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OLOLO - Onotole needs your help |
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Onotole has a lot of pyani. Each pyani has a number, writing on it. Pyanis with equal numbers are indistinguishable. Onotole knows everything, so, he knows that each pyani appeared twice, and only one pyani is unique. He wants to get вздръжни эффект, and he needs the unique pyani. Given the list of pyanis denote which one of them appeared once (it is guaranteed that other pyanis appeared twice). |
Input
First line of input contains number of pyanis N <= 500 000. Next N lines contain a single positive integer 1 <= Pi <= 10^9.
Output
Output one positive integer on pyani, which appeared once.
Example
Input:
3
1
8
1
Output: 8
Onotole has found not optimal AC algorithms, so all solutions will be rejudged. He is watching you.
| Added by: | Efim |
| Date: | 2010-11-04 |
| Time limit: | 1s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 |
hide comments
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2016-12-06 09:24:29
One pass. O(N). |
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2016-12-03 20:14:53
easy question just took o(logn+n) time complexity |
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2016-11-27 00:36:55
Accepted in one go! Did a similar question back on HackerRank. |
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2016-11-12 18:33:20
TLE in O(n) even after using FastReader in JAVA :( |
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2016-11-01 02:24:44
TLE even for O(n) solution...wtf? |
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2016-10-28 19:16:57
O(nlogn+n) gives AC.0(n^2) TLE.Nice question. |
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2016-10-07 09:01:23
Use ios_base::sync_with_stdio(false); Then U will not get exceeded time limit |
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2016-10-05 21:45:45
Logic that gave TLE in python got accepted in C++. Can anyone please explain why this happens? |
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2016-10-03 13:00:18
someone mentioned XOR, and golly, never taught about that! re-read the problem statement and re-read about XOR, too. |
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2016-09-13 16:33:00 Vipin
CPP users using cin and cout use ios_base::sync_with_stdio(false); |
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