HDU 2256 Problem of Precision theory matrix fast Power

The title asks for an integer part of (√2+√3) 2n and then mod 1024.

(√2+√3) 2n= (5+2√6) n

If the direct calculation, with a double value, when n is very large, the loss of precision will become larger, unable to get the desired result.

We find that (5+2√6) n+ (5-2√6) n is an integer (the 2√6 's even power will be offset off or minus), and (5-2√6) n is less than 1. So we just need to ask for a Sn-1. Make

an= (5+2√6) n; bn= (5-2√6) n.

Sn=an+bn sn is an integer.

sn* ((5+2√6) + (5-2√6)) =sn*10

sn*10= (5+2√6) n+1+ (5-2√6) n+1+ (5+2√6) n-1+ (5-2√6) n-1

Sn*10=sn+1+sn-1

Recursive type: sn=10*sn-1-sn-2

Then it transforms to the matrix to find the SN quickly.

#include <iostream>#include<cstdio>#include<cstring>using namespacestd;Const intMod=1024x768;Const intn=2;structmat{intmat[n][n];} A    Mat Multiply (Mat A, Mat b) {mat C; memset (C.mat,0,sizeof(C.mat));  for(intK =0; K <2; ++k) for(inti =0; I <2; ++i)if(A.mat[i][k]) for(intj =0; J <2; ++j)if(B.mat[k][j]) c.mat[i][j]= (C.mat[i][j] +a.mat[i][k] * b.mat[k][j])%Mod; returnC;} Mat Quickpower (Mat A,intk)    {Mat C; memset (C.mat,0,sizeof(C.mat));  for(inti =0; I <2; ++i) c.mat[i][i]=1;  for(; k; k >>=1)    {        if(k&1) C =Multiply (c,a); A=Multiply (a,a); }    returnC;}voidInitmat (Mat &A) {a.mat[0][0]=Ten; a.mat[0][1]=-1; a.mat[1][0]=1; a.mat[1][1]=0;}intMain () {//freopen ("In.txt", "R", stdin);    intT; scanf ("%d",&t);  while(t--)    {        intN; scanf ("%d",&N); if(n==1) printf ("9\n"); Else if(n==2) printf ("97\n"); Else{Initmat (a); A=quickpower (a,n-2); intAns= (a.mat[0][0]*98+a.mat[0][1]*Ten-1)%1024x768;//we're asking for s[n]-1.             while(ans<0) ans+=1024x768; printf ("%d\n", ans); }    }    return 0;}

HDU 2256 Problem of Precision theory matrix fast Power