题意:求\(\sum_{i=0}^{n-1}a_ib_i\)
其中,\(a_i=A_xa_{i-1}+A_y,b_i=B_xb_{i-1}+B_y\)
构造矩阵分别维护\(a_ib_i,a_i,b_i,A_yB_y,A_y,B_y,S_i\)
那么\(S_n\)即为所求
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define println(a) printf("%lld\n",(ll)a)
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
struct Matrix{
ll mt[9][9],r,c;
void init(int rr,int cc,bool flag=0){
r=rr;c=cc;
memset(mt,0,sizeof mt);
if(flag) rep(i,1,r) mt[i][i]=1;
}
Matrix operator * (Matrix rhs){
Matrix ans; ans.init(r,rhs.c);
rep(i,1,r){
rep(j,1,rhs.c){
int t=max(r,rhs.c);
rep(k,1,t){
ans.mt[i][j]+=(mt[i][k]*rhs.mt[k][j])%mod;
ans.mt[i][j]=(ans.mt[i][j])%mod;
}
}
}
return ans;
}
};
Matrix fpw(Matrix A,ll n){
Matrix ans;ans.init(A.r,A.c,1);
while(n){
if(n&1) ans=ans*A;
n>>=1;
A=A*A;
}
return ans;
}
ll a0,ax,ay,b0,bx,by;
ll n;
ll bas2[8],base[8][8];
int main(){
while(cin>>n){
cin>>a0>>ax>>ay;
cin>>b0>>bx>>by;
if(n==0){
println(0);
continue;
}
bas2[1]=(a0%mod)*(b0%mod)%mod;bas2[2]=a0%mod;bas2[3]=b0%mod;bas2[4]=(ay%mod)*(by%mod)%mod;
bas2[5]=ay%mod;bas2[6]=by%mod; bas2[7]=0;
memset(base,0,sizeof base);
base[1][1]=(ax%mod)*(bx%mod)%mod;base[1][2]=(ax%mod)*(by%mod)%mod;base[1][3]=(ay%mod)*(bx%mod)%mod;base[1][4]=1;
base[2][2]=ax%mod;base[2][5]=1;
base[3][3]=bx%mod;base[3][6]=1;
base[4][4]=1;
base[5][5]=1;
base[6][6]=1;
base[7][1]=1;base[7][7]=1;
Matrix A; A.init(7,7);
rep(i,1,7)rep(j,1,7) A.mt[i][j]=base[i][j];
Matrix b; b.init(7,1);
rep(i,1,7) b.mt[i][1]=bas2[i];
Matrix res=fpw(A,n)*b;
println(res.mt[7][1]);
}
return 0;
}