So apperently i'm reading a document written in Chinese about recusion for learning purpose. In that document there is a problem that seems to relate to IOI but i cannot understand what it's saying even after i have it translated to Vietnamese(This used to be "English", typo) with Google translator. I've done a bit of googling about IOI 99 but non of the problems i encountered seems to be what i'm looking for. Could the Chinese community on Codeforces please provide me with a proper translation of the problem ?
Here's the original text: 01序列：近来IOI的专家们在进行一项有关二进制数的研究，研究设计的一个统计问题令他们大伤脑筋。问题是这样的：对于一个自然数n，可以把他转换成对应的二进制数 ，其中：n = a_k*2^k + a_(k-1)*2^(k-1) + ... + a_1*2^1+a_0；而且a_i=0或1(0<=i<k)，a_k=1。如：10=1010；5=101。我们统计一下a_0~a_k，这k+1个数中0的个数和1的个数。如果在这k+1个数中，0的个数比1的个数多，就称为A类数。现在的任务是，对于一个给定的m，求1~m中a类数的个数。