501 - Black Box (by: Alexander Dolin)
Use two heap data structures, one is maximum heap (heap1) and the other is minimum heap (heap2). At the step i, we have to find the number with order statistic i in the final sorted array. So, we keep first i-1 numbers in heap1, other numbers in heap2. Minimum from heap2 will be the answer.
507 - Jill Rides Again
The underlying algorithm for this problem is a "maximum interval sum", and there is a nice linear time DP algorithm to solve this problem (yes only linear time algo can pass the time limit, since the problem size can be as big as 20000 'stops'.
Niceness[i] = Niceness[start] + ... + Niceness[i]
if sum from index 'start' to i is >= 0 or
Niceness[i], set start=i+1 (start new interval)
if sum from index 'start' to 'i' < 0 The simple reasoning of this DP formulation is as follows: if you have positive (or zero) sum, then this current sequence can still be extended to a longer interval with bigger value or at least similar value but longer interval... but if the partial sum is negative... then there is no point to extend it further... Example from sample input: Niceness: -1 6 Sum : -1 6 ^ max sum Niceness: 4 -5 4 -3 4 4 -4 4 -5 Sum : 4 -1 4 1 5 9 5 9 4 ^ ^ stop max sum Niceness: -2 -3 -4 Sum : -2 -3 -4 ^ max sum, but negative... no nice parts So, just do a linear sweep from left to right, accumulate the sum one element by one element, start new interval whenever you encounter partial sum < 0... At the end, output the longest and most nicest, "j-i" interval. 514 - Rails Using only one-end station (Hint: a Stack), you must determine whether it is possible to marshal the coaches in the order required on the corresponding line of the input file. Output "Yes" if it is possible, otherwise output "No". Solution: 1. Use a stack. 2. Trial & Error using a piece of blank paper first, then you'll see the pattern. Common Mistake: 1. Input can be like this 5 1 4 3 2 5 0 0 And the output for this is "Yes". 2. Incorrect stack implementation, an 1000-elements array is sufficient. 516 - Prime Land You are given a "Prime representation" of an integer number X | 2<=32767. Then you have to decrement X by 1, and then output the value of X in its new "Prime representation" Example of "Prime representation": Let X=5, then Prime representation of X is 5^1 (Written "5 1") Let X=10, then Prime representation of X is 5^1 * 2^1 (Written "5 1 2 1") Let X=100, then Prime representation of X is 5^2 * 2^2 (Written "5 2 2 2") So "Prime representation" is the form of product of powers of prime factors. There will only be one way to represent X in its "Prime representation" for all X>1. Solution:
1. Convert "prime representation" to an integer X, multiply the powers of prime factors of X.
2. Decrement X by 1.
3. Turn X into its "Prime representation" again. Use prime factors algorithm
(See my programming page) ~> (Similar to number 583).
524 - Prime Ring Problem (with help from: Arief, Lucas)
This problem can be solved using efficient backtracking. Even though n is "just" 16, finding the combination of "prime ring" can be as big as 16! if you do brute force. Prune whenever you can.
526 - String Distance and Transform Process
Although Edit Distance DP algorithm is quite popular. It's a bit hard to tweak the code to get it accepted by the judge. I can only say good luck in tweaking your Edit Distance / Approximate String Matching algorithm.
529 - Addition Chains
Again, this is another backtracking problem. Always remember the rule of thumb: "Prune whenever you can"
530 - Binomial Showdown (by: Felix Halim)
This is just standard nCr (Combination) calculation, where nCr = n! / (r!*(n-r)!).
But this one uses very large numbers and you are likely to get overflow, or time limit.
It's up to you to design any algorithm that can solve this.
However, the basic idea is how to make algorithm like this:
1. Simplifies n! / (r!*(n-r)!) to simpler form.
Example: 5C2 = 5! / (2!*(5-2)!) = 5! / (2!*3!) = 5*4*3*2*1/2*1*3*2*1 = 5*2
2. And then multiply the simplest form of nCr (5*2) = 10
3. Output the result. Using this trick, you will not get overflow error.
531 - Compromise
If the normal LCS compare characters, this version compare strings..., just re-use your LCS algorithm and adjust it to compare strings... done
532 - Dungeon Master (by: Felix Halim)
2-D maze problems are very familiar. This problem is similar, but in 3-D. Fortunately, you don't need to worry much about the complexity of moving to 3-D space..., you can simply re-use your BFS code without major modifications.
