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LASTDIG - The last digit |
Nestor was doing the work of his math class about three days but he is tired of make operations a lot and he should deliver his task tomorrow. His math’s teacher gives him two numbers a and b. The problem consist of finding the last digit of the potency of base a and index b. Help Nestor with his problem. You are given two integer numbers: the base a (0 <= a <= 20) and the index b (0 <= b <= 2,147,483,000), a and b both are not 0. You have to find the last digit of ab.
Input
The first line of input contains an integer t, the number of test cases (t <= 30). t test cases follow. For each test case will appear a and b separated by space.
Output
For each test case output an integer per line representing the result.
Example
Input: 2 3 10 6 2
Output: 9 6
Added by: | Jose Daniel Rodriguez Morales |
Date: | 2008-12-01 |
Time limit: | 1s |
Source limit: | 700B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: GOSU |
Resource: | Own |
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2018-04-20 12:26:36
@narutohokage_1, thanks for the pointer. |
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2018-04-10 07:41:19
It can be solved in O(1), all you have to do is calculate the last digit of a, a^2, a^3, a^4, after that it just keeps on repeating, you can check it on your own. |
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2018-03-11 10:32:34
ac in 2 go !! gys use string and consider 700 b limitation. |
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2018-03-10 15:21:52
700 B limit is annoying. |
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2018-01-10 22:31:21
Modular exponenention _/\_ thanks for @ajayc1007 and @prasad_235 |
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2017-12-31 08:16:40
AC after two TLE (I was using pow function).Modular exponention does the work.No need to consider when one of them is 0 cases. |
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2017-12-30 20:57:42
Check 10^2=0 10^0=1 0^10=0 Last edit: 2017-12-30 21:02:13 |
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2017-12-30 10:54:11
solved after 5 WA.but got AC in d end.... |
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2017-11-25 13:50:18
should i check for negative values for a and b? Why am I getting WA :/ I am getting correct output using fast modular exponentiation. |
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2017-11-18 12:57:07
In this problem you could solve it by calculating (a ^ b) % 10, using fast modular exponentiation. Time complexity would be O(logb) |