### tianbu's blog

By tianbu, history, 4 weeks ago, A Tutorial
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E2 Code Tutorial of Codeforces Round #729 (Div. 2) Comments (132)
 » Great problems
 » Good stuff, I liked C a lot, it made me think of LCM in a new way
•  » » » yes plz how u thought of lcm i thought of 3 first then thought for 6 then 12 and come up with factorial approach but it hasn't worked.
•  » » » » 4 weeks ago, # ^ | ← Rev. 4 →   I'll try to explain.$n \leq 1e16$, so just starting a loop and trying to find an answer for each $i (1 \leq i \leq n)$ is a bad idea. This is why we need to "group" the answers. It is important to understand that the answers cannot be great. Because if $f(i) = x$, it means that i has divisors $2, 3, 4 \dots x - 1$, so $i \geq lcm (2, 3, 4 \dots x - 1)$Let's try to figure out what $⌊n / lcm (1,2, \dots, i) ⌋$ means. This is the number of elements that are divisible by $2, 3, 4 \dots i$, so we have not counted the answer for them yet. Thus, based on the editiorial "The number of ks such that $f (k) = i$ is $⌊n / lcm (1,2, \dots, i — 1) ⌋ — ⌊n / lcm (1,2, \dots, i)⌋$" It means that $⌊n / lcm (1,2, ..., i — 1) ⌋$ is the number of elements that have "reached" us. And $⌊n / lcm (1,2, ..., i) ⌋$ is the number of elements for which the answer is greater than $i$. Thus, subtracting one from the other, we get the number of elements for which the answer is equal to $i$. And of course, sorry for my bad English, I tried my best.My C++ solution:121264752
•  » » » » » Thanks for the clear explanation. Just one small correction, instead of Because if f(i)=x, it means that i has divisors 2,3,4…i−1, so i≥lcm(2,3,4…i−1) it probably should be Because if f(i)=x, it means that i has divisors 2,3,4… x −1, so i≥lcm(2,3,4… x −1)
•  » » » » » » Oh, my fault. I will edit it immediately.Thank you!
•  » » » » » thanks man, after reading it looks simple logic but why i can't think of these logics in contest :(noob me :{
•  » » » » » » I'm really happy that my comment was useful for somebody.You are not noob) In my opinion it's absolutely normal, when you can't solve the problem during the contest. P.S. I couldn't solve C too))
•  » » » » » 3 weeks ago, # ^ | ← Rev. 6 →   Clear explanation.
•  » » » » » Do you think f(k) = i = 6, is possible ??
•  » » » » » » f(k) always takes the value which are perfect power of primes. Like 2^1,2^2,2^3, 3^1, 3^2, 3^3. Let's say if f(k)=p*q. Where p and q both are primes. Here p
•  » » » » » I was trying to solve this question,everything seems fair to me in my code.But It was giving WA, so I removed only floor and it has passed all TCs. Can you tell me,why?123126566
•  » » » » it didn't work becauselcm of 1,2,3,4 is 12while the factorial is 24
•  » » » 4 weeks ago, # ^ | ← Rev. 2 →   to know how many numbers in range 1 to n are divisable by all numbers from 1 to i it equals to n / lcm(1,2,3,..,i) for example :- if n = 15 and you want to know how many number divisable by 1, 2, 3 and 4 floor(15 / lcm(1,2,3,4)) = 1 there is just one number divisable by 1, 2, 3, and 4 and this number is 12.
•  » » » » tenks man
 » 4 weeks ago, # | ← Rev. 2 →   Thanks for the editorial
•  » » My C++ solution for the problem C (feel free to ignore the template): https://codeforces.cc/contest/1542/submission/121274457. The key idea is for a $2 \leq k \leq n$ to be the minimum positive number that is not a divisor of a number $x$, all 1,...k-1 must be divisor of $x$, that is $lcm(1,...k-1)$ divides $x$. Thus we can iterate over $k$ and each time count how many such $x$ for each $k$, which is $n/lcm(1,..,k-1) - n/lcm(1,..,k)$. See the code for more details.
•  » » » I've seen it,nice solution
•  » » » Nice solution! Very clear.
•  » » » can you elaborate what do you mean by iterate over k, and also if anyone could elaborate on why is the count equal to n/lcm(1,..,k−1)−n/lcm(1,..,k).
•  » » Which problem? I've only got B & D
 » Mathforces!
 » This one requires a lot more math, but it's not friendly to younger participants. And, this is an algorithm competition, not a math competition.
