Obviously, suits are not change the answer. Look at all possible situations:

[0 - 10] — zero ways.
[11] — four ways using aces.
[12 - 19] — four ways using cards from 2 to 9.
[20] — 15 ways, they are described in statement.
[21] — four ways using aces.
[22 - 25] — zero ways.

We can't move to the i + 1-th question before clicking all answers at i-th question. We will move to the very beginning clicking at a_i - 1 answers and only one will move us to the next step. Every way from beginning to i-th question have length equal to i - 1 plus one for click. There is a simple formula for i + 1-th question:

c_i = (a_i - 1)· i + 1, 1-based numeration.

Let's summarize all values of c_i for making the answer.

Look at the formal part of statement. Cthulhu is a simple cycle with length of 3 or more and every vertix of the cycle should be a root of some tree. Notice this two important facts:

a) Cthulhu is a connected graph
b) Cthulhu is a graph with only cycle

If we add only one edge to tree (not a multiedge nor loop) then we will get a graph that a) and b) properties. There is a simple solution: let's check graph for connectivity (we can do it by one DFS only) and check n = m restriction. If all are ok then graph is Cthulhu.

Let's solve the problem when value n is even. Obviously, if k = 1 we should put a bullet at most right position. For example, choose n = 8 and our answer string will be:

.......X

Let's find probability of winning at this situation. We will write zeros at losing positions and ones at winning positions:

10101010

The simple idea is to put every next bullet for non-dropping down our winning chance as long as we can. After that we should put bullets in lexicographically order with growing value of k. Look for the answer for another k's with n = 8:

.....X.X
...X.X.X
.X.X.X.X
.X.X.XXX
.X.XXXXX
.XXXXXXX
XXXXXXXX

When n is odd our reasoning will be the same with a little bit difference at the beginning. Look at the answer with n = 9 and k = 1::

........X
010101010

We may put second bullet as at even case. But there is exists another way, we may put our bullet at 1st position and turn left the cylinder, look here:

X.......X => .......XX
010101010    101010100

Obvioulsy probability of winning was not changed but answer made better. Next steps of putting are equal to even case:

.....X.XX
...X.X.XX
.X.X.X.XX
.X.X.XXXX
.X.XXXXXX
.XXXXXXXX
XXXXXXXXX

There is no difficulties to give answers to queries.

Let's solve this problem by offline. Read all queries and sort them by increasing b. After that there are many effective solutions but I tell you a most simple of them. Look at two following algorithms:

1. For each (a, b)-query calculate the answer by the simple way:

for(int i = a; i < n; i += b)
  ans += a[i];

2. Fix the value of b and calculate DP for all possible values of a:

for(int i = n - 1; i >= 0; i--)
  if (i + b >= n)
    d[i] = a[i];
  else
    d[i] = d[i + b] + a[i];

Obvioulsy, the 1st algorithm is non-effective with little values of b and the 2nd algorithm is non-effecitve with very different values of b. Behold that it's impossible to generate the situation where all b's are so little and different simultaniously. There is a simple idea: we may solve the problem by 1st algo if values of b are so large (if they are greater than c) and solve by 2nd algo if values of b are so small (if they are less of equal to c).

We need to sort our queries because of no reason to calculate DP of any b more than once.

If we choose c equal to that we will get a complexity.

Translated by Borisp.

http://codeforces.me/blog/entry/2539