[897A — Scarborough Fair](http://codeforces.me/contest/897/problem/A) By [user:mcfx,2017-12-04]↵
↵
For every $i$ in range $[l,r]$, if $c_i$ is $c_1$ then change it into $c_2$...↵
↵
Because $n, m$ are all very small, $O( nm )$ can easily pass it.↵
↵
PS. You can use binary search tree to solve it in $O( m\log n )$ time.↵
↵
[897B — Chtholly's request](http://codeforces.me/contest/897/problem/B) By [user:mcfx,2017-12-04]↵
↵
The $k$-th smallest zcy number is $conn(str(k),rev(str(k)))$, where $str$ denotes the decimal representation of a positive integer as a string, $conn$ denotes the concatenation two strings, and $rev$ denotes the reverse of a string.↵
↵
Then go over the smallest $k$ such numbers and sum them up to obtain the answer.↵
↵
[896A — Nephren gives a riddle](http://codeforces.me/contest/896/problem/A) By [user:mcfx,2017-12-04]↵
↵
$f(n)=str_1+f(n-1)+str_2+f(n-1)+str_3$.↵
↵
First we can compute the length of $f(n)$ for all possible $n$.↵
↵
For a pair of $(n,k)$, we can easily determine which part the $k$-th character is in.↵
↵
If it's in $f(n-1)$, we can solve the problem recursively.↵
↵
The complexity of this algorithm is $O(n)$, which is sufficient to pass all tests.↵
↵
Obviously, $length(f(n))\ge length(f(n-1))\cdot 2$, so $length(f(60))\ge k_{\max}$.↵
↵
It means that for all $n>60$, the $k$-th character of $f(n)$ can only be in $str_1$ or the first $f(n-1)$.↵
↵
Then we can answer a query in $O(\log k)$ time.↵
↵
[896B — Ithea Plays With Chtholly](http://codeforces.me/contest/896/problem/B) By [user:dogther,2017-12-04]↵
↵
As the initial sheet "has already" in a non-decreasing order (although it has no numbers), what we should do is just "maintain" this order.↵
↵
We use a simple method to do so: find the first sheet whose number is strictly greater than the given number (or it's an empty sheet) and replace it with the new number.↵
↵
For each round, we either replace an existing number with a strictly smaller one, or fill in an empty sheet. The first case will happen at most $c-1$ times for each sheet, and the second case will happen only once for each sheet. Thus in total, we will modify a sheet for at most $c$ times. Thus, the total rounds won't be more than $n \times c$.↵
↵
To pass all the tests, we only need to maintain 2 similar sequences, one non-decreasing from the first and one non-increasing from the last, which makes a total round of $(n-1)\times \lceil \frac{c}{2} \rceil +1$, precisely, and use binary search or brute force to complete the "finding" process.↵
↵
[896C — Willem, Chtholly and Seniorious](http://codeforces.me/contest/896/problem/C) By [user:ODT,2017-12-04]↵
↵
This is an interesting algorithm which can easily deal with many data structure problems------if the data is random...↵
↵
I initially named it as "Old Driver Tree" ( Which is my codeforces ID ).↵
↵
(But now I call it Chtholly Tree~).↵
↵
We can find that there is an operation that makes a range of number the same.↵
↵
We can use an interval tree (std::set is enough) to maintain every interval that consists of the same number.↵
↵
And for operation $2$, we destory all the intervals in range $[l,r]$ , and put in a new interval $[l,r]$ into the interval tree.↵
↵
For operations $1$, $3$, $4$, we can brute-forcely walk on the tree, find every interval in range $[l,r]$, and do the required operation on it.↵
↵
Proof of time complexity:↵
↵
We suppose that we have a randomly selected range $[l,r]$ now, and we randomly choose which operation it is, suppose that there are $x$ intervals in this range.↵
↵
1/4 possibility we use $O( x )$ time to erase $O( x )$ nodes.↵
↵
2/4 possibility we use $O( x )$ time to erase nothing.↵
↵
1/4 possibility we use $O( x )$ time to erase nothing and add 2 new nodes into the tree.↵
↵
So we are expected to use $O( x )$ time to erase $O( x )$ nodes.↵
↵
By using interval tree to maintain, the time complexity of this problem is $O( m\log n )$.↵
↵
If operation $3$ and $4$ are changed into output the sum of $a_i$ for every $i$ range $[l,r]$, it seems that the time complexity may change into $O( m\log\log n )$ , but I do not know how to prove it...↵