•  » » why is it not friendly? i think everybody should know smth about LCM. Also B was not hard at all. I am a young participant (i'm 13) and ABC weren't hard at all
•  » » » smart russians.
•  » » » Saying ABC weren't "hard at all" won't cut it mate. They were hard, for the average folks. I've been coding for almost 1.5+ years now, and could only solve A. I found a research paper on OEIS which was directly related to C, still couldn't code it up in time. Thank your stars (or rather your smart ass genes lol) that you're able to do such good level math problems at such a tender age. You probably are meant for this, keep solving!
•  » » » » Nobody is meant for anything, it's just the timing when they started. And btw you are comparing your present with his, he also learned to check prime from O(n) to O(sqrt(n)) at some point in time.
•  » » » » » i also started about 1.5 years ago)
•  » » » » » » I guess he is just frustrated by the fact that the contest was just Math+ DP, out of which people struggle at both a bit... ig..PS. like, i like geometry and DS based problems, but it has been a long time i haven't seen such problems in recent contests....Anyhow its pretty Subjective, i felt yesterday's contest was pretty doable at least till C.
•  » » » » » » » sorry, my english is bad. what is DS?
•  » » » » » » » » Data structures
•  » » » Hard is a relative subject, some topics can be hard for someone, and easier for others. Age doesn't matter at all. This was a more math-focused contest (ABC all have math tag), and a mathematical approach is often harder for some.
•  » » » Everyone plays games , and only some reach the top but all other plays because they enjoy playing it .
•  » » Beacuse your ability is weak but not the round is not friendly.And I think problem D is easier than B and C.So why this contest unfriendly?
•  » » » Hello, Magic_Moon can you explain D. maybe using dry run or any way you are comfortable with. I am having a tough time understanding it.
 » Great Round, Great problems. Statements were short and crisp problem "C" is my personal favourite. Looking forward to see more such rounds. Thank you!!!
 » Although I didn't participate, I also think the problems especially C and D are wonderful!
 » When I was solving B, for some reason I typed "a^x" as "a*x" and spent almost all round to understand why extended euclidian algorithm doesn't work. Moral of the story: Don't tunnel vision, do your math.Great round!
 » I spent over an hour on b since I put (n % (a ^ k))%b == 0 instead of (n — (a ^ k))%b == 0.Nonetheless the problems were good, and I enjoyed the round
•  » » I did the same mistake and after seeing other people's code I was like when will these silly mistakes would leave me :(
 » I think B is harder to think than D.
 » Imagine C is harder to think than D...
 » Can someone explain the second paragraph of $B$ tutorial?
•  » » Let us assume n is present in the set. n can be represented as x*a^k + y*b, where x,k,y are whole numbers. Now n%b = (x*a^k + y*b)%b = (x*a^k)%b = x%b * (a^k)%b. x can be any of the previous values in the set, used to generate n. Thus we can assume x = (x')*(a^k') + y*b and x%b = (x')%b * (a^k')%b. Eventually we can see that x%b will be of type a^k where k will be a whole number. Hence we can write n%b = (a^k)%b, given n is present in the set.
•  » » » thanks! your explanation is so easy to understand.
•  » » » I have an easier explanation. All numbers in the set can be represented as a^x + b*y. Thus, if we subtract a^k from n and it gives a multiple of b, then n belongs to the set. You can iterate over k to check if the number belongs to the set.
•  » » » » This explanation is so far the easiest one I have found for this problem, I have been trying to understand the other solution for some time now, glad I stumbled upon yours.