533 - Equation Solver
A bit complex... Given the grammar of the math in BNF, calculate the unknown variable. You can simulate this using elementary school technique..., quite troublesome, I know, but doable...
534 - Frogger
A frog's jump range is must be at least as long as the longest jump occurring in the sequence.
The frog distance (minimax distance) between two stones therefore is defined as the minimum necessary jump range (NOT total jumps) over all possible paths between the two stones.
Example:
2
0 0
3 4
Output -> 5.000, direct jump from stone freddy stone to fiona stone
3
0 0
3 4
3 0
Output -> 4.000
Jump from freddy stone (0,0) to intermediate stone (3,0), range->3.000, then jump from intermediate stone (3,0) to fiona stone (4,0), range->4.000. The longest jump in this sequence is 4.000, therefore the jump range for this sequence is 4.000.
This sequence is smaller than direct jump (example 1 above) which is 5.000, so, for this test case, you output 4.000 (minimum necessary jump range).
Algorithm: All Pairs Shortest Path, for example: Floyd Warshall algorithm.
535 - Globetrotter
Use spherical / geometrical distance formula. Read more here.
536 - Tree Recovery
You are given two representation of a tree, the preOrder and inOrder representation.
You have to re-build the tree and output the tree in postOrder representation.
Use Tree Recursion, recursively partition the string based on this fact:
the first element of preOrder is the root, find this root in inOrder representation, partition the string according to that root, recursively.
537 - Artificial Intelligence?
This problem is basically simple, compute P=U*I or U=P/I or I=P/U. However, parsing the input can be harder than the problem itself :). Master your programming language I/O skill in order to parse the input correctly...
538 - Balancing Bank Accounts
Sort the input and greedily assign the money properly...
539 - The Settlers of Catan
Simple backtracking will solve this problem. Just explore everything... The number of node and edges are small (less than 25).
540 - Team Queue
Even though the problem description is clear and should be easy... The size of input will be the real problem. "In constant time" may be impossible (not sure)... but binary search (log n) is sufficient (I get accepted). This problem can be a good test for testing how efficient your code is (in terms of memory and speed).
541 - Error Correction
Error correction mainly used in Computer system's memory management. There are even parity and odd parity. In this problem, we have to check even parity.
Count the number of ``1'' for each rows and columns, all of them must be even.
If there exist one or more error, do this:
If the error is on the same ROW and COLUMN, then output "change bit (row,col)"
else output "corrupt"
543 - Golbach's Conjecture
Simulate this: "Every even number greater than 4 can be written as the sum of two odd prime numbers."
Store a prime list in array (up to n), find the pair (If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized.)
This line is, however, will not be executed...(If I'm not mistaken) (If there is no such pair, print a line saying ``Goldbach's conjecture is wrong.")
544 - Heavy Cargo
Whenever you encounter a phrase like "maximize the minimum" in a problem statement... you can guess that the problem has to do with All Pairs Shortest Path, Floyd Warshall maximin variant. Try it.
551 - Nesting a Bunch of Brackets
Use a stack, push when you encounter open bracket, pop when you encounter close bracket. Errors will occurs if the popped item is not matching with current close bracket, or when at the end, the stack is not empty...
555 - Bridge Hands
Simple card simulation. Simply deal those card to 4 piles (remember starting position, it can be from North, East, South or West). After that, sort the piles according to this problem rules.
556 - Amazing
A simulation of robot movement..., just do according to problem description.
557 - Burger
Refer to your discrete mathematic books (probability theory)
Probability a child get a hamburger => (1/2)^x,
where x=people-2 because we want to keep Ben & Bill get the same burger
since this flipping is done sequentially, first child get (1/2)^x,
second child get (1/2)^x * (1/2)^x, and so on...
558 - Wormholes
Construct the graph, and then pass this graph to your Bellman Ford Shortest Path algorithm. Bellman Ford can detect the presence of negative cycle, and this is what you want to know...
562 - Dividing Coins (by: Abdullah Al Mamun)
Use one dimensional, left to right traversal, dynamic programming.
