PNVZarif's blog

By PNVZarif, history, 7 days ago, In English
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Let $$$a_1, a_2, ..., a_C$$$ be the separations between consecutive partners.

We need to solve for $$$\;a_1+a_2+...+a_c=N-C\;$$$ where $$$a_i\geq R\; \forall i$$$.

This is the same as $$$\;a'_1+a'_2+...+a'_c=N-C-R*C\;$$$ where $$$a'_i\geq 0\; \forall i\quad$$$ (Substituting $$$a'_i=a_i-R$$$)

Now, notice that we can arrange a particular tuple $$$(a_1,a_2,...a_C)$$$, exactly $$$N$$$ times around the table and each will be treated as a distinct way as long as this tuple is a unique circular permutation. (Eg. (1,1,3) and (1,3,1) or (2,3,4,5) and (3,4,5,2) are not).

One special case being that if all $$$a_i$$$'s are equal (can only happen when $$$C|N$$$) then we cannot arrange $$$N$$$ times around the table but only $$$N/C$$$ times as after that we will get a repetition.

Finally, in every scenario Pixie can be seated in $$$N-C$$$ independent ways, each counted as distinct. We can just multiply the final total with $$$N-C$$$ at the very last to incorporate that.

Therefore an algorithm would be:

  • Find the number of (unique circular permutation) tuples $$$(a_1,a_2,...a_C)$$$ where at least two $$$a_i$$$'s differ. Let this be $$$cnt_1$$$.
  • Find the number of tuples $$$(a_1,a_2,...a_C)$$$ where all $$$a_i$$$'s are same. Let this be $$$cnt_2$$$. (it will either be $$$0$$$ or $$$1$$$)
  • Output $$$ans=(cnt_1*N+cnt_2*\frac{N}{C})*(N-R)\%p$$$

Note that, $$${m+C-1 \choose m}$$$ is the total number of tuples (where $$$m=N-C-R*C$$$). Out of these at most one is of type two (if C divides N) and remaining are exactly all tuples with at least two distinct values, but these are not circularly invarient yet. In fact we have exactly $$$C$$$ copies of each such tuple.

Thus, $$$cnt_2=(N\%C==0)$$$ and $$$cnt_1=\frac{{n+C-1 \choose n}-cnt_2}{C}\;$$$. This makes the final answer simplier — $$$\;ans=\frac{{m+C-1 \choose m}*N*(N-R)}{C}\%p$$$


Test Case 1

$$$a_1+a_2+a_3=9-3=6,\;\; a_1,a_2\geq2\;\;\;\qquad$$$ or $$$\qquad\;\;\;a'_1+a'_2+a'_3=0,\;\; a'_1,a'_2\geq0$$$


$$$cnt_1=0$$$. no such ways.

$$$cnt_2=1$$$ comprises of $$$(2,2,2)$$$ case repeated $$$9/3=3$$$ times. $$$\therefore 1*3=3$$$ ways.

Total $$$(0+3)*(N-C)=3*6=18$$$ ways.

Test Case 2

$$$a_1+a_2=10-2=8,\;\; a_1,a_2\geq1\;\;\;\qquad$$$ or $$$\qquad\;\;\;a'_1+a'_2=6,\;\; a'_1,a'_2\geq0$$$


$$$cnt_1=3$$$ comprises of $$$(1,7),(2,6),(3,5)$$$, each repeated $$$10$$$ times. $$$\therefore 3*10=30$$$ ways.

$$$cnt_2=1$$$ comprises of $$$(4,4)$$$ case repeated $$$10/2=5$$$ times. $$$\therefore 1*5=5$$$ ways.

Total $$$(30+5)*(N-C)=35*8=280$$$ ways.