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BPM2 - Bipartite Permutation (Hard) |
Given a positive integer N, consider any permutation of all the numbers from 1 to N. It is required to create two partitions, P1 and P2, from these numbers such that |sum(P1) – sum(P2)| is minimum, where sum(X) denotes the summation of all the numbers in partition X. A partition is defined to be a non-empty subset of the permutation. In other words, find the minimum absolute difference between the summation of all the numbers in each partition. Note that you cannot leave out any number, every number from 1 to N must be part of exactly one partition.
To add a little bit more of spice, also find the lexicographically smallest partition P1 such that |sum(P1) – sum(P2)| is minimum. To make your life easier, you don’t need to output the entire sequence, only the hash of the sequence as computed by the function below would suffice.
1 ≤ T ≤ 1000
2 ≤ N ≤ 109
N < B ≤ 109
1 ≤ M ≤ 109
Bipartite Permutation
To add a little bit more of spice, also find the lexicographically smallest partition P1 such that |sum(P1) – sum(P2)| is minimum. To make your life easier, you don’t need to output the entire sequence, only the hash of the sequence as computed by the function below would suffice.
def sequence_hash(sequence, B, M): result = 0 for number in sequence: result = (result * B + number) % M return result
Input
The first line contains an integer T, denoting the number of test cases. Each of the next subsequent T lines contain three positive integers, N, B and M.Constraints
Output
For each test case, first print the case number followed by the minimum absolute difference and the hash of the lexicographically smallest partition P1.Sample Input
12 2 10 1000000000 3 10 1000000000 4 10 1000000000 5 10 1000000000 6 10 1000000000 7 10 1000000000 8 10 1000000000 9 10 1000000000 1000 1000000000 1000000 1000000 1003001 998244353 123456789 987654321 666666667 444444444 666666666 888888888
Sample Output
Case 1: 1 1 Case 2: 0 12 Case 3: 0 14 Case 4: 1 124 Case 5: 1 1234 Case 6: 0 1247 Case 7: 0 12348 Case 8: 1 123457 Case 9: 0 1000 Case 10: 0 553178755 Case 11: 1 214459309 Case 12: 0 557434257
Challenge
Don't like challenges? Try the easier version here:Bipartite Permutation
Added by: | sgtlaugh |
Date: | 2019-11-16 |
Time limit: | 10s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All |
Resource: | Own Problem |
hide comments
2021-03-15 08:50:26
For the 4th case 5 10 1000000000 With n = 5, I found P1 = {1, 2 , 5}, P2 = {3, 4} So the smallest lexicographically partition P1 is { 1, 2, 5} The hash will be: H0 = 0; H1 = (0 *10 + 1)%1000000000= 1 H2 = (1 * 10 + 2)%1000000000 = 12 H3 = (12 * 10 + 5)%1000000000 = 125 So why the result in the example is 124 Am I misunderstanding something? [NG]: For n=5, smallest P1 is {1, 2, 4}. Last edit: 2021-03-15 15:14:13 |
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2019-11-19 09:13:42 :D
Great problem. I really like how output is handled. It allowed for big N values and actually creates a sub-problem to solve. -> Thanks! Glad you liked it :D Last edit: 2019-11-23 20:45:43 |