Hint 2
Tutorial
Solution
Feedback

Hint 1
Hint 2
Hint 3
Tutorial
Solution
Feedback

#### E. Power or XOR?

Hint 1
Hint 2
Hint 3
Hint 4
Tutorial
Solution
Feedback

Hint 1
Hint 2
Tutorial
Solution
Feedback

• +93

 » 4 months ago, # | ← Rev. 2 →   +18 I spend at least 5 times as much time on B as I do on C ...
•  » » 4 months ago, # ^ |   0 It is alright mate, I did B in 15mins, but I could not solve C at all...
•  » » » 4 months ago, # ^ |   0 Couldn't solve B but managed C in <40 mins. :(
 » 4 months ago, # | ← Rev. 3 →   -69 .
•  » » 4 months ago, # ^ |   +1 maybe its because U solve it too late , and it gave barely any points
•  » » » 4 months ago, # ^ |   -65 They just could give me 0 pints not -78 !!
•  » » » » 4 months ago, # ^ |   +7 when you perform lower than ur current rating level ur rating will drop
•  » » » » » 4 months ago, # ^ |   -66 Okay , I won't enter any contexts anymore fuck their judgment
•  » » » » » » 4 months ago, # ^ |   +28 You shouldnt think like , contest bring your rating to your real performance , sometimes you can have rating drops but in the end when you look for like a 1 year rating graph , you will see a linear growth
•  » » » » » » 4 months ago, # ^ |   +19 Lol, bye.
•  » » » » » » 4 months ago, # ^ |   +8 Fun fact: For the first five contests you participate, you get an extra few hundred rating points as technically unrated accounts start at 1500 (behind the scene), and after that, there is no such thing.
•  » » » » » » » 4 months ago, # ^ |   0 https://codeforces.com/blog/entry/77890 I think it is 1400.
•  » » » » » » » 4 months ago, # ^ |   +3 You are right, my rating started to fall after my fifth contest. I always wondered why, thanks for this info.
•  » » » » 4 months ago, # ^ |   +6 If CF worked like that I could have been red...
•  » » 4 months ago, # ^ |   0 I like the idea of "I solved A, so I get some points" ;) We good looking people must stick together!
 » 4 months ago, # |   0 I couldnt understand how do we prevent counting the same thing over and over in C
•  » » 4 months ago, # ^ | ← Rev. 3 →   0 This code won't count the same thing for (int i = 1; i <= tot; i++) { for (int j = num[i]; j <= maxn - 1; j++) { dp[j] = (dp[j] + dp[j - num[i]]) % mod; } } but if you swap the the order of i and j, you'll get wrong answer because you count the same thing
•  » » 4 months ago, # ^ |   0 instead of counting for T test cases, you only count once for all the possible values of n and then get the answer for each test case in O(1). So your final answer would be O(498*N + T), not O(T*N*498). I hope that it helps.
 » 4 months ago, # | ← Rev. 2 →   +9 in the editorial of problem C, aren't there only M = 498 palindromes?
•  » » 4 months ago, # ^ |   +9 Yeah, you are right. It was a typing error and has been fixed now.
•  » » » 4 months ago, # ^ |   -18 That's called typo in English
 » 4 months ago, # |   +5 Here are the video Solutions for A-D in case someone is interested.
 » 4 months ago, # | ← Rev. 2 →   -27 /
 » 4 months ago, # |   +12 If you prefer command line, checkout CF Doctor for more flexibility and faster feedback.If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints. If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).
 » 4 months ago, # |   +23 Finally reached green :))
•  » » 4 months ago, # ^ |   0 Hey man do you recommend learning dp to reach pupil? Or should I just focus more on basics? Congratulations btw :)
•  » » » 4 months ago, # ^ | ← Rev. 3 →   0 Thanks! About dp, I didn't really know dp that well, I was usually doing graph and bitwise operation problems because it was hard for me to understand. But I didn't use them xd. I recommend you to do the math and basic problems, basically, anything you don't feel comfortable doing, so you can improve some bit of your skills slowly, but surely. Also don't forget to check out editorials, even if you solve the problem because it happens that you solve it "bad" way that doesn't make you any good, and by seeing editorials you can understand the true idea behind solutions, so you can see how coding works from the perspective of "professional" coders who created the contest. Good luck improving your skills, hopefully, you will reach green soon, but don't forget that it is necessary to be fast, first of all, improve yourself and the new color will come as if nothing happened :)))
