Problem Description
An Arc of Dream is a curve defined by following function:
where a0 = A0 ai = ai-1*AX+AY b0 = B0 bi = bi-1*BX+BY What is the value of AoD(N) modulo 1,000,000,007?
Input
There are multiple test cases. Process to the End of File. Each test case contains 7 nonnegative integers as follows: N A0 AX AY B0 BX BY N is no more than 1018, and all the other integers are no more than 2×109.
Output
For each test case, output AoD(N) modulo 1,000,000,007.
Sample Input
1 1 2 3 4 5 6 2 1 2 3 4 5 6 3 1 2 3 4 5 6
Sample Output
4 134 1902
Author
Zejun Wu (watashi)
Source
2013 Multi-University Training Contest 9
因为：a[i]*b[i]=(a[i-1]*AX+AY)*(b[i-1]*BX+BY)
=(a[i-1]*b[i-1]*AX*BX+a[i-1]*AX*BY+b[i-1]*BX*AY+AY*BY)
构造矩阵：
| 1 0 0 0 0 |
| AX*BY AX 0 AX*BY 0 |
{AoD(n-1),a[i-1],b[i-1],a[i-1]*b[i-1],1}* | BX*AY 0 BX BX*AY 0 | ={AoD(n),a[i],b[i],a[i]*b[i],1}
| AX*BX 0 0 AX*BX 0 |
| AY*BY AY BY AY*BY 1 |
另外注意：
if(n==0){//这个判断条件很重要，没有就会超时
printf("0\n");
continue;
}
1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<math.h>
7 #include<algorithm>
8 #include<queue>
9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 1000000
23 #define inf 1e12
24 ll n;
25 ll A0,Ax,Ay,B0,Bx,By;
26 struct Matrix{
27 ll mp[5][5];
28 };
29 Matrix Mul(Matrix a,Matrix b){
30 Matrix res;
31 for(ll i=0;i<5;i++){
32 for(ll j=0;j<5;j++){
33 res.mp[i][j]=0;
34 for(ll k=0;k<5;k++){
35 res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%MOD+MOD)%MOD;
36 }
37 }
38 }
39 return res;
40 }
41 Matrix fastm(Matrix a,ll b){
42 Matrix res;
43 memset(res.mp,0,sizeof(res.mp));
44 for(ll i=0;i<5;i++){
45 res.mp[i][i]=1;
46 }
47 while(b){
48 if(b&1){
49 res=Mul(res,a);
50 }
51 a=Mul(a,a);
52 b>>=1;
53 }
54 return res;
55 }
56 int main()
57 {
58 while(scanf("%I64d",&n)==1){
59 scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&A0,&Ax,&Ay,&B0,&Bx,&By);
60
61 if(n==0){//这个判断条件很重要，没有就会超时
62 printf("0\n");
63 continue;
64 }
65
66
67 ll a0=A0;
68 ll b0=B0;
69
70 Matrix tmp;
71 memset(tmp.mp,0,sizeof(tmp.mp));
72 tmp.mp[0][0]=1%MOD;
73 tmp.mp[1][0]=Ax*By%MOD;
74 tmp.mp[1][1]=Ax%MOD;
75 tmp.mp[1][3]=Ax*By%MOD;
76 tmp.mp[2][0]=Bx*Ay%MOD;
77 tmp.mp[2][2]=Bx%MOD;
78 tmp.mp[2][3]=Bx*Ay%MOD;
79 tmp.mp[3][0]=Ax*Bx%MOD;
80 tmp.mp[3][3]=Ax*Bx%MOD;
81 tmp.mp[4][0]=Ay*By%MOD;
82 tmp.mp[4][1]=Ay%MOD;
83 tmp.mp[4][2]=By%MOD;
84 tmp.mp[4][3]=Ay*By%MOD;
85 tmp.mp[4][4]=1%MOD;
86
87 Matrix cnt=fastm(tmp,n-1);
88
89 Matrix g;
90 memset(g.mp,0,sizeof(g.mp));
91 g.mp[0][0]=a0*b0%MOD;
92 g.mp[0][1]=a0%MOD;
93 g.mp[0][2]=b0%MOD;
94 g.mp[0][3]=a0*b0%MOD;
95 g.mp[0][4]=1%MOD;
96 Matrix ans=Mul(g,cnt);
97 printf("%I64d\n",ans.mp[0][0]);
98 }
99 return 0;
100 }