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EVENODD - Even and Odd |
Given undirected graph, we are interested to state 2 points
- Is the count of nodes with odd degree, even or odd?
- Is the count of nodes with even degree, even or odd?
Use following pseudocode to construct the graph:
FOR i = 1 to N FOR J = 1 to min(N-i, K) add_graph_edge(i, i+j);
where node degree is defined as: the number of edges incident to the vertex. This graph will be generated through following pseudocode, for a given N (number of nodes) and K (connectivity factor).
For example, given N = 5, K = 3 we have a graph consisting of 2 nodes with odd degree and 3 nodes with even degree.
Input
The first line of input contains an integer T that represents the number of test cases, then follow T lines each contains only two integer numbers N, K where 1 ≤ N ≤ 100,000 and 0 ≤ K ≤ N.
Output
For each test case, output on a single line “Case #K:” where K is the number of the test case, followed by either "even" or "odd" for the count of nodes with odd degree, then single space, then either "even" or "odd" for the count of nodes with even degree. Check Sample below to see the format.
Example
Input: 1 5 3 Output: Case #1: even odd
Added by: | Mostafa Saad Ibrahim |
Date: | 2012-11-05 |
Time limit: | 0.5s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | 11th Egyptian Collegiate Programming Contest |
hide comments
2012-11-09 21:29:12 :D
Sorry, but that problem is very basic. Although test cases aren't so weak to allow O(N), O(1) is still very easy and requires hardly any coding. Moved to tutorial. |