hdu 4686

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2168 Accepted Submission(s): 682

Problem Description

An Arc of Dream is a curve defined by following function:

where
a₀ = A0
a_i = a_i-1*AX+AY
b₀ = B0
b_i = b_i-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

11 2 34 5 621 2 34 5 631 2 34 5 6

Sample Output

41341902

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9 

矩阵快速幂

ai = ai-1 * Ax + Ay;

bi = bi-1 * Bx + By;

ai * bi = ai-1 * bi-1 * (Ax * Bx) + ai-1 * (Ax * By) + bi-1 * (Bx * Ay) + AyBy;

Si = Si - 1 + ai * bi = Si - 1  + ai-1 * bi-1 * (Ax * Bx) + ai-1 * (Ax * By) + bi-1 * (Bx * Ay) + AyBy;

根据SI ， 一定需要  Si - 1，  ai - 1,  bi - 1,  ai - 1 * bi - 1,  常数。

矩阵大小为5，

构造 初始矩阵A 为 记  S0

[1， A0,  B0,  A0 * B0,  A0 * B0]  即 [1, a0, b0, a0 * b0, S0]

[0 …… 0]

转换矩阵 B.   使得  A * B = S1 = [1,  a1,  b1, a1 * b1 , S1]

根据上面4个公式得。

第一列为  1， 0， 0， 0， 0；

第二列为  Ay,  Ax,   0,   0,   0;

第三列为  By,  0,   Bx,  0,   0;

第四列为  AyBy, Ax * By,   Bx * Ay,  Ax * Bx, 0;

第五列为  AyBy, Ax * By,   Bx * Ay,  Ax * Bx, 1;

即

N次递推得

现在要求 Sn - 1 ,   所以  init * Pow ^ n - 1.   然后取第一行， 第5个元素。

#include<stdio.h>#include<string.h>#define mod 1000000007 struct matrix{    __int64 mat[6][6];};matrix multi(matrix a,matrix b,__int64 n){    __int64 i,j,k;    matrix ans;    memset(ans.mat,0,sizeof(ans.mat));    for(i=0;i<n;i++)        for(j=0;j<n;j++)          for(k=0;k<n;k++)           {            ans.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%mod;            ans.mat[i][j]%=mod;           }    return ans;}matrix pow(matrix a,__int64 n,__int64 k){    matrix ans=a;    k--;    while(k)    {        if(k&1)            ans=multi(ans,a,n);        a=multi(a,a,n);        k/=2;    }    return ans;}int main(){    __int64 a0,b0,ax,ay,bx,by,n;    while(scanf("%I64d",&n)!=EOF)    {        matrix a,b,ans;        memset(a.mat,0,sizeof(a.mat));        memset(b.mat,0,sizeof(b.mat));        scanf("%I64d%I64d%I64d",&a0,&ax,&ay);        scanf("%I64d%I64d%I64d",&b0,&bx,&by);        a0%=mod; ax%=mod; ay%=mod; b0%=mod; bx%=mod; by%=mod;        if(n==0)        {//n=0的时候特殊处理一下。            printf("0\n");            continue;        }        if(n==1)        {            printf("%I64d\n",a0*b0%mod);            continue;        }        a.mat[0][0]=1; a.mat[1][0]=ax*bx%mod; a.mat[1][1]=ax*bx%mod;        a.mat[2][0]=ax*by%mod; a.mat[2][1]=ax*by%mod; a.mat[2][2]=ax;        a.mat[3][0]=ay*bx%mod; a.mat[3][1]=ay*bx%mod; a.mat[3][3]=bx;        a.mat[4][0]=ay*by%mod; a.mat[4][1]=ay*by%mod; a.mat[4][2]=ay;        a.mat[4][3]=by;  a.mat[4][4]=1;        b.mat[0][0]=a0*b0%mod; b.mat[0][1]=a0*b0%mod; b.mat[0][2]=a0;        b.mat[0][3]=b0; b.mat[0][4]=1;        ans=multi(b,pow(a,5,n-1),5);        printf("%I64d\n",ans.mat[0][0]%mod);    }    return 0;}