Actually, worse began the journey to red

I heard dreamoon_love_AA said that he has some plans about it =)

Please, don't neglect any Coder. Every coder has a high ambition to become RED. And mind the Quote Failure is the pillar of success

This round will be held at CST-1:30 (1.5 hours before Codeforces Standard Time)

Doesn't being an international grandmaster (like dreamoon_love_AA was before his fall (partially on purpose)) is a sufficient proof :P?

dreamoon_love_AA also solve around 1432 problems here in codeforces ( as like you i don't know others OJ number but i am pretty much sure there must be more than thousands also ) . i always wonder is problem solving is that kind of addiction to him if he won't solve 10 or more problems in a day he can't eat or sleep :D

I wonder how many problem he solved in one day. Did he sleep at night or goes to school? I was trying to see his first solved problem in uva. But now it is still loading for few minutes.

You count Bayan Qualification along with those other contests? >_>

I'd replace it with Codechef Long Challenge, it has more decent problems now.

I was about to say the same :). Maybe I should read E in the beginning and in case I got this coded already, paste my solution, otherwise not give it any thought :P.

the third problem also 1500 , what do you think ?

I agree with 12310720112 -_-

I think no because newly registered guy has a Div1 contest at this time :D

Goodbye blue color :(

Goodbye green color :(

Goodbye yellow color :((

Sayonara purple!

Goodbye Orange, hello Red :D

Goodbye Div.1 :(

Are you happy with math problems?

I thought: "Problem with MOD! I'll do some hacks today!" and instalock it. After few minutes I figured out that my solution is wrong because I forgot to add some MOD...

Exactly though i didn't give the contest because I actually woke after 30 mins past the start time but definiately this is the first problem I have seen of long overflow.really tricky on this part.

Output ((6i + 1)k, (6i + 2)k, (6i + 3)k, (6i + 5)k) for all i = 0, 1, ..., n - 1. (And thus the maximum is (6n - 1)k.)

Uhm, the solution is in the sample. Just use 6i + 1, 6i + 2, 6i + 3, 6i + 5 (assuming k == 1). Upper bound is easily obtained from the observation that at least 3 numbers in each set should be odd.

for (mod = 1; mod < b; mod++) 
res += mod * b * a(a+1)/2  + mod * a

Observe that all numbers satisfying the condition has the form x = (bk + 1)(mod(x, b)). Since mod(x, b) can be any integer between 1 and b - 1 inclusive, k can be any integer between 1 and a inclusive, and all those numbers are valid, you're adding up (b + 1)(1) + (b + 1)(2) + ... + (b + 1)(b - 1) + (2b + 1)(1) + (2b + 1)(2) + ... + (2b + 1)(b - 1) + ... + (ab + 1)(b - 1).

We group every b - 1 terms; these are all in the form for some x. Thus our sum becomes , which can be simplified to .

There you go, an O(1) solution. I wonder why the time limit is 1.5 seconds...

x = q*b+r ---(1)
and q/r = k
 or q = rk;
put this in 1
x = brk+r = r(bk+1)

Now we see that , r can be 1 to b-1. thus for a particular value of k, we have

1(bk+1) + 2(bk+1) +3(bk+1)... (b-1)(bk+1)
=(bk+1)(1+2+..b-1)

so , psudocode is below

sum = 0;
for(k=1;k<=a;++k)
{
   sum+=b*(b-1)/2 * (bk+1)
}

don't forget to mod every expression.

Yeah i did that too. i was just getting confused with all those numbers . (defined by problem and me !)

удален.

Because your algorithm is wrong !!!!!

I think you misinterpreted the problem or something, thinking that you're only counting numbers between 1 and a or something, not counting numbers whose k is between 1 and a.

DP with state pos, no_char_to_delete, where pos is the position of S.

First pre-calculate next[i] for each i, where you can find P in S[i,next[i]], deleting zero or more character.

Then using this info, you can run the dp.

long double,you should use cout<< to output....Ld is not using to output long double

BTW,if you run this code in CF.

#include<stdio.h>
int main()
{
    printf("%d",sizeof(long double));
}

you will find sizeof(long double) is 12 in CF...

a = 4, b = 3

Try for k = 1. We want div(x, 3) = mod(x, 3). Since mod(x, 3) = 1, 2, we thus have two numbers: one with div(x, 3) = mod(x, 3) = 1, and one with div(x, 3) = mod(x, 3) = 2. The numbers satisfying this are 4 and 8 respectively.

Try for k = 2. We want div(x, 3) = 2·mod(x, 3). Since mod(x, 3) = 1, 2, we thus have two numbers: one with div(x, 3) = 2, mod(x, 3) = 1, and one with div(x, 3) = 4, mod(x, 3) = 2. The numbers satisfying this are 7 and 14 respectively.

Trying for k = 3 gives 10, 20, and for k = 4 gives 13, 26.

Adding them all gives the requested 4 + 8 + 7 + 14 + 10 + 20 + 13 + 26 = 102.

Just write a brute force ....

the numbers are : 4 7 8 10 13 14 20 26 sum = 102

((b % mod) * (sum2 % mod) * (sum1 % mod))

If b % mod, sum2 % mod, sum1 % mod are together pretty large (about 10⁹ each), they make the resulting number to get overflow (about 10²⁷).

Try to mod every value RIGHT AFTER multiply. Don't multiply three values in a time, it will be overflow.

First you need to calculate nearest subsequence that match P and start from i for every i<s.size .

Now we can use a simple dp, at each position we have 3 choices :

1- you can skip this position to next one so : rec(pos+1, should_delete).

2- you can delete this position if should_delete!=0 so : rec(pos+1, should_delete-1).

3- if we have match from pos we can use this match, we have the ending position of this match (calculated in first step) so number of characters we need to delete so we can use this match is : end[pos]-pos+1-p.size . now if this number is less than or equal to should_delete then you can use this match so : 1+rec(end[pos]+1, should_delete-(end[pos]-pos+1-p.size)).

The answer for each step is maximum of this three choices.

First, let's calculate for each position i in string s value L[i] — minimum length of substring which ends in i and it is possible to get string p by removing some characters from it. It can be easy calculated using dynamic programming.

Next, let's find dp[i][cut] — answer for prefix of s which ends in i and with cut character removes done. It can be calculated next way:

dp[i][cut] = dp[i-1][cut];
if (cut > 0) 
	dp[i][cut] = max(dp[i][cut], dp[i-1][cut-1]);
for (int k = L[i]; k <= cut + p.length(); ++k) 
	dp[i][cut] = max(dp[i][cut], dp[i - k][cut - k + p.length()] + 1);

This solution gives us O(|s|³) complexity. But you can notice that in every dp visited in the cycle difference between i - k and cut - k + p.length() is fixed: it is equal to i - cut - p.length(). So, you just need to use fenwick trees where in ftr[i + cut][i] you will update the answer.

Then the cycle will look like this: dp[i][cut] = max(dp[i][cut], get(ftr[i - cut - p.length()], i - k) + 1); where get(a, b) function returning maximum on fenwick tree a prefix with length equal to b.

Of course, you can examine my solution: 8203748

Indeed, Codeforces should start giving T-shirts more often. I need Codeforces and Facebook T-shirts to complete my collection.

Then some people would get too many t-shirts :D

Thanks for the nice comment :)

what's wrong?