 » 4 weeks ago, # | ← Rev. 3 →   I had a slightly different approach for problem C. So I will explain my solution and thinking process.Initially I was not able to solve the problem. Then I remembered one thing that factorial of even small numbers is very large (like $50!$). So, this gave me direction to think that $f(n)$ cannot be very large. So I ran a test to see that when we multiply prime numbers up to $50$, it will exceed $10^{16}$.I still had no clue how to solve this problem. Then I randomly wrote a number in prime factorization form. I saw that $f(n)$ can only be some power of only a single prime i.e. $f(n) = p^k$ where $p$ is some prime and $k$ is its power. I was able to prove this. Now, I knew that $f(n)$ can only be some power of prime up to $50$. So, now if $f(n)=p^k$, then $n$ must have exactly $p^{k-1}$ in its prime factorization form because otherwise $p^{k-1}$ will be $f(n)$. Now, I had to think what else should $n$ must have in its prime factorization form. I observed that for some other prime $q$ its power $m$ must be such that $q^{m+1} > p^k$ because otherwise $q^{m+1}$ can be answer. Then I found a number $D$ satisfying the above requirements. Now, the number of multiples of $D$ minus the number of multiples of $D*p$ would give the numbers for which $f(n) = p^k$. Then I would just add contribution to answer. Note that we have to handle a corner case i.e. when $D$ exceeds $n$, then we don't take any contribution from $p^k$. My code#include #include using namespace std; using namespace __gnu_pbds; // Policy based data structure template using oset=tree, rb_tree_tag, tree_order_statistics_node_update>; #define gin(x) cerr<<#x<<" : "< void sout(T var1,Types... var2){ cout<>n; vector pr; for(int i=2;i<=50;i++){ if(isPrime(i)) pr.push_back(i); } if(n<=50){ int ans=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n+1;j++){ if(i%j!=0){ ans=ans + j; break; } } } cout<n) break; int x=1; if(q==p){ while(x*qn){ can=false; break; } } if(!can){ k*=p; continue; } int loc=n/D - n/(D*p); // sout(k,D,loc); loc=loc*k; loc=(loc%mod + mod)%mod; ans=ans + loc; ans=(ans%mod + mod)%mod; k*=p; } } cout<>t; // t=1; while(t--){ solve(); } cerr<<"Time elapsed: "<<(double)(clock()-t1)/1000<<" s"<
•  » »
•  » » I massively overcomplicated my solution (originally in Rust, translated to C++) and reading it again now I realize that it basically amounts to Um_nik's solution except mine is slightly faster because it only evaluates the numbers that change the lcm. Of course, it's nowhere near the TL so that doesn't even matter.
•  » » can someone tell me what's wrong with this approach? if(i==1)sum+=2; else if(i%2)sum+=2; else if(i%2==0 && i%4==0)sum+=3; else if(i%2==0 && i%6==0)sum+=5; feel free to correct me,
 » Any idea why this solution to B gives TLE? Is it some silly mistake? Cause I've pretty much followed the editorial. void solve(int testcase) { int n, a, b; cin >> n >> a >> b; if (a == 1) { cout << ((n - 1) % b == 0 ? "Yes\n": "No\n"); return; } int m=1; bool res=false; while (m <= n) { if (m % b == n % b) { res = true; break; } m *= a; } cout << (res ? "Yes\n": "No\n"); } 
•  » » m should be long long
•  » » m should be long long . if not , m may less than 0, so tle.
 » Can anyone analysis the time complexity of problem C for me pls?
•  » » Someone correct me if I'm wrong, I think it's O(nlogn) because for every i computing lcm involves computing gcd which is logn with Euclidean algorithm
•  » » » 4 weeks ago, # ^ | ← Rev. 3 →   So you mean there are 10^16 log 10^16 operations in worst case @@?
•  » » » » no, max(N) is about 40-50
•  » » » » Sorry that was a horrible mistake. The loop runs while lcm <= n, and I think by lcm(1,...,40) you passed 10^16, so it should be O(40 log(something))
•  » » » » » Ok I get it thanks a lot
 » The introduction of hints was really very good. Thanks for such a great editorial.
 » Can someone please guide me through the equivalence of both formulas in C?
•  » » 3 weeks ago, # ^ | ← Rev. 6 →   I also want to know about this... I've got the first equation but can anyone explain how the SUM [ n/lcm(1,2,...,i) ] + ncomes from SUM [ i { (n/lcm(1,2,...,i−1)−(n/lcm(1,2,...,i) } ) ? I figured out the front part but how the (+ n) coming out from it? tianbu
 » math math math math math maTh MaTH MATH MATH MATH MATH MATH MATH MATH AAAAAAAAAAAAAAAAAAAA!!!!1!!!11!!!
 » 4 weeks ago, # | ← Rev. 2 →   These format of spoilers hurt my eye really bad, can you make it like this ? It will look much better.
 » Really a great contest! Kudos!
 » 4 weeks ago, # | ← Rev. 2 → » Nice contest.- Team Atcoder Fan, 2021
•  » » Can you explain your D? It seems more crisp.
 » in the C Tutorial : "Since f(n)= i means lcm(1,2,...,i−1) ≤n " why? I don't understand this, can you explain it to me?
•  » » because 'i' is the minimum number not divisor of n. and n is divisible by all number (1 to i-1). Like what is the minimum number that divisible by all number (1 to i-1) ??? it's g = Lcm(1,,,,i-1). Now, if f(n) = i, that means (n % g == 0) and g <= n.
