"Matrix fast Power + cyclic section" HDU 5451 best Solver

Channel

Test instructions: Calculation (5+26√)1+2^x .

Idea: The cyclic section is (p+1) * (p-1), then the bare matrix fast power.

Code:

#include <cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespaceStd;typedefLong Longll;Const intN =2;intMOD;structmatrix{ll Mat[n][n]; Matrixoperator*(Constmatrix& m)Const{Matrix tmp;  for(inti =0; i < N; i++){             for(intj =0; J < N; J + +) {Tmp.mat[i][j]=0;  for(intK =0; K < N; k++) {Tmp.mat[i][j]+ = mat[i][k]*m.mat[k][j]%MOD; TMP.MAT[I][J]%=MOD; }            }        }        returntmp; }};ll Pow (Matrix&m, ll K) {    if(k = =1)        return 9; K--;    Matrix ans; memset (Ans.mat,0,sizeof(Ans.mat));  for(inti =0; i < N; i++) Ans.mat[i][i]=1; //printf ("%d\n", k);     while(k) {if(k&1) ans= ans*m; K>>=1; M= m*m; }   //out (ans);ll x = (ans.mat[0][0]*5+ans.mat[0][1]*2)%MOD; return(x *2-1)%MOD;} ll Now;ll Pow (intx) {ll res=1, both =2;  while(X >0) {        if(X &1) Res = (res * both)%Now ; both= (both *)%Now ; X>>=1; }    returnRes;}intMain () {intT;    Matrix m; scanf ("%d", &T); intCAS =0;  while(t-->0) {        intN; scanf ("%d%d", &n, &MOD); now= (MOD +1) * (MOD-1); ll nn= Pow (n) +1 ; m.mat[0][0] =5; m.mat[0][1] = A; m.mat[1][0] =2; m.mat[1][1] =5; printf ("Case #%d:%i64d\n", ++CAs, Pow (M, nn)); }}

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"Matrix fast Power + cyclic section" HDU 5451 best Solver