You are given three integers n, a, and b. Determine if there exist two permutations p and q of length n, for which the following conditions hold:
The length of the longest common prefix of p and q is a.
The length of the longest common suffix of p and q is b.
A permutation of length n is an array containing each integer from 1 to n exactly once. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array), and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
Each test contains multiple test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows.
The only line of each test case contains three integers n, a, and b (1≤a,b≤n≤100).
Output
For each test case, if such a pair of permutations exists, output "Yes"; otherwise, output "No". You can output each letter in any case (upper or lower).
Example
input
4
1 1 1
2 1 2
3 1 1
4 1 1
output
Yes
No
No
Yes
This is a random Codeforces' problem.
I don't understand the example, input and output.
Related
I want to split a vector into several parts separated by the numeric value 0. For each part, cumulatively calculate the sum of the elements encountered so far. Negative numbers do not participate in the calculation.
For example, with the input [0,1,1,1,0,1,-1,1,1], I expect the result to be [0,1,2,3,0,1,1,2,3].
How to implement this in DolphinDB?
Use the DolphinDB built-in function cumPositiveStreak(X). Treat the negative elements in X as NULL values.
Script:
a = 1 2 -1 0 1 2 3
cumPositiveStreak(iif(a<0,NULL,a))
Execution result:
1 3 3 0 1 3 6
You’re given a chess board with dimension n x n. There’s a king at the bottom right square of the board marked with s. The king needs to reach the top left square marked with e. The rest of the squares are labeled either with an integer p (marking a point) or with x marking an obstacle. Note that the king can move up, left and up-left (diagonal) only. Find the maximum points the king can collect and the number of such paths the king can take in order to do so.
Input Format
The first line of input consists of an integer t. This is the number of test cases. Each test case contains a number n which denotes the size of board. This is followed by n lines each containing n space separated tokens.
Output Format
For each case, print in a separate line the maximum points that can be collected and the number of paths available in order to ensure maximum, both values separated by a space. If e is unreachable from s, print 0 0.
Sample Input
3
3
e 2 3
2 x 2
1 2 s
3
e 1 2
1 x 1
2 1 s
3
e 1 1
x x x
1 1 s
Sample Output
7 1
4 2
0 0
Constraints
1 <= t <= 100
2 <= n <= 200
1 <= p <= 9
I think this problem could be solved using dynamic-programing. We could use dp[i,j] to calculate the best number of points you can obtain by going from the right bottom corner to the i,j position. We can calculate dp[i,j], for a valid i,j, based on dp[i+1,j], dp[i,j+1] and dp[i+1,j+1] if this are valid positions(not out of the matrix or marked as x) and adding them the points obtained in the i,j cell. You should start computing from the bottom right corner to the left top, row by row and beginning from the last column.
For the number of ways you can add a new matrix ways and use it to store the number of ways.
This is an example code to show the idea:
dp[i,j] = dp[i+1,j+1] + board[i,j]
ways[i,j] = ways[i+1,j+1]
if dp[i,j] < dp[i+1,j] + board[i,j]:
dp[i,j] = dp[i+1,j] + board[i,j]
ways[i,j] = ways[i+1,j]
elif dp[i,j] == dp[i+1,j] + board[i,j]:
ways[i,j] += ways[i+1,j]
# check for i,j+1
This assuming all positions are valid.
The final result is stored in dp[0,0] and ways[0,0].
Brief Overview:
This problem can be solved through recursive method call, starting from nn till it reaches 00 which is the king's destination.
For the detailed explanation and the solution for this problem,check it out here -> https://www.callstacker.com/detail/algorithm-1
I have read this: https://www.topcoder.com/community/competitive-programming/tutorials/binary-search.
I can't understand some parts==>
What we can call the main theorem states that binary search can be
used if and only if for all x in S, p(x) implies p(y) for all y > x.
This property is what we use when we discard the second half of the
search space. It is equivalent to saying that ¬p(x) implies ¬p(y) for
all y < x (the symbol ¬ denotes the logical not operator), which is
what we use when we discard the first half of the search space.
But I think this condition does not hold when we want to find an element(checking for equality only) in an array and this condition only holds when we're trying to find Inequality for example when we're searching for an element greater or equal to our target value.
Example: We are finding 5 in this array.
indexes=0 1 2 3 4 5 6 7 8
1 3 4 4 5 6 7 8 9
we define p(x)=>
if(a[x]==5) return true else return false
step one=>middle index = 8+1/2 = 9/2 = 4 ==> a[4]=5
and p(x) is correct for this and from the main theory, the result is that
p(x+1) ........ p(n) is true but its not.
So what is the problem?
We CAN use that theorem when looking for an exact value, because we
only use it when discarding one half. If we are looking for say 5,
and we find say 6 in the middle, the we can discard the upper half,
because we now know (due to the theorem) that all items in there are > 5
Also notice, that if we have a sorted sequence, and want to find any element
that satisfies an inequality, looking at the end elements is enough.
MSSQL: i have this example data:
NAME AValue BValue
A 1 11
B 1 11
C 2 11
D 2 21
E 3 21
F 3 21
G 4 31
H 4 31
I 5 41
J 5 NULL
...
I am looking for algorhitm which looks for all the Names closed by values by different seed (AValue and Bvalue, in this case seed is given by 2 for AValue and by 3 for Bvalue, but this can be skipped and given later and so on, not only looking for smallest multiple). In this case output should be 1,2,3,4,11,21,31 as a first group/result. Then all the Names with these values can be updated etc.
I need to find out all the Names in "closed circle" of values by different seed.
EDIT:
(try of simplier example)
Imagine that you have list of names. Each name is given two numbers. In most cases these numbers are given by some seed (in this example AValue is given twice, BValue three times) but some numbers can be skipped, so you cannot just count smallest multiple of these different seeds(in this case it would be 2x3, ever 6 names you have closed group where no Name contains AValue or BValue from next/different group). For example Name A have 1 and 11. 1 is given for A and B, 11 for A, B, C. These Names have 1,2,11,21. So you check for 2 and 21 and then you get E and F in addition and then the loop of checking should continue, but as long as no more Names are contained there should be output 1,2,3,11,21. "Closed circle"
Could you please help me to understand this problem:
Convert the input sequence of N (1 ≤ N ≤ 20) input numbers so that
the subsequences of the same numbers are replaced with the first
numbers of the subsequences. Each input number is in the range [1, 2
000 000 000].
For example, the input sequence 1 2 2 3 1 1 1 4 4 is converted into
1 2 3 1 4.
Input: First, the number T of test cases is given. Each test case is
specified using two lines. The first one contains the number N and the
second one contains the numbers of the sequence.
Output: The converted sequence. The result for each test case should
be printed in a separate line.
For example, the input sequence 1 2 2 3 1 1 1 4 4 is converted into 1 2 3 1 4.
It looks like the idea is to remove duplicate numbers that occur adjacent to each other when creating the output.
You can do that by just keeping a state variable recording what the previous value was. When you get a new value, compare it to the state value. If it's the same, skip. If different, output it and update the state variable. Remember to initialize the state variable to a value not found in the input stream (e.g. -1 should work in this case).