Spoj 8059. Stocks prediction (matrix embedding matrix)

Question:

Returns a recursive relationship.

This recursive relationship can be used to obtain S_1 S_2 S_3... S_m

Then, tell a K and N

Obtain n terms of segma (S_k, S_2 * k, S_3 * k.

Train of Thought Analysis:

The recursive relationship is given first.

So we can use a matrix to deliver s [I]...

But what he wants is the value of every K.

Defines the recursive matrix of S as.

Sum = S_k + S_2 * k +... + s_n * k
Sum = S_k + A ^ K * S_k + A ^ 2 K * S_k...

Sum = (E + A ^ K + A ^ 2 * k + A ^ 3 * k... a ^ (n-1) * S_k ..

So we use the KK matrix to represent a ^ K

Then the problem is transformed

Calculate the sum of E + A ^ K + A ^ 2 * k + A ^ 3 * k.

Therefore, we need to define a matrix. Each element of this matrix is a small matrix.

Then, continue the recursive evaluation.

The recursive matrix is

E

0 a ^ K

#include <cstdio>#include <iostream>#include <cstring>#include <iostream>using namespace std;typedef long long LL;LL mod=1000000007;LL N;struct matrix//N*N{    LL data[10][10];    friend matrix operator * (const matrix A,const matrix B)    {        matrix res;        memset(res.data,0,sizeof res.data);        for(int i=0;i<N;i++)        for(int j=0;j<N;j++)        for(int k=0;k<N;k++)        {            res.data[i][j]+=(A.data[i][k]*B.data[k][j])%mod;            res.data[i][j]%=mod;        }        return res;    }    friend matrix operator + (const matrix A,const matrix B)    {        matrix res;        for(int i=0;i<N;i++)        for(int j=0;j<N;j++)        {            res.data[i][j]=(A.data[i][j]+B.data[i][j])%mod;            res.data[i][j]%=mod;        }        return res;    }    friend matrix operator - (const matrix A,const matrix B)    {        matrix res;        for(int i=0;i<N;i++)        for(int j=0;j<N;j++)        {            res.data[i][j]=((A.data[i][j]-B.data[i][j])+mod)%mod;            res.data[i][j]%=mod;        }        return res;    }    void print()    {        for(int i=0;i<N;i++)        {            for(int j=0;j<N;j++)            printf("%lld ",data[i][j]);            puts("");        }    }}E,zero;struct supermax{    matrix ret[10][10];    friend supermax operator * (supermax A,supermax B)    {        supermax res;        for(int i=0;i<2;i++)        for(int j=0;j<2;j++)        res.ret[i][j]=zero;        for(int i=0;i<2;i++)        for(int j=0;j<2;j++)        for(int k=0;k<2;k++)        {            res.ret[i][j]=res.ret[i][j]+(A.ret[i][k]*B.ret[k][j]);            for(int p=0;p<N;p++)            for(int q=0;q<N;q++)            res.ret[i][j].data[p][q]%=mod;        }        return res;    }};matrix matmod(matrix origin,LL n){    matrix res=E;    while(n)    {        if(n&1)        res=res*origin;        n>>=1;        origin=origin*origin;    }    return res;}supermax Do(supermax origin,LL n)//2*2{    supermax res;    for(int i=0;i<2;i++)    for(int j=0;j<2;j++)    res.ret[i][j]=zero;    for(int i=0;i<2;i++)    res.ret[i][i]=E;    while(n)    {        if(n&1)        res=res*origin;        n>>=1;        origin=origin*origin;    }    return res;}LL S[10];LL a_[10];int main(){    memset(zero.data,0,sizeof zero.data);    memset(E.data,0,sizeof E.data);    LL n,r,k;    int CASE;    scanf("%d",&CASE);    while(CASE--)    {        scanf("%lld%lld%lld",&n,&r,&k);        N=r;        for(int i=0;i<10;i++)        E.data[i][i]=1;        for(int i=1;i<=r;i++)        {            scanf("%lld",&S[i]);            S[i]%=mod;        }        for(int i=1;i<=r;i++)        {            scanf("%lld",&a_[i]);            a_[i]%=mod;        }        matrix init;        for(int i=0;i<r;i++)            init.data[i][0]=S[r-i];        matrix fib;        fib=zero;        LL fans=0;        LL fuck=1;        while(fuck*k<=r)        {            fans+=S[fuck*k];            fuck++;        }        LL b=fuck*k;        for(int i=0;i<r;i++)        {            fib.data[0][i]=a_[i+1];            fib.data[i+1][i]=1;        }        matrix st = matmod(fib,b-r);        st=st*init;        matrix K=matmod(fib,(LL)k);        supermax o;        o.ret[0][0]=E;        o.ret[0][1]=E;        o.ret[1][0]=zero;        o.ret[1][1]=K;        n=(n*k-b)/k+1;        supermax final=Do(o,n);        matrix tmp=(final.ret[0][0]*zero)+(final.ret[0][1]*E);        matrix B=E;        matrix ans = tmp*st;        printf("%lld\n",(fans+ans.data[0][0])%mod);    }    return 0;}/*510 5 71 2 3 4 51 2 3 4 510 5 75 4 3 2 15 4 3 2 110 5 7123 456 789 987 6546 78 9 7 610 5 734587 98237598 123134 523454 52432598 73897 54897 8978979 312425421 5 71 2 3 4 51 2 3 4 5510 7 1004890 78678 6876 54465 798798 567576 894123545 979123124 789475 32789 6786786 5675675 89789*/