•  » » » » 4 months ago, # ^ |   0 Thanks mate! And about the editorial thing, yes it's so true! I mean in this contest itself I solved B in a weird way. When I looked at the editorial, only then I came to know that it could have been solved much more easily :)
•  » » » » 7 weeks ago, # ^ |   0 Hey man! Just came here to let you know that I finally reached green hehe :)) And I still don't know dp lmao
•  » » » » » 7 weeks ago, # ^ | ← Rev. 5 →   0 Hey, I just found this book about CP, it is a good one, has lots of stuff in it explained well, you should check it out if you have not heard of it before, it is called "Competitive Programming Handbook", you can search it up on google, it is a book and has lots of in it that can help you improve. It also has a site where you can check your solutions to problems from the book. Go to cses.fi -> courses -> CSES Problem Set and then you can practice problems on topics of your choice. Good luck!
•  » » » » » » 7 weeks ago, # ^ |   0 Yup I know about that book. I learnt some basic stuff from there, but most of the stuff from the book is just too advanced for me lol.
 » 4 months ago, # |   +6 Thanks for the fast editorial!
 » 4 months ago, # |   0 For problem C, I don't get why the second term of the dp is $dp_{k-p_m, m}$ but not $dp_{k-p_m, m-1}$, because we have the mth palindrome number fixed and then we have m-1 other palindromes that need to be chosen to give our answer
•  » » 4 months ago, # ^ |   0 Because if we can use the same number more than once, so even if we use the mth palindrome number ending at k, we can still use all other m palindrome numbers
•  » » 4 months ago, # ^ |   0 Because you still need to use pm to construct k-pm
 » 4 months ago, # |   0 will I reach pupil before I die?
 » 4 months ago, # |   +81 Problem FHint1 says "try to find the one which has the least sum of road lengths."I want to know the proof that 47616 is the least sum.
•  » » 4 months ago, # ^ | ← Rev. 4 →   +8 Update: My bad, the solution below is worse than the standard solution. It should be $x_k=16x_{k-1}+3\times 2^k$. Apologies for incovenience.I don't know whether my solution reaches the least sum possible, but it seems much better than the one given in the tutorial.Let's recursively construct the matrix $A$ mentioned in the tutorial. We are going to construct a $2^k\times 2^k$ matrix $A_k$ for each $k\in [0,5]$, thus giving a solution to the original problem by taking the first $n$ rows and columns of $A_5$.The base case $A_0$ is simple, with a single $0$. Here is my way to get $A_k(k\geq 1)$ by $A_{k-1}$: flip $A_{k-1}$ to its right side to form a $2^{k-1}\times 2^k$ matrix (let's say, $B_k$), then flip $B_k$ to its down side to form a $2^k\times 2^k$ matrix $C_k$. Finally, we multiply each element in $C_k$ by $4$, and add $0,1,2,3$ to the up-left, up-right, down-left, and down-right part of the matrix respectively to form $A_k$ (by the up-left part I mean the submatrix consisting of elements in both the first $2^{k-1}$ rows and first $2^{k-1}$ columns, the other three are defined in similar ways).Let $x_k$ be the sum of length of edges in $A_k$. One can find that $x_0=0$, and $x_k=4x_{k-1}+3\times 2^k$. By using the calculator it turns out that $x_5=2976$, far lower than the problem's original constraints.
•  » » » 4 months ago, # ^ | ← Rev. 2 →   0 Do you have an implementation?
 » 4 months ago, # |   +8 Finally able to use some standard dp I learnt in a rated contest. Love it! XD.
 » 4 months ago, # |   +23 Is C too standard?
 » 4 months ago, # | ← Rev. 2 →   0 loved D ;D
