CF574

It will be fixed.

Generally,div.1 is harder than div.2

div.2 C=div.1 A

div.2 D=div.1 B ....

There are three divisions (the more digit the easier problems):

• div3 — usually for pupils
• div2 — students
• div1 — high-skilled students, engineers etc.

you're sorry that you can take place!!!!???? Hey everybody, I'm sorry that I'm alive lol :P

Absolutely yes!

And if you are lucky enough,you can even become an expert.

I think Summer informatics School like Codeforces Round 530 (Div. 2).

SIS is for russian ЛКШ. It's a summer training camp in Russia

Coordinator finds mistakes in problems statements

I agree that Educational Codeforces Round 68 had very strong pretests, which made hacking very difficult.

Oh, you are strong enough, and you won't be afraid of the strong pretests.

i will be master today....waiting for div1 vovuh (^.^)

Challenge accepted :)

SIS stands for summer computer school, so SIS team contest means team contest held in SIS.

Easy div1 :)

Just one more div2D and this would've been a beautiful round with 2 divisions.

rating count begins from 1500 , if you perform well ,you can directly reach blue after first round, else you would be green

Welcome to FairyForces :P

I didn't even understand the problem. Why the english is hard

Exactly .... B should have been A , A question is supposed to be cakewalk right and I wasn't able to understand what it was saying for the first 10 mins ....

Maybe there isn't too much difference if U think a lot about them:)

So it means we can reduce the problem to querying the number of different directions of the directed arrows in a certain range. Didn't figure out this until the last minute of the contest.

data structures

Link

Pretty much the same as https://discuss.codechef.com/t/chsqarr-editorial/12662

I think it is using trees, but I failed to pass it in time.

Argh, deque. The tutorial is everywhere, but it's not that hard to not figure out, so just blame myself.

Thanks abhayps! ;)

In fact only use Sparse Table can solve it. My submission http://mirror.codeforces.com/contest/1195/submission/57270378

Simply too costly, keep in mind that $$$n$$$ and $$$m$$$ may both reach $$$3000$$$.

-First i solved examples on paper, then recognized pattern. Result=summation(f(a[i],a[i])*n

My solution to D1 is:

1. Work out how many times each digit appears in each position.
2. Add the digits in each position.
3. Work out the total by adding these up multiplied by suitable powers of 10

I didn't get a solution for D2. I think the same process should work, but step 1 is more difficult.

For $$$f(x, a)$$$ and $$$f(x, b)$$$, if $$$a$$$ and $$$b$$$ have the same number of digits, the effect of $$$x$$$ on the answer is always the same. The same holds for $$$f(a, x)$$$ and $$$f(b, x)$$$. In addition, if $$$a$$$ has at least the same amount of digit as $$$x$$$, the effect of $$$x$$$ on answer is always the same. We only need to store the frequency of number of digits.

The contribution of $$$x$$$ would be like follows (take $$$x=1234567$$$):

$$$12345670 - \gt 123456070 - \gt 1234506070 - \gt ... - \gt 10203040506070$$$ if it is $$$f(x, a)$$$.

$$$1234567 - \gt 12345607 - \gt 123450607 - \gt ... - \gt 1020304050607$$$ if it is $$$f(a, x)$$$.

$$$a$$$ has number of digits $$$1 - \gt 2 - \gt ... - \gt numberOfDigitsOfx$$$.

To insert a zero between the digits, you can just use the *, /, % operators.

change int to long long

 vector<long long> a(n,0),b(n,0),dp1(n,0),dp2(n,0);

Лол, оказывается, я эту задачу уже сдавал(во время контеста я ее, конечно же, не решил)

Где можно найти решения этих задач?

-I solved with Dynamic Programming. for (i=1;i<=n;i++) { dn1[i]=max(dn1[i-1],a[i]+dn2[i-1]); dn2[i]=max(dn2[i-1],b[i]+dn1[i-1]); } cout<<max(dn1[n],dn2[n])<<endl;

The main idea is DP. Let dp[i][0] be the best sum if we only take players up to position i and take the player in the top row at position i. dp[i][1] is defined similarly, except we take the player in the bottom row at position i. dp[i][2] is also similar, except we take no player in position i. We can then fill this dp table in O(N) time, as the transitions are relatively simple.

use dp. let dp[i][j] be the max sum u can take till column i and u choose jth row {0 , 1} in current column. now transition will be either from i-1 or from i-2 because no point in skipping 2 elements.

Thanks you guys, I will think how to do it with your tutoral. Hope me I will get accept :D