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LASTDIG2 - The last digit re-visited |
Pappu 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 two numbers a and b. The problem consist in find the last digit of the potency of base a and index b. Help Pappu with his problem. You are given two integer numbers: the base a (number of digits d, such that 1 <= d <= 1000) and the index b (0 <= b <= 922*10^15). You have to find the last digit of a^b.
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: 3 3 10 6 2 150 53 Output: 9 6 0
Source limit is 700 Bytes.
Added by: | Hari |
Date: | 2009-12-27 |
Time limit: | 1s |
Source limit: | 700B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: NODEJS OBJC PERL6 SQLITE VB.NET |
Resource: | harrydcoder |
hide comments
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2015-06-08 10:31:57 Ankush
Applied same algorithm which I applied in http://www.spoj.com/problems/LASTDIG/ . Modular Exponentiation and Python \m/ :D |
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2015-06-05 19:15:55 Manish Sombansh
AC in one go! |
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2015-05-26 19:32:26 Tejas Meshram
what is the answer for a=0 and b=0 ... i mean 0^0. |
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2015-05-06 19:51:45 [Mayank Pratap]
Taking a as string and b as long long int made it easier...rest simple maths... |
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2015-05-02 16:50:47 tech96
Only two lines for the actual function in Python |
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2015-04-20 10:34:29 casualcoder
Good practice for boundary cases. |
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2015-03-06 15:14:55 Sue
use math only :)) |
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2015-03-03 08:06:29 Amogh
silly mistakes due to modulo caused me 3 WAs :( exponent will fit in to long long |
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2015-02-28 13:24:41 Ankit
an accepted solution of the problem gives 100^0 as 0. Shouldnt the answer be 1 ? |
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2015-02-08 15:22:21 Zed
variable (string) which stores the base has to have length longer than 101, this has cost me 2WA. Also, exponent easily fits into long long unsigned. |