567 - Risk
This is an All Pairs Shortest Path problem. However, since total vertex is small (at most 20 countries), a simple brute force DFS/BFS will do...
571 - Jugs
Another backtracking problem. You have 6 branching factors (6 type of moves). Perform this backtracking by disallowing repeated cycles (by storing a flag in memory that you already visit a similar jugs configuration before). The main problem here is just the Time Limit...
572 - Oil Deposits
Another backtracking problem... (There are a lot of backtracking problem in this volume). Starting from a particular '@' cell, flood fill it to 8 directions..., then find the number of components.
573 - The Snail
A simple problem, but there are several traps:
1. Beware when fatique<0, the snail will not fall down again 2. If the snail already manage to get out, don't come back !!! 574 - Sum It Up Again..... another backtracking problem. Backtrack, prune, backtrack, prune... 575 - Skew Binary Base number, but 'skewed'... so, use the new rule to convert the binary -> decimal.
576 - Haiku Review
Hm... just follow the rules... I can't tell much...
579 - Clock Hands
You have to determine the angle between 2 clock hands.
Use this simple algorithm:
hAngle=h*30+(m/60)*30; // Angle from 12o'clock to hour hand
mAngle=m*6; // Angle from 12o'clock to minute hand
angle=abs(hAngle-mAngle);
if (angle>180) angle=360-angle;
Common Mistake
1. Forget to subtract the angle with 180 if it is larger 180 (They want the smallest angle)
2. This is 12 hour clock !!! Clock with hands usually 12-hour clock !!!
580 - Critical Mass (by: Rupam)
n L's or U's can stacked up 2^n ways. Say, there is x ways of arranging stacks in which, there are no more than 2 consecutive U's. Then, number of ways stacks can be arranged in which there is at least one occurrence of three consecutive U's, can be written as:
C(n) = 2^n - x.
Now, x can be denoted as A(n), number of ways of arranging stacks of n L's or U's in which, there are no more than 2 consecutive U's and lets name this type of arrangement: B type.
Then n things of B type can be seen as:
_______________ = ________________L + _______________LU + ________________LUU
n things of B type n-1 things of B type n-2 things of B type n-3 things of B type
The right hand side permutations makes n things of B type as well, so both sides are equal.
Therefore, A(n) = A(n-1) + A(n-2) + A(n-3)
where, A(1) = 2, A(2) = 4, A(3) = 7
these are base cases for B type when n = 1,2, and 3
In mathematic, there is a special series called "Tribonacci series", where
T(n) = T(n-1) + T(n-2) + T(n-3)
with base cases:
T(1) = 1, T(2) = 1, T(3) = 2
Our series A(n) can be transformed to Tribonacci series => A(n) = T(n+2)
Therefore:
C(n) = 2^n - x
C(n) = 2^n - A(n)
C(n) = 2^n - T(n+2)
So, just implement this C(n) formula :)
Actually, whoever understands the problem this way, before knowing that the solution is 2^n - T(n+2) can go easily without knowing Tribonacci series, using A(n).
583 - Prime Factors
This problem wants us to convert a number to its Prime factors. Very similar to 516, but this one is simpler. Solution:
1. Take the input X.
2. Start with a counter=2.
3. If counter can properly divide X then print counter and divide X by counter.
4. If counter>=sqrt(X), stop and print X directly, remember divisibility property.
5. If X=1 then stop else go back to step 3.
Common Mistake
1. Remember what to print for negative numbers and when to print " x " symbol.
2. Time limit. Remember this divisibility property: No number > sqrt(X) can properly divide X.
584 - Bowling
Another simulation problem. There are many ways to solve this problem, pick the one that is easiest for you. Familiarize yourself with bowling scoring rule as described in problem description. There is no trap in this problem. As long as you can model the scoring rule in your code, you'll get accepted.
587 - There's treasure everywhere!
Simply move according to the input data. At the final destination, just compute the distance between final destination and point (0,0) using standard phytagoras calculation.
Start with initial value x=10e-12 && y=10e-12, I don't know why, but using this trick your program will get accepted, otherwise, you will possible got WA.
589 - Pushing Boxes (by: Arif Uzzaman)
Two bfs functions are needed. One for you and one for box. But you have to be careful on some point.
