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Home/ Questions/Q 8158841
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T17:44:32+00:00 2026-06-06T17:44:32+00:00

I was trying to solve this problem on SPOJ (http://www.spoj.pl/problems/REC/) F(n) = a*F(n-1) +

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I was trying to solve this problem on SPOJ (http://www.spoj.pl/problems/REC/)

F(n) = a*F(n-1) + b where we have to find F(n) Mod (m)
where

0 <= a, b, n <=  10^100
       1 <= M <= 100000
F(0)=1

I am trying to solve it with BigInteger in JAVA but if I run a loop from 0 to n its getting TLE. How could I solve this problem? Can anyone give some hint? Don’t post the solution. I want hint on how to solve it efficiently.

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  1. Editorial Team
    Editorial Team
    2026-06-06T17:44:33+00:00Added an answer on June 6, 2026 at 5:44 pm

    Note that the pattern of residues mod (m) should have a repeating pattern in a linear recurrence, and with length <= m by the pigeonhole principle. You need only calculate the first m entries, then figure out which of those entries will apply to F(n) for the actual value of n.

    It also helps to solve a simpler problem. Let’s pick really small values, say a=2, b=1, m=5, n=1000.

    • F(0) = 1
    • F(1) = 2*F(0) + 1 = 2*1 + 1 = 3 -> 3 Mod 5 = 3
    • F(2) = 2*F(1) + 1 = 2*3 + 1 = 7 -> 7 Mod 5 = 2
    • F(3) = 2*F(2) + 1 = 2*7 + 1 = 15 -> 15 Mod 5 = 0
    • F(4) = 2*F(3) + 1 = 2*15 + 1 = 31 -> 31 Mod 5 = 1
    • F(5) = 2*F(4) + 1 = 2*31 + 1 = 63 -> 63 Mod 5 = 3
    • etc.

    Notice that the residues are [1, 3, 2, 0, 1, 3, …], which will repeat forever. So from this example, how would you determine F(1000) Mod 5 without looping all the way to the 1000th entry?

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