HDU 5451 Generalized Fibonacci sequence

This topic can be translated first:

Make f (1) = 5+2√6

F (2) = f (1) * (5+2√6)

...

F (n) = f (n-1) * (5+2√6)

F (n) = f (n-1) * (10-(5-2√6)) = 10*f (n-1)-(5-2√6) f (n-1) = 10*f (n-1)-10/(5+2√6) f (n-1) = 10*f (n-1)-10/(5+2√6) * (5+2√6) F (n -2)

= 10*f (n-1)-F (n-2)

Then it can be written in the form of matrix multiplication.

(f (n), F (n-1)) = (f (n-1), F (n-2)) (10, 1

-1, 0)

But the 2^x+1 here is still very large, and here we use the generalized Fibonacci sequence to find the idea of a cyclic festival.

Cycle section length = (mod-1) * (mod+1)

The specific proof can be referenced here: Generalized Fibonacci sequence

Then just ask for a power operation on the length of the die cycle.

But here F (i) is a decimal with a square root, and here we choose to replace it with an approximate integer.

5+2√6 = 9.89 ...

F (0) = (5+2√6) ^0 = 1

F (1) = (5+2√6) ^1 = 5+2√6

/* Embarrassed for a long while I still do not know why F (0) replaced by 2, F (1) with 10 instead of a guarantee that the last to be taken is the top * *

1#include <bits/stdc++.h>2 using namespacestd;3 #defineN 1000104 #definell Long Long5 intn,q;6 ll MOD;7 structmatrix{8     intm[2][2];9     voidInit () {m[0][0]=m[1][1]=1; m[0][1]=m[1][0]=0;}TenMatrixoperator*(ConstMatrix &p)Const{ One Matrix ret; A          for(intI=0; i<2; i++) -              for(intj=0; j<2; J + +){ -ret.m[i][j]=0; the                  for(intk=0; k<2; k++){ -RET.M[I][J] = (ret.m[i][j]+ ((LL) m[i][k]*p.m[k][j])%mod)%MOD; -                 } -             } +         returnret; -     } + }; A  at intQpow (intb) - { -ll ret=1, a=2; -      while(b) { -         if(b&1) ret = ret*a%MOD; -A = a*a%MOD; inb>>=1; -     } to     returnret; + } -  theMatrix Qpow (Matrix A,intb) * { $ Matrix ret;Panax Notoginseng ret.init (); -      while(b) { the         if(b&1) ret = ret*A; +A = A *A; Ab>>=1; the     } +     returnret; - } $  $ intMain () - { -    //freopen ("a.in", "R", stdin); the     intT, cas=0; -scanf"%d", &T);Wuyi      while(t--) the     { -scanf"%d%d", &n, &q); WuMOD = (q1) * (q+1); -n =Qpow (n); AboutMOD =Q; $ Matrix A; -a.m[0][0]=Ten, a.m[1][0]=-1, a.m[0][1]=1, a.m[1][1]=0; -A =Qpow (A, n); -ll val = (ll)Ten*a.m[0][0]+ (LL)2*a.m[1][0]; Aval = ((val%mod) +mod)%MOD; +printf"Case #%d:%i64d\n", ++cas, (val+mod-1)%MOD); the     } -     return 0; $}

HDU 5451 Generalized Fibonacci sequence