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MODPOW - Calculating very big numbers very quickly |
Given numbers A, B, C, calculate (A to the power of B) mod C.
Input
The first line will contain an integer x, the number of test cases. x lines follow, each with three integers A, B, and C, the A, B, and C mentioned above.
Conditions:
A is no more than 1e5
B is no more than 1e18
C is no more than 1e7
Output
For each test case, print (A^B) mod C.
Example
Input: 1
1 1 2
Output: 1
Added by: | hua |
Date: | 2011-09-12 |
Time limit: | 1.465s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | Alex Anderson |
hide comments
2011-09-13 19:22:14 hendrik
yet another a^b%c. There are too many of them. See for example FASTPOW. |