 » 4 months ago, # |   +71 I have a solution for problem F that constructs the same grid but with a completely different intuition. So let's say for a fixed row, all the values should be the same, that way you can know if you have passed that row before. Since if you ever make a "redundant" move, you will have to cross it again and the xor will cancel out. The same logic applies to the columns. So then you try to use 5 bits to manipulate the information for rows and 5 bits for the columns. Think about the line that evenly splits the rows into two halves. If you put a line of all of the same bit there, simply by checking if that bit is on, you can see which halves of the partition the rows is a part of. Then you have a smaller problem of half the rows to deal with. Finally, when there is only 1 row left, you can just leave the last bit on to check if it has crossed that row or not. The same logic applies to the columns. Now you also have to pick the bits rows to manipulate and the bits the columns manipulate separately. I found that the choice of rows[] = {1, 4, 16, 64, 512} col[] = {2, 8, 32, 128, 256} fits within the limit of $48000$. To construct the answer itself, you check if a certain bit is on, and if it is, you update the respective row/column value and also xor x to make sure that the contribution of the prefix is considered.My implementation of the idea is here.
•  » » 4 months ago, # ^ |   0 Have the same logic in my solution.
 » 4 months ago, # |   0 Solution to bonus task of problem C : 155417311 My first DP problem solved in a contest even though it was similar to standard coin change problem.
 » 4 months ago, # |   0 I like B
•  » » 3 months ago, # ^ |   0 Could you explain the initial thought process to solve B. I have read the editorial yet don't understand how should one begin to think about the problem to reach the solution.
 » 4 months ago, # | ← Rev. 2 →   +6 In editorial for problem Bs will be perfectly balanced if and only if si,si+1,…,si+k are all pairwise distinctshoudn't it be si,si+1,…,si+k-1 since there are k distinct chars
•  » » 4 months ago, # ^ |   0 Yes, it has been corrected now.
 » 4 months ago, # |   +3 My A problem was garbled but I passed the in-match test and got RE after the match because the array was opened too small.How happy I was at the end of the race and how sad I was after the retest.｡ﾟ(ﾟ´Д｀ﾟ)ﾟ｡
 » 4 months ago, # |   +6 Here is another way to calculate $\binom{n}{m} \bmod 2$. We know that $\binom{n}{m} = \frac{n!}{m!(n - m)!}$. Define $P(n) = \max\{k : 2^k | n!\} = \sum_{i = 1}^{+\infty} \lfloor \frac{n}{2^i} \rfloor$. Then $\binom{n}{m} \bmod 2 = 1$ iff $P(n) - P(m) - P(n - m) = 0$.Obviously, we can calculate $P(n)$ in $O(\log n)$ time, but here is a more efficient way. Suppose $n = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_m}$. Then \begin{aligned} \sum_{i = 1}^{+\infty} \lfloor \frac{n}{2^i} \rfloor &= \sum_{j = 1}^m\sum_{i = 1}^{+\infty} \lfloor \frac{2^{a_j}}{2^i} \rfloor \\\\ &= \sum_{j = 1}^m\sum_{i = 1}^{a_j} 2^{a_j - i}\\\\ &= \sum_{j = 1}^m (2^{a_j} - 1)\\\\ &= n - m \end{aligned}Obviously, $m = \text{popcount}(n)$, the number of $1$(s) in binary form of $n$, which can be calculated fast in C++ with __builtin_popcount(n) or __builtin_popcountll(n) (when n is long long type). This improvement helped me from TLE to AC, and shows that __builtin_popcount really has very small constant. :P
•  » » 4 months ago, # ^ |   +26 It's Legendre formula.
•  » » 3 months ago, # ^ |   +5 easer way is user lucus thero, (n+m, n)(i.e., n+mCn) is even if (n^m == n&m) so here ncr is even iff (n-r)^r == (n-r)&r https://math.stackexchange.com/questions/11002/cn-p-even-or-odd
 » 4 months ago, # |   0 I have coded almost the same solution (with same time complexity) for 1673C - Палиндромный базис but still TLE! Can someone help me understand why? int checkPalindrome(int n) { int original = n; int remainder; int reversed = 0; while (n != 0) { remainder = n % 10; reversed = reversed * 10 + remainder; n /= 10; } if (original == reversed) return 1; return 0; } int mod = 1e9 + 7; vector arr; int n; void helper() { for (int i = 0; i <= 40000; i++) { if (checkPalindrome(i)) { arr.push_back(i); } } n = arr.size(); } void solve() { int sum; cin >> sum; vector> dp(n + 1, vector(sum + 1, 0)); for (int i = 0; i <= n; i++) { dp[i][0] = 1; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= sum; j++) { if (arr[i] <= j) { dp[i][j] = (dp[i][j - arr[i]] + dp[i - 1][j]) % mod; } else { dp[i][j] = dp[i - 1][j]; } } } cout << dp[n][sum] << endl; } 
•  » » 4 months ago, # ^ |   +14 You recompute $dp$ for every testcase, so the number of operations is on the order $10^4 \times 4 \cdot 10^4 \times 500 = 2 \cdot 10^{11}$.Instead precompute the $dp$ table once before all testcases, then simply output $dp[n][sum]$ in each test case.
