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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 |
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2020-09-27 09:36:39
@ks_r 7 Last edit: 2020-09-27 09:40:50 |
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2020-09-21 21:08:28
x1 = ((a[0][0]*b[0][0])%mod + (a[0][1]*b[1][0])%mod)%mod is this is the correct way of modulo in matrix multiplication |
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2020-07-27 08:52:50
can any one tell what should be ans if i/p is 1 1 1000000000 |
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2020-07-22 23:57:40
f(2 * k) = f(k) * f(k) + f(k - 1) * f(k - 1) f(2 * k + 1) = f(k) * f(k + 1) + f(k - 1) * f(k) Cant we use these recurrence relations ? Its O(Log N) time. |
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2020-04-15 07:16:27
How negative values are possible? :). |
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2020-01-18 07:24:54
Finally AC after 9 WA,3TLE Done it using matrix exponentiation and modulo Use modulo in multiplication of each step of matrix Last edit: 2020-01-18 07:25:25 |
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2019-10-26 11:03:38
use (a%mod -b%mod +mod)%mod) instead of (a-b)%mod |
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2019-09-25 20:42:35
The case of negative modulos is interesting.... |
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2019-09-17 18:58:23
normal iteration won't work, will obviously give TLE use matrix exponentiation(O(log(n)) and use modulo cleverly, good luck :) Last edit: 2019-09-17 18:59:07 |
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2019-07-21 21:34:22
Take Care of Data Type...! Got 2 WA for using int and long int. |
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