题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4686
因为ai = ai-1*AX+AY ，bi = bi-1*BX+BY ，那么ai*bi=AX*BX*A*ai-1*bi-1+AX*BY*ai-1+BX*AY*bi-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。。。
1 //STATUS:C++_AC_1296MS_232KB
2 #include <functional>
3 #include <algorithm>
4 #include <iostream>
5 //#include <ext/rope>
6 #include <fstream>
7 #include <sstream>
8 #include <iomanip>
9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 #include <map>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,102400000")
25 //using namespace __gnu_cxx;
26 //define
27 #define pii pair<int,int>
28 #define mem(a,b) memset(a,b,sizeof(a))
29 #define lson l,mid,rt<<1
30 #define rson mid+1,r,rt<<1|1
31 #define PI acos(-1.0)
32 //typedef
33 typedef __int64 LL;
34 typedef unsigned __int64 ULL;
35 //const
36 const int N=10000010;
37 const int INF=0x3f3f3f3f;
38 const int MOD=1000000007,STA=8000010;
39 const LL LNF=1LL<<60;
40 const double EPS=1e-8;
41 const double OO=1e15;
42 const int dx[4]={-1,0,1,0};
43 const int dy[4]={0,1,0,-1};
44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
45 //Daily Use ...
46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
56 //End
57
58 LL n,a0,ax,ay,b0,bx,by;
59
60 const int size=7;
61
62 struct Matrix{
63 LL ma[size][size];
64 Matrix friend operator * (const Matrix a,const Matrix b){
65 Matrix ret;
66 mem(ret.ma,0);
67 int i,j,k;
68 for(i=0;i<size;i++)
69 for(j=0;j<size;j++)
70 for(k=0;k<size;k++)
71 ret.ma[i][j]=(ret.ma[i][j]+a.ma[i][k]*b.ma[k][j]%MOD)%MOD;
72 return ret;
73 }
74 }A;
75
76 Matrix mutilpow(LL k)
77 {
78 int i;
79 Matrix ret;
80 mem(ret.ma,0);
81 for(i=0;i<size;i++)
82 ret.ma[i][i]=1;
83 for(;k;k>>=(1LL)){
84 if(k&(1LL))ret=ret*A;
85 A=A*A;
86 }
87 return ret;
88 }
89
90 int main(){
91 // freopen("in.txt","r",stdin);
92 int i,j;
93 LL ans;
94 LL B[size];
95 Matrix F;
96 while(~scanf("%I64d",&n))
97 {
98 scanf("%I64d%I64d%I64d",&a0,&ax,&ay);
99 scanf("%I64d%I64d%I64d",&b0,&bx,&by);
100 if(n==0){printf("0\n");continue;}
101
102 a0%=MOD;ax%=MOD;ay%=MOD;
103 b0%=MOD;bx%=MOD;by%=MOD;
104 mem(A.ma,0);
105 A.ma[0][0]=A.ma[0][5]=1;
106 A.ma[1][1]=ax;A.ma[1][2]=ay;
107 A.ma[2][2]=1;
108 A.ma[3][3]=bx;A.ma[3][4]=by;
109 A.ma[4][4]=1;
110 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;
111 A.ma[6][6]=1;
112 F=mutilpow(n-1);
113 B[0]=a0*b0%MOD;
114 B[1]=(a0*ax%MOD+ay)%MOD;B[2]=1;
115 B[3]=(b0*bx%MOD+by)%MOD;B[4]=1;
116 B[5]=B[1]*B[3]%MOD;B[6]=1;
117 ans=0;
118 for(i=0;i<size;i++){
119 ans=(ans+F.ma[0][i]*B[i]%MOD)%MOD;
120 }
121
122 printf("%I64d\n",ans);
123 }
124 return 0;
125 }
来源：https://www.cnblogs.com/zhsl/p/3271284.html