[BZOJ2326] [HNOI2011] Math assignment-matrix multiplication

Math Homework

Description

Suddenly found that matrix multiplication is not well-written ...
(This ZZ overloaded operator does not play return debug full 30min)

Find a very simple recursive type:
F[n]=f[n−1]∗ (Floor (LG10) +1) +n f[n]=f[n-1]* (Floor (LG10) +1) +n
So it is obvious that this floor (LG10) +1 floor (LG10) +1 will be the same for a long time, so this value can be divided into blocks.
Then consider a block of values by floor (LG10) floor (LG10), as in shape 10. 999,100 ... 999, the same without repeated calculations.
Consider using matrix multiplication to optimize this process:
Make floor (LG10) +1=x floor (LG10) +1=x
So for each piece you can construct the following transfer matrices:

X 0 0
1 1 0
0 1 1

The direct chunking matrix can be quickly idempotent.

#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <

Algorithm> using namespace std;
typedef long Long LL;

const int n=5;

ll N,MD;

    struct Matrix {ll a[n][n],lena,lenb;
    void E () {for (int i=0;i<lena;i++) a[i][i]=1;
        } matrix (int _a,int _b,ll f=-1) {lena=_a;
        Lenb=_b;
        memset (A,0,sizeof (a));
            if (f!=-1) {e ();
            A[0][0]=f;
        A[1][0]=a[2][1]=1;

        }} Matrix operator * (Matrix O) {matrix res (LENA,O.LENB);
                    for (int i=0;i<lena;i++) for (int. j=0;j<o.lenb;j++) for (int k=0;k<lenb;k++)

        (RES.A[I][J]+=A[I][K]*O.A[K][J]%MD)%=MD;
    return res;
                } void Out () {for (int i=0;i<lena;i++,puts (""))) for (int j=0;j<lenb;j++)
    printf ("%lld", A[i][j]);

}
}; Inline Matrix Qpow (MatRix A,ll b) {matrix ret (3,3);
    RET.E ();
        while (b) {if (b&1) ret=ret*a;
        A=a*a;
    b>>=1;
} return ret;
    } int main () {scanf ("%lld%lld", &AMP;N,&AMP;MD);
    Matrix ans (3,3);
    ANS.E ();
    ll I;
    for (i=10;i<=n;i=i*10) Ans=ans*qpow (Matrix (3,3,I%MD), I-I/10);
    Ans=ans*qpow (Matrix (3,3,I%MD), n-i/10+1);
    printf ("%lld\n", (Ans.a[1][0]+ans.a[2][0])%MD);
return 0; }