↵
[896D — Nephren Runs a Cinema](http://codeforces.me/contest/896/problem/D) By [user:dogther,2017-12-04]↵
↵
First let's consider a simpler problem that there are no customers with VIP cards and there are no 50-$yuan$ notes left. For convinence, we suppose that $n$ is an even number. The situation that $n$ is an odd number will be similar.↵
↵
By defining points (number of customers currently, number of 50-$yuan$ note left) on a 2d-plane, the answer to our second question is the ways of drawing lines from (0,0) to (n,0), such that two adjacent points' y-axis have a difference of 1, and that all the points are above the x-axis.↵
↵
The total routes will be $C_{n}^{n/2}$, but some of them are invalid. Consider another route starting from (0,-2).↵
↵
For each invalid way in the previous route, consider the first point (x,y) that y<0 (y=-1).↵
↵
By creating a symmetry route with y=-1 for the route before this point, this route will become exactly one route starting from (0,-2), and every route starting from (0,-2) will become an invalid route in a similar way.↵
↵
So the number of invalid routes is $C_{n}^{n/2-1}$ (that is the number of routes from (0,-2) to (n,0)). Thus the answer will be $C_{n}^{n/2}-C_{n}^{n/2-1}$.↵
↵
Similarly if there are [l,r] 50-$yuan$ notes left, the answer will be $C_{n}^{n/2-r/2}-C_{n}^{n/2-l/2-1}$.↵
↵
Now let's enumerate how many customers are there with VIP cards. If there are $i$ of them, the answer will time a factor $C_{n}^{i}$. ↵
↵
One last question is about the modulo number. First separate it into forms like $(p_1^{a_1})*(p_2^{a_2})$... where $p_1...$ are primes. We can calculate how many factor $p_i$ are there in $(j!)$, and the modulo value of the remaining ones.↵
↵
Each time we take out a facter $p_i$ in $(j!)$, and it becomes some product of numbers that are not divisble by $p_i$ as well as a remaining part $(j/p_i)!$. For example, we want to calculate the number of factor 3 in (16!), and the product of numbers that are not divisble by 3 in (16!) mod (3^2). Then we have:↵
↵
16! = (1 * 2 * 4 * 5 * 7 * 8 * 10 * 11 * 13 * 14 * 16) * (1 * 2 * 3 * 4 * 5) * (3^5)↵
↵
The first part are not divisble by 3, so we can calculate their value (mod 3^2) in advance, the second part is a smaller problem (5!), so we can solve it recursively. For the number of factor 3, just add 5 in this case and solve it recursively.↵
↵
After calculating how many factor $p_i$ in $(j!)$ and the modulo value of the remaining ones, we can calculate the combnation numbers correctly. Finally use Chinese Remainder Algorithm to combine them.↵
↵
[896E — Welcome home, Chtholly](http://codeforces.me/contest/896/problem/E) By [user:ODT,2017-12-04]↵
↵
I'm really sorry for letting the brute force algorithm pass the tests...↵
↵
My code uses about 800ms, in order to let some algorithms with large constant or larger time complexity ( like $O( m\sqrt{n\log n} )$ ) pass, I set the time limit to 3000ms.↵
↵
The most naive brute forces uses about 8000ms to 9000ms, I added some tricks and the fastest can pass the tests in 5600ms.↵
↵
In all the contests I've attended, pragma GCC was not allowed to be used...↵
↵
But on codeforces, this can optimize brute force algorithm from 5600ms to about 2500ms...↵
↵
<img src='http://i4.bvimg.com/599369/8c9586c17c01019b.png'>↵
↵
Thanks to [user:MrDindows,2017-12-03] and [user:Shik,2017-12-03] for teaching me this lesson...↵
↵
My solution to this problem:↵
↵
Split the array into $O( \sqrt{n} )$ blocks, each containing $O( \sqrt{n} )$ numbers.↵
↵
In each block, for example block $x$, use $f[x][v]$ to represent the number of $v$ in block $x$.↵
↵
For each number $i$, $belong[i]$ is the the block that $i$ is in.↵
↵
We need to maintain each number in the block.↵
↵
This can be maintained by using DSU or linked list.↵
↵
By maintaining this, we can get the value of every number in a block in $O( \sqrt{n} )$ time.↵
↵
Notice that this two operations are the same:↵
↵
