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.
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.
For each test case you have to output a single line containing the answer for the task.
Input: 3 0 3 3 5 10 19 Output: 4 10 10857
- T <= 1000
- 0 <= N <= M <= 109
Oh god the mod..! Cost me 9 WAs! You don't have to apply mod to just the answer. You have to apply mod to the matrix at each single step(each time you multiply matrices)!! No idea why even after using long long int!!
if u get negative ans just add 1000000007 to the ans!!
can anyone post a tricky test case which involves negative modulo ??
Used Dijkstra's formula...with memorization...
I mistakingly printed 1 for 0 0 case . -_- . costed me 3 WA!!
@buttman , Though given n<=m .But you are taking mod with 10^9+7;So there might be a problem of a negative mod.Take this case ,suppose (N!)=(10^9+6) and (M)!=(10^9)+8, (M!)%mod=1 and (N!)%mod=(10^9)+6;
Last edit: 2016-05-14 05:32:30
@ abhi_vicky, thanks, got A.C 0.0
It is possible in C #Vmcode by using matrix exponantion and sum of fibonacci series!!!!AC in 0.01s