How to solve the boundary problem with android integer dichotomy template

This article mainly explains "how to solve the boundary problem with android integer dichotomy template". Interested friends may wish to have a look. The method introduced in this paper is simple, fast and practical. Next let the editor to take you to learn "android integer dichotomy template how to solve the boundary problem" bar!

1. Interval

/ / the interval is divided into [lmidmid] and [mid+1,r], as follows, x1; if (a [mid] > = x) else midmidline 1;} / / interval is divided into [lmaine midmid 1] and [mid,r], as follows, x > = a [mid] judgment condition such that x is either in [mid,r] while (l > 1; if (Amid] > x) The second template is to consider when to compress while to the left and when to compress else l=mid+1 to the right. / because mid is rounded down, mid will never get the initial right boundary / / similarly, the second template will never take the initial left boundary if (check (mid)) ringing midway / compress else l=mid+1 to the left if the condition is satisfied. / / compress} cout to the right