$1.$For every number that is bigger than $x$, decrease it by $x$↵
↵
$2.$Decrease every number by $x$, and for every number that is less than $1$, increase it by $x$↵
↵
For operation $1$:↵
↵
We get the value of each number in block $belong[l]$ and $belong[r]$ using the DSU or linked list, then for every number that should change, we change them.↵
↵
Then we build block $belong[l]$ and $belong[r]$ up again.↵
↵
For blocks numbered from $belong[l] + 1$ to $belong[r] - 1$:↵
↵
If $x \times 2$ $\leq $ max value in block $p$↵
We merge all the numbers in range $[1,x]$ to $[x+1,x\times2]$, and add $x$ to $tag[p]$ , $tag[p]$ means that all the numbers in block $p$ has decreased by $tag[p]$.↵
↵
If $x \times 2$ $\textgreater$ max value in block $p$↵
We merge all the numbers in range $[x+1,max value]$ to $[1,max value - x]$.↵
↵
For operation $2$:↵
↵
We get the value of each number in block $belong[l]$ and $belong[r]$ using the DSU or linked list.↵
↵
We only need to traverse all the numbers in blocks $belong[l]$ and $belong[r]$, and traverse all the blocks between $belong[l]$ and $belong[r]$.↵
↵
For block $i$ in range $[l,r]$, $f[i][x + tag[i]]$ is the number of $x$ in block $i$, so we just need to add this into the answer↵
↵
Proof of time complexity:↵
↵
There are $O( \sqrt{n} )$ blocks.↵
↵
The difference between the max number and the min number in each block is initially $n$.↵
So the sum of this in every block is $O( n\sqrt{n} )$.↵
↵
For each operation $1$, we use $O( x )$ time or $O( max - x )$ time to make the difference of max and min element $O( x )$ or $O( max - x )$ smaller.↵
↵
For each operation $2$, we traverse $O( \sqrt{n} )$ numbers and $O( \sqrt{n} )$ blocks.↵
↵
So the total time complexity if $O( n\sqrt{n} + m\sqrt{n} )$↵
↵
There seems to be another algorithm with the same time complexity, and has a smaller constant, but I couldn't prove its complexity so I used this algorithm instead.↵
↵
<img src='http://i4.bvimg.com/599369/2bc9aa596d3ada38.png'>↵
↵
(The missing picture on Div. 1E)↵
↵
If you don't like it, just skip it please.
↵
For every $i$ in range $[l,r]$, if $c_i$ is $c_1$ then change it into $c_2$...↵
↵
Because $n, m$ are all very small, $O( nm )$ can easily pass it.↵
↵
PS. You can use binary search tree to solve it in $O( m\log n )$ time.↵
↵
[897B — Chtholly's request](http://codeforces.me/contest/897/problem/B) By [user:mcfx,2017-12-04]↵
↵
The $k$-th smallest zcy number is $conn(str(k),rev(str(k)))$, where $str$ denotes the decimal representation of a positive integer as a string, $conn$ denotes the concatenation two strings, and $rev$ denotes the reverse of a string.↵
↵
Then go over the smallest $k$ such numbers and sum them up to obtain the answer.↵
↵
[896A — Nephren gives a riddle](http://codeforces.me/contest/896/problem/A) By [user:mcfx,2017-12-04]↵
↵
$f(n)=str_1+f(n-1)+str_2+f(n-1)+str_3$.↵
↵
First we can compute the length of $f(n)$ for all possible $n$.↵
↵
For a pair of $(n,k)$, we can easily determine which part the $k$-th character is in.↵
↵
If it's in $f(n-1)$, we can solve the problem recursively.↵
↵
The complexity of this algorithm is $O(n)$, which is sufficient to pass all tests.↵
↵
Obviously, $length(f(n))\ge length(f(n-1))\cdot 2$, so $length(f(60))\ge k_{\max}$.↵
↵
It means that for all $n>60$, the $k$-th character of $f(n)$ can only be in $str_1$ or the first $f(n-1)$.↵
↵
Then we can answer a query in $O(\log k)$ time.↵
↵
[896B — Ithea Plays With Chtholly](http://codeforces.me/contest/896/problem/B) By [user:dogther,2017-12-04]↵
↵
As the initial sheet "has already" in a non-decreasing order (although it has no numbers), what we should do is just "maintain" this order.↵
↵
We use a simple method to do so: find the first sheet whose number is strictly greater than the given number (or it's an empty sheet) and replace it with the new number.↵
↵
For each round, we either replace an existing number with a strictly smaller one, or fill in an empty sheet. The first case will happen at most $c-1$ times for each sheet, and the second case will happen only once for each sheet. Thus in total, we will modify a sheet for at most $c$ times. Thus, the total rounds won't be more than $n \times c$.↵
