### a_gt3_rs's blog

By a_gt3_rs, history, 3 months ago,

When I was doing 1207F - Задача об остатках, I don't know why my program got TLE. Then this ONE trick cut 1 second off my program. You can see this video for more information: https://www.youtube.com/watch?v=ssDBqQ5f5_0. When the compiler compiles x/9, the code roughly converts to ((long long)954437177*x)>>33. Division is a much more expensive operation than multiplies and shifts. But what's the special constant 954437177? It's $\lceil \frac{2^{33}}{9} \rceil$. So we have this identity which is vaild for every $0\leq a < 2^{31}, 0 < x < 2^{31}$:

${\lfloor \frac{a}{x} \rfloor} ={ \lfloor{\frac{a*\lceil{2^{31+\lfloor{log_2(x)}\rfloor}/x}\rceil}{2^{31+\lfloor{log_2(x)}\rfloor}}}\rfloor}.$

You can clearly see how this speeds up 1207F: 244531668, 244540601. Note that gcc also does this trick on 64 bits.

After brute forcing with $a = 2^{31}-1$ I got this new, fixed identity. This should work for all $a$ (while the old one does not).

• +46

 » 3 months ago, # |   0 very good
 » 4 weeks ago, # |   0 thank you
 » 4 weeks ago, # |   0 Really cool trick! Sadly g++ doesn’t optimise divisions with numbers not known at compile-time this way.
 » 4 weeks ago, # |   0 You can make GCC use this trick in just one line:#pragma GCC unroll B where B is your block size!With B = 700 it is 1000ms faster: 262433429, 262433450Compile to assembly and you can see fixed point multiplication being used.
 » 4 weeks ago, # | ← Rev. 3 →   +4 just remove #define int long longhttps://codeforces.com/contest/1207/submission/262501542