[An Ac a day ^_^] hdu 4565 mathematical deduction + matrix fast Power

From today onwards, there are network races.

Today is the CCPC National game online game.

I hope everything goes well. I hope I can go to Dalian once again.

Test instructions

It's clear you're asking for sn=[a+sqrt. (b) ^n]%m

Ideas:

At first, I thought it was a water problem, a violent one.

Surf the internet to know that is the fast power

And the derivation of the characteristic equation is subtle.

In particular, the conjugate-phase offset structure is really too much to see.

(Transferred from a great God blog)

Characteristic equation is c^2=-2*a*c+ (a*a-b)

And then solve it with a fast power

Temporarily learned the next matrix fast power

I can tell from this question.

Get ACM really want to learn

This is not a high-number line-generation probability score for school cognition.

But a mathematical mindset.

It's the most important thing to get a formula that can be deduced.

The scores on the school test papers can only prove your ability to test.

It's useless to have a higher score without mathematical thinking.

From such a question can see a person's mathematics in the end is good or not

My road is far away ...

1#include <stdio.h>2#include <iostream>3#include <algorithm>4#include <math.h>5#include <string.h>6#include <string>7#include <map>8#include <Set>9#include <vector>Ten#include <queue> One #defineM (A, B) memset (A,b,sizeof (a)) A using namespacestd; -typedefLong Longll; - ll M; the structmatrix{ -ll mt[2][2]; -Matrix Mul (Matrix A,matrix b) {//matrix multiplication - Matrix C; +M (C.MT,0); -          for(intI=0;i<2; i++) +              for(intj=0;j<2; j + +) A                  for(intk=0;k<2; k++) atc.mt[i][j]+= (A.mt[i][k]*b.mt[k][j])%m;//This is going to be modulo m not because of the overflow WA -         returnC; -     } -Matrix Fast_pow (Matrix A,ll MoD) {//Fast Power - matrix result; -          for(intI=0;i<2; i++) in              for(intj=0;j<2; j + +) -Result.mt[i][j]= (i==j); to          while(MoD) { +             if(mod%2) result=Mul (result,a); -A=Mul (a,a); theMod/=2; *         } $         returnresult;Panax Notoginseng     } - }mt; the intMain () { + ll A,b,n; A      while(SCANF ("%i64d%i64d%i64d%i64d", &a,&b,&n,&m)! =EOF) { themt.mt[0][0]=2*A; +mt.mt[0][1]=1; -mt.mt[1][0]=b-a*A; $mt.mt[1][1]=0; $Mt=mt.fast_pow (mt,n-1); -ll Ans= ((2*a*mt.mt[0][0]+2*mt.mt[1][0])%m+m)%m;//take two times here because the first modulo may be negative . -printf"%i64d\n", ans); the     } -     return 0;Wuyi } the /* -  Wu 2 3 1 - 2 3 2 About 2 2 1 $  - */

[An Ac a day ^_^] hdu 4565 mathematical deduction + matrix fast Power