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FIBOSUM - Fibonacci Sum |
The Fibonacci sequence is defined by the following relation:
- F(0) = 0
- F(1) = 1
- F(N) = F(N - 1) + F(N - 2), N >= 2
Your task is very simple. Given two non-negative integers N and M, you have to calculate the sum (F(N) + F(N + 1) + ... + F(M)) mod 1000000007.
Input
The first line contains an integer T (the number of test cases). Then, T lines follow. Each test case consists of a single line with two non-negative integers N and M.
Output
For each test case you have to output a single line containing the answer for the task.
Example
Input: 3 0 3 3 5 10 19 Output: 4 10 10857
Constraints
- T <= 1000
- 0 <= N <= M <= 109
Added by: | David Gómez |
Date: | 2010-12-04 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | My Own |
hide comments
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2018-03-14 19:57:24
My 101th :) |
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2018-01-10 19:48:53
Matrix Exponentiation it is! :) |
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2017-12-24 21:40:59
Learnt matrix expo: but not tried with f(n+2)-1 :) |
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2017-12-19 13:08:46
AC in one go:-) |
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2017-12-14 21:46:06
Matrix exponentiation and f(1)+f(2)+f(3)+... + f(n) = f(n+2)-1 did this :D !! 82nd AC |
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2017-10-08 19:06:09
Matrix Expo :) |
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2017-08-23 14:38:43
AC after 1 wa....take care of negative modulus.. |
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2017-07-14 08:55:24
Ac in one go :) Didn't use Matrix Exponention, Dijkstra's formula was good enough. And yes negative mod must be taken into account. Thanks @sagnik_66 for the hint |
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2017-06-23 10:31:35
learn something new !! Matrix exponentiation !! care about mod !! |
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2017-06-14 22:46:36
use long long !! Cost me 3WA's |