Zonal Computing Olympiad 2013, 10 Nov 2012
N teams participate in a league cricket tournament on Mars, where each pair of distinct teams plays each other exactly once. Thus, there are a total of (N × (N-1))/2 matches. An expert has assigned a strength to each team, a positive integer. Strangely, the Martian crowds love one-sided matches and the advertising revenue earned from a match is the absolute value of the difference between the strengths of the two matches. Given the strengths of the N teams, find the total advertising revenue earned from all the matches.
For example, suppose N is 4 and the team strengths for teams 1, 2, 3, and 4 are 3, 10, 3, and 5 respectively. Then the advertising revenues from the 6 matches are as follows:
Thus the total advertising revenue is 23.
Line 1 : A single integer, N.
Line 2 : N space-separated integers, the strengths of the N teams.
A single integer, the total advertising revenue from the tournament.
4 3 10 3 5
In all subtasks, the strength of each team is an integer between 1 and 1,000 inclusive.
- Subtask 1 (30 marks) : 2 ≤ N ≤ 1,000.
- Subtask 2 (70 marks) : 2 ≤ N ≤ 200,000.
Live evaluation data
- Subtask 1: Testcases 0,1,2.
- Subtask 2: Testcases 3,4,5.
The answer might not fit in a variable of type int. We recommend that type long long be used for computing all advertising revenues. If you use printf and scanf, you can use %lld for long long.
|Time Limit:||1 sec|
|Source Limit:||50000 Bytes|
|Languages:||C, C99 strict, CPP 4.3.2, CPP 4.9.2, CPP14, JAVA, PYTH, PYTH 3.4|
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