// time n over space m recognition: n/m details: (m + n/(1+m less lgm)) lgm
/// (time n over space m is search theory)
num age=zero;// count all finds one by one (page alts)
inline num map(nums& p,num& z,num& dep,num& lga)
{num d=zero;for(num h=zero;h<dep;h++)d=(d<<lga)+p.v[z+h];return d;}

num skips (nums& string, nums& book, num& lga, numnums* myn) {
/// allocate proper memory space of empty sets O(constant)
num m = string.size(); num deep = zero; num d = one; while (d < m) { d = d << lga; deep++; }
for (num g = (*myn).size(); g < d; g++)(*myn).add(new nums);
/// the m part of partten are mapped precisely O(mlgm)
for (num g = zero; g + deep < m; g++)(*(*myn)[map(string, g, deep, lga)]).add(g);
/// map part of the book into pattern O(nlgm/(m+1-lgm))
num mm = minus(m + one, deep); for (num j = mm; j + deep < book.size(); j = j + mm) {
nums* d = (*myn)[map(book, j, deep, lga)]; for (num o = zero; o < (*d).size(); o++)
{num be = minus(j, (*d)[o]); num h = one; for (num g = zero; g < m; g++) {
if (book[be + g] == string[g])continue; h = zero; break; } if (h)age++; } }
/// release resources from step two O(mlgm)
for (num g = zero; g + deep < m; g++)(*(*myn)[map(string, g, deep, lga)]).clear(); return age; }