In bfs function for the box:
- in general bfs algorithm one node is visited only once but in this case a node can visited more than once.
- the box can visit a node from east once, from west once, from south once and from north once. so a node can be visited by a box 4 times except the initial node for the box, it can be visited at most five times.
Some critical inputs:
12 11
###########
#.#.......#
#.###..####
#...#...#.#
#####.##..#
#.....#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
12 11
###########
#.#.......#
#.###..####
#...#...#.#
#####.##..#
#...#.#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
12 11
##.....####
#.........#
#.###.#.###
#...#...#.#
#####.##..#
#...#.#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
output:
Maze #1
wnNwwwnneeeSnwwwsseeEEE
Maze #2
wnNNNNNennwSSSSSesWWWswwnEEEEEE
Maze #3
wnNNNNNeennwwSSSSSesWWWswwnEEEEEE
591 - Box of Bricks
There are a lot of programming problem similar to this one, memorize this useful technique (If you want).
Example: 5 2 4 1 7 5
Sum all items => 5+2+4+1+7+5=24
Find the average value => 24/6=4
Do a looping from first item, count the differences from the average, get the absolute value
5-4 = 1 => 1
2-4 = -2 => 2
4-4 = 0 => 0
1-4 = -3 => 3
7-4 = 3 => 3
5-4 = 1 => 1
Sum the absolute difference => 1+2+0+3+3+1=10
Divide by 2 (because you don't have to do it twice, think about it) => 10/2=5
Output the result = 5
594 - One Little, Two Little, Three Little Endians
You need to swap bits!
The input is an integer N, convert this to 32-bit integer. You have to swap 8 bits of Least Significant Bit to Most Significant Bit. Partition them into this:
X1 X2 X3 X4
Where X1,2,3, & 4 is 8-bit from the complete 32-bit integer.
Then you need to swap it so the position is like this:
X4 X3 X2 X1
So, use bitwise manipulation
<< Shift Left >> Shift Right
& bitwise And
The implementation is up to you.
598 - Bundling Newspaper
This is a simple backtracking enumeration problem. The only problem that you may encounter is in reading the multiple input format precisely. Other than that, this is just a simple brute-force enumeration problem using backtrack. Btw, no need to sort the newspaper names. The terms "lexicographic" in the problem refer to the notation A,B,C,D given in the problem, that is, you enumerate using the order given in the input, not based on the newspaper name.
Use two heap data structures, one is maximum heap (heap1) and the other is minimum heap (heap2). At the step i, we have to find the number with order statistic i in the final sorted array. So, we keep first i-1 numbers in heap1, other numbers in heap2. Minimum from heap2 will be the answer.
507 - Jill Rides Again
The underlying algorithm for this problem is a "maximum interval sum", and there is a nice linear time DP algorithm to solve this problem (yes only linear time algo can pass the time limit, since the problem size can be as big as 20000 'stops'.
Niceness[i] = Niceness[start] + ... + Niceness[i]
if sum from index 'start' to i is >= 0 or
Niceness[i], set start=i+1 (start new interval)
if sum from index 'start' to 'i' < 0 The simple reasoning of this DP formulation is as follows: if you have positive (or zero) sum, then this current sequence can still be extended to a longer interval with bigger value or at least similar value but longer interval... but if the partial sum is negative... then there is no point to extend it further... Example from sample input: Niceness: -1 6 Sum : -1 6 ^ max sum Niceness: 4 -5 4 -3 4 4 -4 4 -5 Sum : 4 -1 4 1 5 9 5 9 4 ^ ^ stop max sum Niceness: -2 -3 -4 Sum : -2 -3 -4 ^ max sum, but negative... no nice parts So, just do a linear sweep from left to right, accumulate the sum one element by one element, start new interval whenever you encounter partial sum < 0... At the end, output the longest and most nicest, "j-i" interval. 514 - Rails Using only one-end station (Hint: a Stack), you must determine whether it is possible to marshal the coaches in the order required on the corresponding line of the input file. Output "Yes" if it is possible, otherwise output "No". Solution: 1. Use a stack. 2. Trial & Error using a piece of blank paper first, then you'll see the pattern. Common Mistake: 1. Input can be like this 5 1 4 3 2 5 0 0 And the output for this is "Yes". 2. Incorrect stack implementation, an 1000-elements array is sufficient. 516 - Prime Land You are given a "Prime representation" of an integer number X | 2
1. Convert "prime representation" to an integer X, multiply the powers of prime factors of X.
2. Decrement X by 1.
3. Turn X into its "Prime representation" again. Use prime factors algorithm
(See my programming page) ~> (Similar to number 583).