•  » » I think what they are actually trying to say is that it doesn't make sense to go to such an i where lcm(1,2,...,i-1)>n since the contribution for such an i would be zero.
 » why chinese people think,that math problems are more interesting than algorithmic or data structure?
 » Can someone help me understand how they thought of this DP (and it's states) in problem D? The current editorial focuses more on implementation, less on the idea.
•  » » watch galen colin video.
•  » » When you have these expected value problems and you have to compute the total value (or expected value) of all sums, you can generally compute the total contribution of each term and add it separately. I think this can somewhat be seen by putting all the subsets of sums with the # of times they get counted vertically. When you add all of them up, you just get one equation in the $n$ variables (for whatever $n$ is, here it's the number of terms that start with $+$). The coefficient of each variable here is the number of times it's counted across all subsets -- which is the same as its independent contribution. Then, you can proceed as above.
 » Look at my code for B, it's very funny compared to the original solution, but still my code passed all the tests. Here is the link: https://codeforces.cc/contest/1542/submission/121210576
 » What are $p1$ and $q1$ in E1 editorial? And what do you mean by enumerating $p1$ and $q1$? tianbu
•  » » $p_1$ is the first element of $p$
 » Hi, I don't understand the editorial of problem B. Can someone please explain it to me?Thanks
•  » » Here is my approach — Every element of the set can be expressed in the form ${a^p + qb}$ where p and q are whole numbers. You can try writing the elements of the set in a tabular form to see it for yourself, intuitively. $1,1+b,1+2b,1+3b...$$a,a+b,a+2b,a+3b...$ $a^2,a^2+b,a^2+2b,a^2+3b...$We want to check if n is expressible in this form. To do this, iterate through p from 0 to any large enough value. For each value of p, check if $n = a^p + qb$, that is, ($n - a^p$)%b == 0. If this is true, the answer is YES. The loop terminates when $a^p$ becomes greater than n itself.
•  » » » Thanks a lot! Got it now :)
 » Does anyone have a different solution for E1?
 » 4 weeks ago, # | ← Rev. 2 →   froggyzhang had given a solution for e2 by using generating functions which i think is much easier than the official tutorial(which is also great!). Unluckily, it's in Chinese. And if anyone want to read it you can find it here.
•  » » My English is very bad, can anyone help me translate it into English? awa
•  » » » I used a browser extension that automatically translated and it looks like this image.
•  » » » I translated it into English here
 » 4 weeks ago, # | ← Rev. 5 →   .
 » Is there any recursive approach to solve D
•  » » I solved with recursive.
 » In B, is it true that if $n$ is equal to some power of $a$ then $n$ belongs to the set?I had the following condition in my code: if( n % a == 0) { cout << "Yes\n"; continue; } and for some reason this gives a wrong answer. My reasoning was the following. If $1$ and $1 \cdot a$ are in the set then $1 \cdot a \cdot a$ must also be in the set and other powers e.g. $1 \cdot a \cdot a \cdot a$ as well. So if $a$ divides $n$ i.e. $n \pmod a == 0$ then $n$ should be in the set. Am I missing something trivial ????
•  » » n = 10, a = 5
•  » » » Hmm... Ehh... Thank you! Now it looks so easy...
 » Can someone explain in better/different words what the states in the DP for D actually represent? The editorial is not at all clear to me.
•  » » For each number x, to determine whether it's in the final set or not, we only need to consider those numbers that are smaller than x. Because those larger ones will always be popped after x is popped. If there are more than one x, we assume the one that appears first is smaller.So, for each number x, define f[i][j] represent: after considering the ith operation, how many subsequences are there satisfies that there are j numbers in the set are smaller than x. (Remember to ignore the operation for the number x).Then do the dynamic programming as the tutorial shows.
•  » » » "If there are more than one x, we assume the one that appears first is smaller." Can you explain why so?
•  » » » » Actually, this is just an assumption to prevent some number being counted twice.For example, the whole sequence is +1 +1 -.After processing the whole sequence +1 +1 -, the final set is 1.When we count the first +1, we will not count the sequence +1 +1 -, cause the first +1 is smaller, according to the assumption. The 1 in the final set is the second +1, we'll count this one when processing the second +1.If you don't set the rule for equal elements, +1 +1 — will be counted for both +1, which is not what we want.
•  » » » » » Now I got the intution behind it! Thnx!
•  » » » Thanks got it!