↵
To pass all the tests, we only need to maintain 2 similar sequences, one non-decreasing from the first and one non-increasing from the last, which makes a total round of $(n-1)\times \lceil \frac{c}{2} \rceil +1$, precisely, and use binary search or brute force to complete the "finding" process.↵
↵
[896C — Willem, Chtholly and Seniorious](http://codeforces.me/contest/896/problem/C) By [user:ODT,2017-12-04]↵
↵
This is an interesting algorithm which can easily deal with many data structure problems------if the data is random...↵
↵
I initially named it as "Old Driver Tree" ( Which is my codeforces ID ).↵
↵
(But now I call it Chtholly Tree~).↵
↵
We can find that there is an operation that makes a range of number the same.↵
↵
We can use an interval tree (std::set is enough) to maintain every interval that consists of the same number.↵
↵
And for operation $2$, we destory all the intervals in range $[l,r]$ , and put in a new interval $[l,r]$ into the interval tree.↵
↵
For operations $1$, $3$, $4$, we can brute-forcely walk on the tree, find every interval in range $[l,r]$, and do the required operation on it.↵
↵
Proof of time complexity:↵
↵
We suppose that we have a randomly selected range $[l,r]$ now, and we randomly choose which operation it is, suppose that there are $x$ intervals in this range.↵
↵
1/4 possibility we use $O( x )$ time to erase $O( x )$ nodes.↵
↵
2/4 possibility we use $O( x )$ time to erase nothing.↵
↵
1/4 possibility we use $O( x )$ time to erase nothing and add 2 new nodes into the tree.↵
↵
So we are expected to use $O( x )$ time to erase $O( x )$ nodes.↵
↵
By using interval tree to maintain, the time complexity of this problem is $O( m\log n )$.↵
↵
If operation $3$ and $4$ are changed into output the sum of $a_i$ for every $i$ range $[l,r]$, it seems that the time complexity may change into $O( m\log\log n )$ , but I do not know how to prove it...↵
↵
[896D — Nephren Runs a Cinema](http://codeforces.me/contest/896/problem/D) By [user:dogther,2017-12-04]↵
↵
First let's consider a simpler problem that there are no customers with VIP cards and there are no 50-$yuan$ notes left. For convinence, we suppose that $n$ is an even number. The situation that $n$ is an odd number will be similar.↵
↵
By defining points (number of customers currently, number of 50-$yuan$ note left) on a 2d-plane, the answer to our second question is the ways of drawing lines from (0,0) to (n,0), such that two adjacent points' y-axis have a difference of 1, and that all the points are above the x-axis.↵
↵
The total routes will be $C_{n}^{n/2}$, but some of them are invalid. Consider another route starting from (0,-2).↵
↵
For each invalid way in the previous route, consider the first point (x,y) that y<0 (y=-1).↵
↵
By creating a symmetry route with y=-1 for the route before this point, this route will become exactly one route starting from (0,-2), and every route starting from (0,-2) will become an invalid route in a similar way.↵
↵
So the number of invalid routes is $C_{n}^{n/2-1}$ (that is the number of routes from (0,-2) to (n,0)). Thus the answer will be $C_{n}^{n/2}-C_{n}^{n/2-1}$.↵
↵
Similarly if there are [l,r] 50-$yuan$ notes left, the answer will be $C_{n}^{n/2-r/2}-C_{n}^{n/2-l/2-1}$.↵
↵
Now let's enumerate how many customers are there with VIP cards. If there are $i$ of them, the answer will time a factor $C_{n}^{i}$. ↵
↵
One last question is about the modulo number. First separate it into forms like $(p_1^{a_1})*(p_2^{a_2})$... where $p_1...$ are primes. We can calculate how many factor $p_i$ are there in $(j!)$, and the modulo value of the remaining ones.↵
↵
Each time we take out a facter $p_i$ in $(j!)$, and it becomes some product of numbers that are not divisble by $p_i$ as well as a remaining part $(j/p_i)!$. For example, we want to calculate the number of factor 3 in (16!), and the product of numbers that are not divisble by 3 in (16!) mod (3^2). Then we have:↵