•  » » » 4 months ago, # ^ |   0 Thank you so much!
•  » » 4 months ago, # ^ |   0 Don't calculate dp array every time. It is enough to calculate it only one time at the very beginning.
 » 4 months ago, # |   +2 i solved A and B after contest
 » 4 months ago, # | ← Rev. 2 →   0 Hey if some one can tell me what is the max size of vector we can have of long long int in c++? AS for problem B I had something of 1e6 but neither it gave MLE nor any other error.
•  » » 4 months ago, # ^ |   0 One long long int occupies 8 bytes. The array of 1M values itself occupies 8 Mbytes of memory. vector ocupies more but not twice. How many arrays do you create?
•  » » » 4 months ago, # ^ |   0 actually it was vector not array and its maximum size will be 26*2*1e5 155418148you can check here . I am doing same with array and its giving run time error with array in my own system but for vector its accepting I don't know why though
 » 4 months ago, # | ← Rev. 2 →   0 I need a help, in problem B firstly we need to find out that how many distinctive characters are there. So, i wrote a function in C to find out the number of elements which are distinctive. So my question is: What is the more efficient way to find out the distinctive elements in a string? here is my code in C: long int DistinctChars(char *s) { long int len = strlen (s); long int i, j, d = 0; for (i = 0; i < len; i++) { if (s[i] != '^') { for (j = i + 1; j < len; j++) { if (s[i] == s[j]) { s[j] = '^'; } } d++; } } return d; } 
•  » » 4 months ago, # ^ |   +3 There are only 26 letters. You could make an array of 26 zeroes and mark with 1 every letter found. The sum of the array is an answer.
•  » » » 4 months ago, # ^ |   0 wow, amazing idea, thanks Sir.
 » 4 months ago, # |   +3 The Bonus Tasks are really educational !! Thanks a lot
 » 4 months ago, # |   -6 In editorial for B, do you mean $s_i, s_{i+1}, \dots s_{i+k-1}$
 » 4 months ago, # |   0 For the editorial of Problem D:Could you explain why For a particular value of p, there are r/p possible values of a satisfying conditions 2 and 3 and r/p possible values of l satisfying conditions 2 and 4?
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 Given p is a factor of r which we take as common difference of array A. Then let elements of C be x1 x2 x3 x4....xn the number of terms of A b/w xi and xi+1 will be (x2-x1)/p-1= r/p-1before x1 there could be thus (r/p-1) +1(no number to left of x1) = r/p numbers this because the number x0=x1-rx0 is not in C which implies it is not in A since x0 = x1-k*q (r=k*q) ( Not taking infinity case into consideration here)
•  » » 4 months ago, # ^ |   0 Let p is the step of sequence A. Sequence A contains z — the first value of C. Sequence A can start exactly from z, from z-p, from z-2*p, ..., from z — (r/p-1)*p. Total is r/p variants. Absolutely the same situation at the end of seqence. It can stop at f, at f+p, at f+2*p, ..., at f+(r/p-1)*p. So every divider p adds (r/p)*(r/p) variants.
•  » » 4 months ago, # ^ |   0 Thanks, I got it!
 » 4 months ago, # |   0 Can someone explain how to extend 2^k*2^k gray code to 2^k*2^(k+1) please?
 » 4 months ago, # |   +37 I have a very simple solution for problem F:First, lets look at the 1d version of the problem. Recall that the action of $+1$ or $-1$ is just flipping the first few bits of the number in binary representation. Firstly, 0-index the buildings. For edge between $i$ and $i+1$, we will assign the length as the highest power of 2 that divides $i+1$. For each query, we will check through the bits of $re$ (the number returned by the interactor). if the $ith$ bit is on, we will filp the first $i$ bits of the current $x$ coordinate.The sum of length, if $n$ is a perfect power of 2, is $n/2\times\log(n)$. Proof is left to the reader.By implementing for the $x$ coordinate on odd bit positions and $y$ coordinate on even bit positions, the sum is exactly $47616$. Proof is again, very simple and left to the reader. Implementation, which is quite short and pretty: here
 » 4 months ago, # |   0 Can anyone explain how the dp is working for problem C Palindrome Basis?? If anyone can share some resources then also it will be helpful.
