proof - Codeforces

#	User	Rating
1	tourist	4009
2	jiangly	3831
3	Radewoosh	3646
4	jqdai0815	3620
4	Benq	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	gamegame	3386
10	ksun48	3373

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
7	adamant	156
8	TheScrasse	154
8	nor	154
10	Dominater069	153

proof of this

Theorem: Let $$$p=\lceil\frac km\times2^{64}\rceil$$$, when $$$a_i\le \frac{2^{64}}m$$$, the computation of the lower rounding is exact.

Proof: Let $$$\frac p{2^{64}}=\frac km+\epsilon$$$, where $$$\epsilon\in[0,\frac1{2^{64}})$$$.

$$$ \begin{aligned} \lfloor\{a_i\times \frac p{2^{64}}\}\times m\rfloor&=\lfloor\{a_i\times (\frac km+\epsilon)\}\times m\rfloor\\ &=\lfloor\{a_i\times \frac km+a_i\epsilon\}\times m\rfloor \end{aligned} $$$

Here if $$$\lfloor a_i\times\frac km+a_i\epsilon\rfloor>\lfloor a_i\times \frac km\rfloor$$$ must be wrong, $$$a_i\times \frac km$$$ is at least $$$\frac1m$$$ away from $$$\lfloor a_i\times \frac km\rfloor+1$$$, so as long as $$$a_i\epsilon<\frac1m$$$ it's OK, let's continue the derivation.

$$$ \begin{aligned} &=\lfloor\{a_i\times \frac km+a_i\epsilon\}\times m\rfloor\\ &=\big\lfloor\left(\{a_i\times \frac km\}+a_i\epsilon\right)m\big\rfloor\\ &=\lfloor\{a_i\times \frac km\}\times m+a_im\epsilon\rfloor \end{aligned} $$$

Since $$$\{a_i\times\frac km\}\times m$$$ is an integer, the result is exact as long as $$$a_im\epsilon<1$$$.

The $$$a_i\epsilon<\frac1m$$$ and $$$a_im\epsilon<1$$$ are the same, and then the condition can be rewritten as $$$a_i\le\frac{2^{64}}m$$$ according to $$$\epsilon<\frac1{2^{64}}$$$.

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