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
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
");
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 "); 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 ",ans.mp[0][0]); 98 } 99 return 0; 100 }