•  » » 4 months ago, # ^ |   +3 I think the editorial has skipped a possible intuition for the dp states and transitions under the cover of an existing standard problem.During my upsolving I thought like this :-Since the problem states that sequences are differentiated solely on the basis of frequencies, why not think in terms of sorted sequences.I find out the list of palindromes in sorted order first.It's easy to see that if two sorted sequences are the same then they will have the same frequencies. My states for dp are :- $dp_{value,last}$basically the current sequence has a sum = value and we had appended some element x as the latest element which can be <= last.$dp_{value,last} = dp_{value,last−1} + dp_{value−p_{last},last}$So we can either have a sequence whose sum is = value and the latest element appended is <= last $-$ 1 and we do not append anything or we can have a sequence where we append $p_{last}$, in that case, our sequence's former last element should have been <= last and also its sum should be = $value - p_{last}$so transitions are just the same as in the editorialThis is one way to think 'intuitively' about the states perhaps there are other ways to go about it too.
•  » » » 4 months ago, # ^ |   0 Thanks for the analysis. I liked yours way of thinking . It felt easier to understand.
 » 4 months ago, # |   +10 My solution for B was this:Assume $S$ — set of distinct letters in string $s$. Let's check every substring $s_iAs_j$, where $s_i = s_j = c,\,c \not\in A$ for every $c \in S$. Note that $A$ can be empty ($A = \emptyset$).We give answer NO if there is letter $\hat c \ne c$ such that $\hat c \in S$ and $\hat c \not\in A$. That means that substring $s_i A s_j$ has two letters $c$ and zero letters $\hat c$. Otherwise the answer is YES.My submission: 155416323P.S. I guess there might be a problem with time complexity since we have two nested cycles for letters $a...z$. But I think either we find NO fast, either the second cycle will run only each $k$ iterations, where $k$ -- number of distinct letters in $s$. Correct me if I'm wrong.P.P.S In my code I should break the cycle if I find NO, it was my fault :D
 » 4 months ago, # |   0 This is a very great editorial, thanks!
 » 4 months ago, # |   0 Can someone explain, or at least give a hint, about the reason why the formula for problem F is true? This: $\dfrac{3}{2}(4^m)(2^m - 1)$
 » 4 months ago, # |   -21 Problem C What is the problem in this code: (memory limit exceeds) unordered_map dp[502]; vector pd; long long rec(int i, int sum){ // cout<=pd.size()) return -1; if(sum == 0){ return 1; } if(dp[i][sum] !=0) return dp[i][sum]%modulo; if(pd[i]>sum) return dp[i][sum] = -1; if(rec(i+1,sum)==-1) return dp [i][sum] = rec(i, sum-pd[i])%modulo; if(rec(i, sum - pd[i])==-1) return dp[i][sum] = rec(i+1,sum)%modulo; if(rec(i, sum - pd[i])==-1 && rec(i+1, sum) ==-1) return -1; return dp[i][sum] = (rec(i, sum - pd[i]) + rec(i+1, sum) )%modulo; } int reverse(int n) { int r=0; while(n>0) { r=r*10+n%10; n/=10; } return r; } bool palindrome(int n) { return (reverse(n)==n); } void solve(){ int sum; cin>>sum; long long x = rec(0 ,sum); if(x==-1) { cout<<"0\n"; return; } //rec(1,3); //cout<>t; while(t--){ solve(); // cout< pd; long long rec(int i, int sum){ // cout<=pd.size()) return -1; if(sum == 0){ return 1; } if(dp[i][sum] !=0) return dp[i][sum]%modulo; if(pd[i]>sum) return dp[i][sum] = -1; if(rec(i+1,sum)==-1) return dp [i][sum] = rec(i, sum-pd[i])%modulo; if(rec(i, sum - pd[i])==-1) return dp[i][sum] = rec(i+1,sum)%modulo; if(rec(i, sum - pd[i])==-1 && rec(i+1, sum) ==-1) return -1; return dp[i][sum] = (rec(i, sum - pd[i]) + rec(i+1, sum) )%modulo; } int reverse(int n) { int r=0; while(n>0) { r=r*10+n%10; n/=10; } return r; } bool palindrome(int n) { return (reverse(n)==n); } void solve(){ int sum; cin>>sum; long long x = rec(0 ,sum); if(x==-1) { cout<<"0\n"; return; } cout<>t; while(t--){ solve(); // cout<
 » 4 months ago, # |   0 Can someone kindly explain the implementation of B.
 » 3 months ago, # |   0 Very nice contest! I love super-super ad-hoc problems, they're the most interesting to me. Unfortunately I couldn't quite solve E in time, because I initially misread "at least $k$" as "exactly $k$", and then panicked :(
»
3 months ago, # |
Rev. 3   0

