Solution for this problem consists of two stages. First stage - counting the numbers with 1 as their first digit in the [

To solve the first sub-problem one can generate all segments of good numbers and look how many numbers from them lie in the [

So, we've learnt how to calculate the probability

Now we can go to the second stage of the solution: solving the problem with

After that answer to the problem will be a sum of

*L*;*R*] segment. Second stage - using this information (in fact, by given probabilities that*i*-th quantity will be good) solve the problem about*K*percents.To solve the first sub-problem one can generate all segments of good numbers and look how many numbers from them lie in the [

*L*;*R*] segment. Segments of good numbers are of form [1;1] , [10;19] , [100;199] and so on, that is, [10^{ i};2· 10^{ i}- 1] . After noticing that, calculating their intersection with [*L*;*R*] segment is quite easy.So, we've learnt how to calculate the probability

*p*[*i*] that*i*-th quantity is correct: this probability*p*[*i*] equals to a fraction of the number of good numbers and*R*[*i*] -*L*[*i*] + 1 .Now we can go to the second stage of the solution: solving the problem with

*N*quantities and*K*percents. Now we know the probabilities*p*[*i*] that this or that quantity is good, and want to find the probability that*K*percents of them will be good. This can be done using dynamic programming: let's*D*[*i*][*j*] - probability that among first*i*quantities exactly*j*will be good. The starting state is*D*[0][0] = 1 , and calculation of other states can be done as following:*D*[*i*][*j*] =*p*[*i*- 1]·*D*[*i*- 1][*j*- 1] + (1 -*p*[*i*- 1])·*D*[*i*- 1][*j*].After that answer to the problem will be a sum of

*D*[*n*][*j*] over all*j*such, that*j*/*n*≥*k*/ 100 .
why you dont put any tutorial about problem A and problem B ?

although these are simple(maybee!) but i hope you put them ,

by the way , thanks.....