Bzoj 3665:maths

Sol

Matrix multiplication + Fast Power + Euler theorem.

The first observation of the topic can be found that \ (a_n\) can be expressed in the form of several powers of \ (k\) and \ (a_0\).

\ (a_0\) is relatively simple \ (m^n\) so the first part \ (ans1=a_0^{m^n}\).

Look at \ (k\) Find the law can be found, \ (k\) The power of the form is \ (m^{n-1}+2m^{n-2}+3m^{n-3}+...+nm^0\).

This thing can construct a matrix to hand out. a \ (3\) variable \ (ans,b,1\) is required in the matrix.

where \ (ans\) is the current answer, \ (b\) is the loop variable, \ (1\) is the value of \ (b\) increment each time.

\[\begin{bmatrix}ans\\ b\\ 1\end{bmatrix}\]

That's the kind of a matrix that we want it to become

\[\begin{bmatrix}ans*m+b\\ b+1\\ 1\end{bmatrix}\]

Obviously, the matrix is very well constructed.

\[\begin{bmatrix}m&1&0\\0&1&1\\0&0&1\end{bmatrix}\]

This matrix can be quickly idempotent, but note is the power of \ (n-1\).

Then the problem has a pit point is \ (p\) is not a prime, but to ensure that coprime, if it is \ (k\) or \ (a_0\) we only need to mold about \ (p\) The inverse of the recurses can be, here can be directly used Euler theorem.

\[a^{\varphi (n)}\equiv 1 (mod n) \]

Then it became a template problem ... Slowly all is the template ...

MD!SBT Card constant. Write a quick multiply. Turn into \ (log^2\) T to die. This constant needs to be optimized, and a \ (log\) is optimized.

Fast multiplication can be segmented to write first before \ (10^6\) number and multiply after \ (10^6\) number on it.

I did not want to quickly ride this \ (log\) feel can be too ... Then for the card constant ... Everything has been written ...

Code

/************************************************************** problem:3665 User:beiyu language:c++ Result : Accepted time:12284 ms memory:20836 kb****************************************************************/#include &L t;cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector># Include<iostream>using namespace Std; typedef long LONG Ll;//typedef vector<ll> vec;//typedef vector<vec> Mat;    struct mat{LL a[3][3]; Mat () {memset (a,0,sizeof (a));}}; LL T,p,m,n,k,a0,phi; LL Ans1,ans2; Char *ps= (char *) malloc (20000000); inline ll in (ll X=0) {for (;*ps> ' 9 ' | |    *ps< ' 0 ';p s++); for (; *ps>= ' 0 ' &&*ps<= ' 9 ';p s++) x= (x<<3) + (x<<1) +*ps-' 0 '; return x;    }inline void Out (LL x) {int L=0;char ch[65];    if (!x) {Putchar (' 0 '); return;}    if (x<0) Putchar ('-'), x=-x;    while (x) ch[++l]=x%10+ ' 0 ', x/=10; for (int i=l;i;i--) Putchar (Ch[i]);}   inline ll Getphi (ll p) {ll res=p,m=sqrt (p) +0.5; for (int i=2;i<=m;i++) if (p%i==0) {res=res/i* (i-1);    while (p%i==0) p/=i; } if (p>1) res=res/p* (p-1); return res;}    inline ll Mul (ll a,ll b,ll p) {ll t1=b/1000000,t2=b%1000000; Return ((t1*a)%p*1000000%p+t2*a%p)%p;//if (p<=1000000000) return a*b%p;//return (a*b-(LL) (A/long double) p*b+1e-3 ) *p+p)%p;//for (; b;b>>=1,a= (a+a)%p) if (b&1) res= (res+a)%p;return Res;} inline ll Pow (LL a,ll b,ll p,ll Res=1) {for (; B;b>>=1,a=mul (a,a,p)) if (b&1) Res=mul (res,a,p); return res;}    MAT operator * (const MAT &a,const mat &b) {Mat C; for (int i=0;i<3;i++) for (int j=0;j<3;j++) for (int k=0;k<3;k++) c.a[i][j]= (C.a[i][j]+mul (a.a[i][k],b.a[k][j    ],phi))%phi; return C;}    Mat operator ^ (Mat A,ll b) {mat res;    for (int. i=0;i<3;i++) for (int j=0;j<3;j++) res.a[i][j]= (i==j)? 1:0;    for (; b;b>>=1,a=a*a) if (b&1) res=res*a; return res;}  int main () {//Freopen ("Maths.in", "R", stdin);//Freopen ("Maths.out", "w", stdout);//Ios::sync_with_stdio (FALSE);    Fread (Ps,1,20000000,stdin);    Mat A; For (T=in (), P=in (), Phi=getphi (p);         t--;) {m=in ()%phi,a0=in ()%p,k=in ()%p,n=in (); Cout<<mul (N,pow (n,5)) <<endl;//cout<< "0" <<endl;//cout<<m<< "" &LT;&L         t;a0<< "" <<k<< "" <<n<<endl;        cout<<ans1<<endl;         Ans1=pow (A0,pow (M,n,phi), p);//cout<<ans1<<endl;                 cout<< "1" <<endl;        a.a[0][0]=m,a.a[0][1]=1,a.a[0][2]=0;        A.a[1][0]=0,a.a[1][1]=1,a.a[1][2]=1;        A.a[2][0]=0,a.a[2][1]=0,a.a[2][2]=1;         a=a^ (n-1);                 cout<< "2" <<endl;         Ans2=pow (k, ((A.a[0][0]+2*a.a[0][1])%phi+a.a[0][2])%phi,p);        cout<< "3" <<endl;//cout<<ans1<< "" <<ans2<<endl;    Out (Mul (ans1,ans2,p)), Putchar (' \ n ');//printf ("%i64d\n", Mul (ans1,ans2,p)); } return 0;}

　　

Bzoj 3665:maths