524 - Prime Ring Problem (with help from: Arief, Lucas)
This problem can be solved using efficient backtracking. Even though n is "just" 16, finding the combination of "prime ring" can be as big as 16! if you do brute force. Prune whenever you can.
526 - String Distance and Transform Process
Although Edit Distance DP algorithm is quite popular. It's a bit hard to tweak the code to get it accepted by the judge. I can only say good luck in tweaking your Edit Distance / Approximate String Matching algorithm.
529 - Addition Chains
Again, this is another backtracking problem. Always remember the rule of thumb: "Prune whenever you can"
530 - Binomial Showdown (by: Felix Halim)
This is just standard nCr (Combination) calculation, where nCr = n! / (r!*(n-r)!).
But this one uses very large numbers and you are likely to get overflow, or time limit.
It's up to you to design any algorithm that can solve this.
However, the basic idea is how to make algorithm like this:
1. Simplifies n! / (r!*(n-r)!) to simpler form.
Example: 5C2 = 5! / (2!*(5-2)!) = 5! / (2!*3!) = 5*4*3*2*1/2*1*3*2*1 = 5*2
2. And then multiply the simplest form of nCr (5*2) = 10
3. Output the result. Using this trick, you will not get overflow error.
531 - Compromise
If the normal LCS compare characters, this version compare strings..., just re-use your LCS algorithm and adjust it to compare strings... done
532 - Dungeon Master (by: Felix Halim)
2-D maze problems are very familiar. This problem is similar, but in 3-D. Fortunately, you don't need to worry much about the complexity of moving to 3-D space..., you can simply re-use your BFS code without major modifications.
533 - Equation Solver
A bit complex... Given the grammar of the math in BNF, calculate the unknown variable. You can simulate this using elementary school technique..., quite troublesome, I know, but doable...
534 - Frogger
A frog's jump range is must be at least as long as the longest jump occurring in the sequence.
The frog distance (minimax distance) between two stones therefore is defined as the minimum necessary jump range (NOT total jumps) over all possible paths between the two stones.
Example:
2
0 0
3 4
Output -> 5.000, direct jump from stone freddy stone to fiona stone
3
0 0
3 4
3 0
Output -> 4.000
Jump from freddy stone (0,0) to intermediate stone (3,0), range->3.000, then jump from intermediate stone (3,0) to fiona stone (4,0), range->4.000. The longest jump in this sequence is 4.000, therefore the jump range for this sequence is 4.000.
This sequence is smaller than direct jump (example 1 above) which is 5.000, so, for this test case, you output 4.000 (minimum necessary jump range).
Algorithm: All Pairs Shortest Path, for example: Floyd Warshall algorithm.
535 - Globetrotter
Use spherical / geometrical distance formula. Read more here.
536 - Tree Recovery
You are given two representation of a tree, the preOrder and inOrder representation.
You have to re-build the tree and output the tree in postOrder representation.
Use Tree Recursion, recursively partition the string based on this fact:
the first element of preOrder is the root, find this root in inOrder representation, partition the string according to that root, recursively.
537 - Artificial Intelligence?
This problem is basically simple, compute P=U*I or U=P/I or I=P/U. However, parsing the input can be harder than the problem itself :). Master your programming language I/O skill in order to parse the input correctly...
538 - Balancing Bank Accounts
Sort the input and greedily assign the money properly...
539 - The Settlers of Catan
Simple backtracking will solve this problem. Just explore everything... The number of node and edges are small (less than 25).
540 - Team Queue
Even though the problem description is clear and should be easy... The size of input will be the real problem. "In constant time" may be impossible (not sure)... but binary search (log n) is sufficient (I get accepted). This problem can be a good test for testing how efficient your code is (in terms of memory and speed).
541 - Error Correction
Error correction mainly used in Computer system's memory management. There are even parity and odd parity. In this problem, we have to check even parity.