•  » » You can have a look here and the parent comment, I tried to explain it with a picture: » What a nice math contest!(C is really nice)
 » Brilliant Problems!! Thanks for such a nice contest.
 » 4 weeks ago, # | ← Rev. 2 →   Could anyone explain why I got a time limit exceed in problem B test 2?Thank you so much. #include using namespace std; int main() { int tcs, tc, n, a, b, i; bool f; cin >> tcs; for (tc = 0; tc < tcs; tc++) { cin >> n >> a >> b; if (a == 1) cout << (n % b == 1 ? "Yes" : "No") << endl; else { i = 1; f = false; while (i <= n) { if ((n - i) % b == 0) {f = true; break;} i *= a; } cout << (f ? "Yes" : "No") << endl; } } return 0; } 
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   Because you used int instead of long long.
•  » » » thank you, but I got a wrong answer in test 2 now.
•  » » » » 4 weeks ago, # ^ | ← Rev. 2 →   Try this data, the answer is Yes. 1 2 1 1 
•  » » » » » thx, I didn't think about this. You help me a lot
 » Someone please explain solution of D. I am unable to comprehend the tutorial solution.
 » Can anyone explain me div 2 D solution
 » when coming up with dp solutions of taskE1 would you write the initial statements first(for(int i=1;i<=n*(n-1)/2;i++)s[i]=1;) or figure it out after the recursive statements? Because I had a hard time writing the initial statements,not knowing how many elements in the array should I initialize(Like why is s=1 not needed). Please share some tips! and sorry for my bad English
 » I'M THE ONLY ONE FST ON D!!!!!!!!!!!!! :(
•  » » Vegetable!
 » Let me introduce froggyzhang's idea of problem E in English. You can find the original edition here(in Chinese).I recommend you read the first and second paragraph of the official tutorial of E2 and the official tutorial of E1 before reading the following part of this article.Assume $p_{i+1}=a$ and $q_{i+1}=b$, $a,b$ are their rank in their permutations. So $inv(p)> inv(q)\Leftrightarrow inv(p[i+2,\dots n])-inv(q[i+2\dots n]\ge b-a+1$. So we just want to find the number of the pairs of permutations $p,q$ of length $n-i-1$ and $inv(p)-inv(q)\ge b-a$.Precalculate $f(i,j)$ the number of the pairs of permutations $p,q$ and $inv(p)-inv(q)=j$. You can insert the numbers to the permutation from small to big and you can do it in $O(n^4)$ or $O(n^5)$ by using brute force(which is also mentioned in the official tutorial of E2).You will find inserting number $i$'s generating function is as follow: $\sum_{j=0}^{i-1}x^j\sum_{j=0}^{i-1}x^{-j}=\frac{x^i-1}{x-1}\cdot\frac{x^{-i}-1}{x^{-1}-1}=\frac{x^{i+1}+x^{-i+1}-2x}{(x-1)^2}$About the above part we can write it in to DP functions($a\to b$ means we add the value of $a$ to $b$). $f[i-1][j]\to f[i][j+i+1]\\ f[i-1][j]\to f[i][j-i+1]\\ -2f[i-1][j]\to f[i][j+1]\\$And $\frac{1}{(x-1)^2}$ means we want to roll back twice. Just think what do you do when you $\times (x-1)^2$. So we can calculate $f$ in $O(n^3)$(both time and memory). We can only keep two layers of transition to optimize the memory to $O(n^2)$. And we can calculate suffix sum and the answer of the same prefix in $O(n^2)$.You can read froggyzhang's code in his blog. You can find other details of coding in his code if you need.
 » In the implementation of problem D, why is the additional if (i!=t) : f[i][j]=(f[i][j]+f[i-1][j])%mod there at the end of innermost for loop? Isn't the case where we skip the ith element already considered in each case?
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   Please read the definition of the dp statement again. It means the number of sequences for $x$ exit at last if we add option $t$ to the sequences. So that we needn't consider option $t$.
•  » » » Understood. Thanks!
