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### ammar_hammoud's blog

By ammar_hammoud, history, 10 months ago,

Hello CF community, I was trying to solve this problem 584B - Kolya and Tanya , I saw that my logic gives wrong answers but I don't know why, my friends' solutions was like find the total cases, subtract the invalid cases but I was thinking of finding the valid cases immediately, here's how I think:
Let's take $n=2$ for example:
I can either make $a_0, a_2, a_4$ valid or $a_1, a_3, a_5$ valid, I know that for every $3$ vertexes (which are to be valid) I have $3^3 - 7 = 20$ valid cases, as the invalid cases are:
1. ($1, 2, 3$)
2. ($1, 3, 2$)
3. ($2, 1, 3$)
4. ($2, 3, 4$)
5. ($3, 1, 2$)
6. ($3, 2, 1$)
7. ($2, 2, 2$)
and for the rest of the vertexes the values I can take are either $1, 2,$ or $3$, which means $3^{3n-3}$ options, so I have $n$ options ($a_0$, $a_1$ in case $n=2$) with $20$ valid cases, and for every options of this I have $3^{3n-3}$ options $\rightarrow$ $(20\times 3^{3n-3})\cdot n$.
Can anyone tell me what's wrong?

• -2

By ammar_hammoud, history, 10 months ago,

Hello CF community, I'm trying to solve this problem, the problem tells me that I have to calculate

$C_n^r = \frac{N!}{M!\cdot (N-M)!}; 5\leq N\leq 100, M\leq N$

in the numerator it will always be $\prod _{i = N-M+1}^{n} i$ and in the denominator will be $\prod _{i=1}^{min(N, N-M)}i$ as the rest will cancel out
and this is my code:

but it gives me WA, I'm not really sure why I'm getting this. Any ideas?

• +1

By ammar_hammoud, history, 11 months ago,

Hello CF community, I'm trying to solve this problem 810B - Летняя распродажа, and here's my submission 216849580, Although the code gives AC for the first $15$ test, it gives a WA on the $16$-th test, I'm not able to identify the issue, any ideas?. I don't want another solution, I just want to know what's wrong in my solution.

• +4