↵
16! = (1 * 2 * 4 * 5 * 7 * 8 * 10 * 11 * 13 * 14 * 16) * (1 * 2 * 3 * 4 * 5) * (3^5)↵
↵
The first part are not divisble by 3, so we can calculate their value (mod 3^2) in advance, the second part is a smaller problem (5!), so we can solve it recursively. For the number of factor 3, just add 5 in this case and solve it recursively.↵
↵
After calculating how many factor $p_i$ in $(j!)$ and the modulo value of the remaining ones, we can calculate the combnation numbers correctly. Finally use Chinese Remainder Algorithm to combine them.↵
↵
[896E — Welcome home, Chtholly](http://codeforces.me/contest/896/problem/E) By [user:ODT,2017-12-04]↵
↵
I'm really sorry for letting the brute force algorithm pass the tests...↵
↵
My code uses about 800ms, in order to let some algorithms with large constant or larger time complexity ( like $O( m\sqrt{n\log n} )$ ) pass, I set the time limit to 3000ms.↵
↵
The most naive brute forces uses about 8000ms to 9000ms, I added some tricks and the fastest can pass the tests in 5600ms.↵
↵
In all the contests I've attended, pragma GCC was not allowed to be used...↵
↵
But on codeforces, this can optimize brute force algorithm from 5600ms to about 2500ms...↵
↵
<img src='http://i4.bvimg.com/599369/8c9586c17c01019b.png'>↵
↵
Thanks to [user:MrDindows,2017-12-03] and [user:Shik,2017-12-03] for teaching me this lesson...↵
↵
My solution to this problem:↵
↵
Split the array into $O( \sqrt{n} )$ blocks, each containing $O( \sqrt{n} )$ numbers.↵
↵
In each block, for example block $x$, use $f[x][v]$ to represent the number of $v$ in block $x$.↵
↵
For each number $i$, $belong[i]$ is the the block that $i$ is in.↵
↵
We need to maintain each number in the block.↵
↵
This can be maintained by using DSU or linked list.↵
↵
By maintaining this, we can get the value of every number in a block in $O( \sqrt{n} )$ time.↵
↵
Notice that this two operations are the same:↵
↵
$1.$For every number that is bigger than $x$, decrease it by $x$↵
↵
$2.$Decrease every number by $x$, and for every number that is less than $1$, increase it by $x$↵
↵
For operation $1$:↵
↵
We get the value of each number in block $belong[l]$ and $belong[r]$ using the DSU or linked list, then for every number that should change, we change them.↵
↵
Then we build block $belong[l]$ and $belong[r]$ up again.↵
↵
For blocks numbered from $belong[l] + 1$ to $belong[r] - 1$:↵
↵
If $x \times 2$ $\leq $ max value in block $p$↵
We merge all the numbers in range $[1,x]$ to $[x+1,x\times2]$, and add $x$ to $tag[p]$ , $tag[p]$ means that all the numbers in block $p$ has decreased by $tag[p]$.↵
↵
If $x \times 2$ $\textgreater$ max value in block $p$↵
We merge all the numbers in range $[x+1,max value]$ to $[1,max value - x]$.↵
↵
For operation $2$:↵
↵
We get the value of each number in block $belong[l]$ and $belong[r]$ using the DSU or linked list.↵
↵
We only need to traverse all the numbers in blocks $belong[l]$ and $belong[r]$, and traverse all the blocks between $belong[l]$ and $belong[r]$.↵
↵
For block $i$ in range $[l,r]$, $f[i][x + tag[i]]$ is the number of $x$ in block $i$, so we just need to add this into the answer↵
↵
Proof of time complexity:↵
↵
There are $O( \sqrt{n} )$ blocks.↵
↵
The difference between the max number and the min number in each block is initially $n$.↵
So the sum of this in every block is $O( n\sqrt{n} )$.↵
↵
For each operation $1$, we use $O( x )$ time or $O( max - x )$ time to make the difference of max and min element $O( x )$ or $O( max - x )$ smaller.↵
↵
For each operation $2$, we traverse $O( \sqrt{n} )$ numbers and $O( \sqrt{n} )$ blocks.↵
↵
So the total time complexity if $O( n\sqrt{n} + m\sqrt{n} )$↵
↵
There seems to be another algorithm with the same time complexity, and has a smaller constant, but I couldn't prove its complexity so I used this algorithm instead.↵
↵
<img src='http://i4.bvimg.com/599369/2bc9aa596d3ada38.png'>↵
↵
(The missing picture on Div. 1E)↵
↵
If you don't like it, just skip it please.
