HDU 1757 matrix multiplication, fast power template problem

HDU 1757

Test Instructions : If x < Ten, f (x) = x;

If x >=, f (x) = a0 * F (X-1) + A1 * F (x-2) + A2 * F (x-3) + ... + A9 * F (x-10);

Give the K and MoD and ask F (k).

Summary : 1 , especially note that matrix multiplication does not satisfy the commutative law, that is, a*b! = b*a.   2 , the sense-push equation is a bit difficult. 3 , matrix initialization note.

F (x-10)    0 0 0 0 0 0 0 0 0        (First matrix )       
F (x-9)     0 0 0 0 0 0 0 0 0
F (x-8)     0 0 0 0 0 0 0 0 0
F (x-7)     0 0 0 0 0 0 0 0 0
F (x-6)     0 0 0 0 0 0 0 0 0
F (x-5)    ;  0 0 0 0 0 0 0 0 0
F (x-4)     0 0 0 0 0 0 0 0 0
F (x-3)     0 0 0 0 0 0 0 0 0
F (x-2)     0 0 0 0 0 0 0 0 0
F (x-1)     0 0 0 0 0 0 0 0 0

*

  0 1 0 0 0 0 0 0 0 0 (temp matrix)
  0 0 1 0 0 0 0 0 0 0
  0 0 0 1 0 0 0 0 0 0
  0 0 0 0 1 0 0 0 0 0
&N Bsp 0 0 0 0 0 1 0 0 0 0
  0 0 0 0 0 0 1 0 0 0
  0 0 0 0 0 0 0 1 0 0
  0 0 0 0 0 0 0 0 1 0
  0 0 0 0 0 0 0 0 0 1

A9 8 7 6 5 4 3 2 1 0 (this line is AI)

=

F (x-9) 0 0 0 0 0 0 0 0 0
F (x-8) 0 0 0 0 0 0 0 0 0
F (x-7) 0 0 0 0 0 0 0 0 0
F (x-6) 0 0 0 0 0 0 0 0 0
F (x-5) 0 0 0 0 0 0 0 0 0
F (x-4) 0 0 0 0 0 0 0 0 0
F (x-3) 0 0 0 0 0 0 0 0 0
F (x-2) 0 0 0 0 0 0 0 0 0
F (x-1) 0 0 0 0 0 0 0 0 0
F (x) 0 0 0 0 0 0 0 0 0

#include <iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<map>#include<bitset>#include<vector>#include<Set>using namespacestd;#pragmaComment (linker, "/stack:102400000,102400000")#defineF (I,A,B) for (int i=a;i<b;i++)#defineFF (I,A,B) for (int i=a;i<=b;i++)#defineMes (A, b) memset (A,b,sizeof (a))#defineINF 0x3f3f3f3ftypedefLong Longll;Const intN = 1e5+Ten, MAXN = the; ll K,m= -;structmat{ll MAT[MAXN][MAXN]; Mat () {Mes (Mat,0); F (i,0, MAXN) mat[i][i]=1; }} E; Mat First, temp; Matoperator*(Mat A, Mat b) {mat ans; memset (Ans.mat,0,sizeof(Ans.mat));  for(inti =0; i < MAXN; i++)         for(intj =0; J < Maxn; J + +)             for(intK =0; K < MAXN; k++) Ans.mat[i][j]= (Ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%m; returnans;} Matoperator^(Mat a,ll x) {mat P= E, q =A;  while(x) {if(x&1) p = p*Q; X>>=1; Q= q*Q; }    returnp;}intMain () {mes (First.mat,0); F (i,0,Ten) first.mat[i][0]=i;  while(~SCANF ("%lld%lld", &k, &m) {mes (Temp.mat,0); FF (i,0,8) temp.mat[i][i+1]=1; F (i,0,Ten) {scanf ("%lld", &temp.mat[9][9-i]); }        if(k<Ten) {printf ("%lld\n", K); Continue; } Mat ans= (temp^ (k-9)) *first;//notice, the order here is not reversed.printf"%lld\n", ans.mat[9][0]); }    return 0;}

View Code

Matrix templates

Const intMOD, MAXN;//mod for remainder, MAXN as matrix rangestructmat{ll MAT[MAXN][MAXN]; //open a long longMat () {Mes (Mat,0); F (i,0, MAXN) mat[i][i]=1;//the diagonal is initialized to 1, the other 0}} E; //Unit MatrixMat First, temp; Matoperator* (Mat A, Mat b)//overloading *, especially note that matrix multiplication does not satisfy the commutative law, i.e. a*b! = B*a{Mat ans; memset (Ans.mat,0,sizeof(Ans.mat));  for(inti =0; i < MAXN; i++)         for(intj =0; J < Maxn; J + +)             for(intK =0; K < MAXN; k++) Ans.mat[i][j]= (Ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%MOD; returnans;} Matoperator^ (Mat a,ll x)////Heavy Duty ^{Mat P= E, q =A;  while(x) {if(x&1) p = p*Q; X>>=1; Q= q*Q; }    returnp;} Mat Mul (Mat A,mat b)//function *{Mat ans; inti,j,k;  for(i=0; i<maxn;i++){         for(j=0; j<maxn;j++) {Ans.mat[i][j]=0;  for(k=0; k<maxn;k++) {Ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j]%m; ANS.MAT[I][J]%=m; }        }    }    returnans;} Mat Pow_mod (Mat S,intb) {//function RadicalMat ans;  for(intI=0; i<maxn; ++i) ans.mat[i][i]=1;  while(b) {if(b&1) ans=Mul (ans, s); S=Mul (S, s); b>>=1; }    returnans;}voidPrmat (Mat a)//Print Matrix{    inti,j;  for(i=0; i<maxn;i++)    {         for(j=0; j<maxn;j++) printf ("%d", A.mat[i][j]); Puts (""); } puts (""); return ;}

HDU 1757 matrix multiplication, fast power template problem