EDITORIAL: In order to protect copyright is Rui topic, here and hold surface title, write only problem solution.
"I am Rui never guaranteed simulation game title must be original."
"I did not write copy, there is a problem to find a teacher to meow." - Tsai
- A
R Ye overturned again, do six years ago out of a CF questions.
\(100pts:\)
However, instead of the original title is also very simple, optimized slope board, monotonous queue and do some get away.
May wqs half, the complexity can be done \ (O (m \ log m) \) , \ (and \) P \ (irrelevant. So R Ye almost to \) P $ out to \ (10 ^ 5 \) .
- B
This question fucks practice very much,So R Ye permission to use the card fucks.
However, R Ye still very conscience, to be able to leave the AC fucks \ (50pts \) .
"The hamster pull out." - money grandpa
\(100pts:\)
Do backwards, which states considered to reach the final state, there is no watch in which the initial state.
Each time find a peer with no other columns blank spaces, while the peer-column grid with all assigned to "superposition state."
Since no other peers in the same column space, so it will not lead to no solution can be greedy.
Superposition may also be used as a space.
There will be a number of rows, columns, all superposition states, the last sentence about each row must be at least a hamster.
- C
Sensibility to understand what \ (F (the X-) \) , found its meaning is \ (x \) able to open the biggest \ (k \) root of the result after the square root.
\(54pts:\)
Classic anti-Mo.
\[ans=\sum_{i=2}^n F(n)~~~~~~~~~~~~\]
\[=\sum_{i=2}^n \prod_{j} p_j^{\frac{q_j}{\gcd(q)}}\]
\[~~~~~~~~~~~~~~~~~~~~=\sum_{k}\sum_{i=2}^n \sqrt[k]{i}\cdot [\gcd(q)=k]\]
\[~~~~~~~~~~~~~~~~=\sum_{k}\sum_{i=2}^{\sqrt[k]{n}}i\cdot [\gcd(q)=1]\]
\[~~~~~~~~~~~~=\sum_{k}\sum_{i=2}^{\sqrt[k]{n}}i\sum_{d|\gcd(q)}\mu(d)\]
\[~~~~~~~~~=\sum_{k}\sum_d \mu(d)\sum_{i=2}^{\sqrt[kd]{n}}i^d\]
\[~~~~~~~~=\sum_{T}\sum_{d|T}\sum_{i=2}^{\sqrt[T]{n}}\mu(d)i^d\]
Summation just behind the pile of interpolation just fine.
\(100pts:\)
Precision open \ (K \) th root, the pressure may be binary bits, down to \ (2 ^ {30} \) about the card can be in the past.
Incidentally precision open \ (K \) method is to first two times the root of a sub-length, then the specific binary value. Every time a high-precision rapid power to determine the size.
It is really an anti-AK good question. / Cy / qiang