Count the number of ``1'' for each rows and columns, all of them must be even.
If there exist one or more error, do this:
If the error is on the same ROW and COLUMN, then output "change bit (row,col)"
else output "corrupt"
543 - Golbach's Conjecture
Simulate this: "Every even number greater than 4 can be written as the sum of two odd prime numbers."
Store a prime list in array (up to n), find the pair (If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized.)
This line is, however, will not be executed...(If I'm not mistaken) (If there is no such pair, print a line saying ``Goldbach's conjecture is wrong.")
544 - Heavy Cargo
Whenever you encounter a phrase like "maximize the minimum" in a problem statement... you can guess that the problem has to do with All Pairs Shortest Path, Floyd Warshall maximin variant. Try it.
551 - Nesting a Bunch of Brackets
Use a stack, push when you encounter open bracket, pop when you encounter close bracket. Errors will occurs if the popped item is not matching with current close bracket, or when at the end, the stack is not empty...
555 - Bridge Hands
Simple card simulation. Simply deal those card to 4 piles (remember starting position, it can be from North, East, South or West). After that, sort the piles according to this problem rules.
556 - Amazing
A simulation of robot movement..., just do according to problem description.
557 - Burger
Refer to your discrete mathematic books (probability theory)
Probability a child get a hamburger => (1/2)^x,
where x=people-2 because we want to keep Ben & Bill get the same burger
since this flipping is done sequentially, first child get (1/2)^x,
second child get (1/2)^x * (1/2)^x, and so on...
558 - Wormholes
Construct the graph, and then pass this graph to your Bellman Ford Shortest Path algorithm. Bellman Ford can detect the presence of negative cycle, and this is what you want to know...
562 - Dividing Coins (by: Abdullah Al Mamun)
Use one dimensional, left to right traversal, dynamic programming.
567 - Risk
This is an All Pairs Shortest Path problem. However, since total vertex is small (at most 20 countries), a simple brute force DFS/BFS will do...
571 - Jugs
Another backtracking problem. You have 6 branching factors (6 type of moves). Perform this backtracking by disallowing repeated cycles (by storing a flag in memory that you already visit a similar jugs configuration before). The main problem here is just the Time Limit...
572 - Oil Deposits
Another backtracking problem... (There are a lot of backtracking problem in this volume). Starting from a particular '@' cell, flood fill it to 8 directions..., then find the number of components.
573 - The Snail
A simple problem, but there are several traps:
1. Beware when fatique<0, the snail will not fall down again 2. If the snail already manage to get out, don't come back !!! 574 - Sum It Up Again..... another backtracking problem. Backtrack, prune, backtrack, prune... 575 - Skew Binary Base number, but 'skewed'... so, use the new rule to convert the binary -> decimal.
576 - Haiku Review
Hm... just follow the rules... I can't tell much...
579 - Clock Hands
You have to determine the angle between 2 clock hands.
Use this simple algorithm:
hAngle=h*30+(m/60)*30; // Angle from 12o'clock to hour hand
mAngle=m*6; // Angle from 12o'clock to minute hand
angle=abs(hAngle-mAngle);
if (angle>180) angle=360-angle;
Common Mistake
1. Forget to subtract the angle with 180 if it is larger 180 (They want the smallest angle)
2. This is 12 hour clock !!! Clock with hands usually 12-hour clock !!!
580 - Critical Mass (by: Rupam)
n L's or U's can stacked up 2^n ways. Say, there is x ways of arranging stacks in which, there are no more than 2 consecutive U's. Then, number of ways stacks can be arranged in which there is at least one occurrence of three consecutive U's, can be written as:
C(n) = 2^n - x.
Now, x can be denoted as A(n), number of ways of arranging stacks of n L's or U's in which, there are no more than 2 consecutive U's and lets name this type of arrangement: B type.
Then n things of B type can be seen as:
_______________ = ________________L + _______________LU + ________________LUU
n things of B type n-1 things of B type n-2 things of B type n-3 things of B type
The right hand side permutations makes n things of B type as well, so both sides are equal.
