HDU-4686 Arc of Dream 构造矩阵

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4686

　　因为a_i = a_i-1*AX+AY ，b_i = b_i-1*BX+BY ，那么ai*bi=AX*BX*A*a_i-1*b_i-1+AX*BY*a_i-1+BX*AY*b_i-1+AY*BYAY。令Sn为ai*bi前n项的和，Sn=Sn-1 + an*bn，因此我们可以构造一个如下的转移矩阵：

          

　　然后矩阵乘法优化就可以了。。。

　　注意此题n=0的情况！

　　其实矩阵大小只要5就可以了，那几个常数项可以合并到一列。。。

//STATUS:C++_AC_1296MS_232KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=10000010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

LL n,a0,ax,ay,b0,bx,by;

const int size=7;

struct Matrix{
    LL ma[size][size];
    Matrix friend operator * (const Matrix a,const Matrix b){
        Matrix ret;
        mem(ret.ma,0);
        int i,j,k;
        for(i=0;i<size;i++)
            for(j=0;j<size;j++)
                for(k=0;k<size;k++)
                    ret.ma[i][j]=(ret.ma[i][j]+a.ma[i][k]*b.ma[k][j]%MOD)%MOD;
        return ret;
    }
}A;

Matrix mutilpow(LL k)
{
    int i;
    Matrix ret;
    mem(ret.ma,0);
    for(i=0;i<size;i++)
        ret.ma[i][i]=1;
    for(;k;k>>=(1LL)){
        if(k&(1LL))ret=ret*A;
        A=A*A;
    }
    return ret;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j;
    LL ans;
    LL B[size];
    Matrix F;
    while(~scanf("%I64d",&n))
    {
        scanf("%I64d%I64d%I64d",&a0,&ax,&ay);
        scanf("%I64d%I64d%I64d",&b0,&bx,&by);
        if(n==0){printf("0
");continue;}

        a0%=MOD;ax%=MOD;ay%=MOD;
        b0%=MOD;bx%=MOD;by%=MOD;
        mem(A.ma,0);
        A.ma[0][0]=A.ma[0][5]=1;
        A.ma[1][1]=ax;A.ma[1][2]=ay;
        A.ma[2][2]=1;
        A.ma[3][3]=bx;A.ma[3][4]=by;
        A.ma[4][4]=1;
        A.ma[5][1]=ax*by%MOD;A.ma[5][3]=bx*ay%MOD;A.ma[5][5]=ax*bx%MOD;A.ma[5][6]=ay*by%MOD;
        A.ma[6][6]=1;
        F=mutilpow(n-1);
        B[0]=a0*b0%MOD;
        B[1]=(a0*ax%MOD+ay)%MOD;B[2]=1;
        B[3]=(b0*bx%MOD+by)%MOD;B[4]=1;
        B[5]=B[1]*B[3]%MOD;B[6]=1;
        ans=0;
        for(i=0;i<size;i++){
            ans=(ans+F.ma[0][i]*B[i]%MOD)%MOD;
        }

        printf("%I64d
",ans);
    }
    return 0;
}