hi everyone, my solution to problem b is different and i think it's better using the concept to sliding window and maps, just wanted to share it:->

# include <bits/stdc++.h>

using namespace std;

# define endl " \n"

signed main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int t; cin>>t; while(t--){ string a; cin>>a; map<char,int> x; for(int i=0;i<a.length();i++){ x[a[i]]++; } int flag=0; int j=0; int i=0; map<char,int> ans; while(j<a.length()){ ans[a[j]]++; if(ans[a[j]]>=2){ if(ans.size()!=x.size()){ flag=1; break; } ans[a[i]]--;

if(ans[a[i]]==0){

flag=1;
break;
}
i++;
}
j++;
}
if(flag==1){
cout<<"No"<<endl;
}
else{
cout<<"yes"<<endl;
}

}

return 0;

}

SORRY for the mess, this was my first time posting, here is the link to the submission https://codeforces.com/contest/1673/submission/155436027

 » 3 months ago, # |   0 I am trying to solve E. wrote the same code as in editorial but still getting Wrong answer on test case 9. This is my code. Plz, let me know where I am going wrong. // Online C++ compiler to run C++ program online #include using namespace std; #define loop(i, s, e) for(long long int i = s; i < e; i++) #define lli long long int const int pow20 = 1048576; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); // Write C++ code here lli n, k; cin >> n >> k ; vector Arr(n); vector Ans(pow20+1,'0'); loop(i, 0, n) cin >> Arr[i]; // loop(i, 0, n) cout << Arr[i] << " "; // cout<=20) break; else pow *= ((lli)1<= pow20) break; int remainingN = n-j+i-3; int remainingK = k-2; if(i == 0) { remainingN++; remainingK++; } if(j == n-1){ remainingN++; remainingK++; } // cout << "=("<< remainingN <<","<< remainingK << ","<< pow <<") " ; lli x = remainingN-remainingK; lli y = remainingK-1; lli z = remainingN-1; if(remainingN>=remainingK && (remainingN ==0 || (remainingK>0 && (z == y|x)))) Ans[pow] = '0'+'1'-Ans[pow]; // Ans[pow] = Ans[pow] ^ 1; } } bool start=false; for(int i=pow20-1;i>=0;i--) { if(start && Ans[i]=='0' ) cout << 0; else if(Ans[i]=='1') { cout << 1; start=true; } } if(!start) cout << 0; cout << endl; return 0; } 
 » 2 months ago, # |   0 I got this as Checker log wrong answer expected YES, found NO [731st token]can you give me test case for 731 token ? 159332450
 » 7 weeks ago, # |   0 Problem F was very good. Thank you!
 » 4 days ago, # |   0 How to solve problem C in space complexity to O(n)?Please help!