Therefore, A(n) = A(n-1) + A(n-2) + A(n-3)
where, A(1) = 2, A(2) = 4, A(3) = 7
these are base cases for B type when n = 1,2, and 3
In mathematic, there is a special series called "Tribonacci series", where
T(n) = T(n-1) + T(n-2) + T(n-3)
with base cases:
T(1) = 1, T(2) = 1, T(3) = 2
Our series A(n) can be transformed to Tribonacci series => A(n) = T(n+2)
Therefore:
C(n) = 2^n - x
C(n) = 2^n - A(n)
C(n) = 2^n - T(n+2)
So, just implement this C(n) formula :)
Actually, whoever understands the problem this way, before knowing that the solution is 2^n - T(n+2) can go easily without knowing Tribonacci series, using A(n).
583 - Prime Factors
This problem wants us to convert a number to its Prime factors. Very similar to 516, but this one is simpler. Solution:
1. Take the input X.
2. Start with a counter=2.
3. If counter can properly divide X then print counter and divide X by counter.
4. If counter>=sqrt(X), stop and print X directly, remember divisibility property.
5. If X=1 then stop else go back to step 3.
Common Mistake
1. Remember what to print for negative numbers and when to print " x " symbol.
2. Time limit. Remember this divisibility property: No number > sqrt(X) can properly divide X.
584 - Bowling
Another simulation problem. There are many ways to solve this problem, pick the one that is easiest for you. Familiarize yourself with bowling scoring rule as described in problem description. There is no trap in this problem. As long as you can model the scoring rule in your code, you'll get accepted.
587 - There's treasure everywhere!
Simply move according to the input data. At the final destination, just compute the distance between final destination and point (0,0) using standard phytagoras calculation.
Start with initial value x=10e-12 && y=10e-12, I don't know why, but using this trick your program will get accepted, otherwise, you will possible got WA.
589 - Pushing Boxes (by: Arif Uzzaman)
Two bfs functions are needed. One for you and one for box. But you have to be careful on some point.
In bfs function for the box:
- in general bfs algorithm one node is visited only once but in this case a node can visited more than once.
- the box can visit a node from east once, from west once, from south once and from north once. so a node can be visited by a box 4 times except the initial node for the box, it can be visited at most five times.
Some critical inputs:
12 11
###########
#.#.......#
#.###..####
#...#...#.#
#####.##..#
#.....#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
12 11
###########
#.#.......#
#.###..####
#...#...#.#
#####.##..#
#...#.#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
12 11
##.....####
#.........#
#.###.#.###
#...#...#.#
#####.##..#
#...#.#...#
##.##...###
#....B..T##
#...#.#####
#####.S..##
##...######
###########
output:
Maze #1
wnNwwwnneeeSnwwwsseeEEE
Maze #2
wnNNNNNennwSSSSSesWWWswwnEEEEEE
Maze #3
wnNNNNNeennwwSSSSSesWWWswwnEEEEEE
591 - Box of Bricks
There are a lot of programming problem similar to this one, memorize this useful technique (If you want).
Example: 5 2 4 1 7 5
Sum all items => 5+2+4+1+7+5=24
Find the average value => 24/6=4
Do a looping from first item, count the differences from the average, get the absolute value
5-4 = 1 => 1
2-4 = -2 => 2
4-4 = 0 => 0
1-4 = -3 => 3
7-4 = 3 => 3
5-4 = 1 => 1
Sum the absolute difference => 1+2+0+3+3+1=10
Divide by 2 (because you don't have to do it twice, think about it) => 10/2=5
Output the result = 5
594 - One Little, Two Little, Three Little Endians
You need to swap bits!
The input is an integer N, convert this to 32-bit integer. You have to swap 8 bits of Least Significant Bit to Most Significant Bit. Partition them into this:
X1 X2 X3 X4
Where X1,2,3, & 4 is 8-bit from the complete 32-bit integer.
Then you need to swap it so the position is like this:
X4 X3 X2 X1
So, use bitwise manipulation
<< Shift Left >> Shift Right
& bitwise And
The implementation is up to you.
598 - Bundling Newspaper
This is a simple backtracking enumeration problem. The only problem that you may encounter is in reading the multiple input format precisely. Other than that, this is just a simple brute-force enumeration problem using backtrack. Btw, no need to sort the newspaper names. The terms "lexicographic" in the problem refer to the notation A,B,C,D given in the problem, that is, you enumerate using the order given in the input, not based on the newspaper name.