 » Can someone explain to me how the formula $f(n,k)=\sum_{|i|  » Can anyone explain problem D. I didn't understand the solution in editorial.  » 4 weeks ago, # | ← Rev. 2 → This is my code for Div2 D Spoiler#include using namespace std; typedef long long int lli; typedef long double lld; const lli M1 = 1e9 + 7; const lli M = 998244353; const lli M2 = 1000000000000000001; lli mod(lli x){ return (x%M);} lli mod_minus(lli a, lli b){ lli ans= (mod(a)-mod(b)); if(ans<0) ans=mod(ans+M); return ans;} lli mod_mul(lli a,lli b){ return mod(mod(a)*mod(b));} lli mod_add(lli a,lli b){ return mod(mod(a)+mod(b));} #define FOR(i,l,u) for(lli i=l;i<=u;i++) #define FAST ios_base :: sync_with_stdio (false); cin.tie (NULL) void solve(){ lli n;cin>>n; lli A[n]; FOR(i,0,n-1){ char c;cin>>c; if(c=='-'){ A[i] = -1; } else{ cin>>A[i]; } } lli ans = 0; FOR(i,0,n-1){ if(A[i]!=-1){ vector> dp(n , vector(n+10,0)); if(i==0) dp = 1; else{ if(A==-1 || A>A[i]) dp = 2; else if(A<=A[i]) { dp = 1; dp = 1; } } FOR(j, 1 , n-1){ FOR(k, 0 , n){ if(A[j] == -1){ dp[j][k] = mod_add(dp[j - 1][k] , dp[j - 1][k + 1]); if(k == 0 && j < i) dp[j][k] = mod_add(dp[j][k] , dp[j - 1][k]); } else if((A[j]>A[i])||(A[j]==A[i] && j < i)){ dp[j][k] = mod_add(dp[j - 1][k] , dp[j - 1][k]); } else if(A[j]i)){ dp[j][k] = mod_add(dp[j][k] , dp[j - 1][k]); if(k>=1){ dp[j][k] = mod_add(dp[j][k] , dp[j - 1][k - 1]); } } else if(A[j] == A[i] && j == i){ dp[j][k] = mod(dp[j - 1][k]); } } } //cout<>t; while(t--) solve(); } Someone please provide me any small testcase for debugging. •  » » test5 + 1 + 2 - + 1 -  •  » » » Thanks for the help. I got AC  » Thank you! This is very helpful!  » In Question B it is clear that every number x which is in the set is of the form 'x mod b = a^k mod b' where k is a non negative integer but how have we proven that every number of such form will be in the set?That is how we know that the condition 'x mod b = a^k mod' is necessary and sufficient and not just necessary?  » 3 weeks ago, # | ← Rev. 2 → Hey I don't know why I am wrong answer on test 2 of Problem B. This is my submission: https://codeforces.cc/contest/1542/submission/121246560 Can you tell me what I am wrong. Thanks you •  » » In your recursion$n-c$become negative so just firstly check if$c>n$then return false if not then check for$n-c\$Just modified your code a bit https://codeforces.cc/contest/1542/submission/121831825
•  » » » Okay thanks you so much
 » the D is pretty good
 » the solution of D: #include #include #include #include #include using namespace std; typedef long long ll; const int mod = 998244353; const int maxn = 505; int dp[maxn][maxn]; int num[maxn]; void ini() { memset(dp, 0, sizeof(dp)); dp = 1; } int main() { int n; char ope; ll ans = 0; scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf(" %c", &ope); if(ope == '-') num[i] = -1; else scanf("%d", &num[i]); } for(int i = 1; i <= n; i++) { if(num[i] == -1) continue; ini(); for(int z = 1; z <= n; z++) { if(i == z) { for(int g = 0; g <= z; g++) dp[z][g] = dp[z-1][g]; } else if(num[z] == -1) { if(z < i) dp[z] = (0ll + dp[z-1]*2 + dp[z-1])%mod; else dp[z] = (0ll + dp[z-1]+dp[z-1])%mod; for(int g = 1; g <= z; g++) { dp[z][g] = (0ll + dp[z-1][g+1] + dp[z-1][g])%mod; } } else if(num[z] < num[i] || (num[z] == num[i] && z < i)) { dp[z] = dp[z-1]; for(int g = 1; g <= z; g++) dp[z][g] = (0ll + dp[z-1][g] + dp[z-1][g-1])%mod; } else { for(int g = 0; g <= z;g++) dp[z][g] = (2ll*dp[z-1][g])%mod; } } for(int g = 1; g <= n; g++) dp[n][g] = (dp[n][g] + dp[n][g-1])%mod; ans = (1ll*dp[n][n]*num[i] + ans)%mod; } printf("%lld", ans); return 0; } 
 » 这代码写的好称头哦
 » In C it was possible calculate everything before sending solution as in my solution ,then only modulo by every long long int in vector :-)
 » i submitted 123918769 for problem[problem:1542A]. This submission is giving run time error. Can anyone please tell my mistake. Thanks in advance.