HDU 4686 Arc of Dream（矩阵加速递推）

题目大意：就是给你你个有两个递推公式乘起来的式子，让你求出第n项的结果。

注意这种递推的需要把式子乘起来然后再构造矩阵。

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2092    Accepted Submission(s): 664

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

 

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-10///#define M 1000100#define LL __int64///#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)#define mod 1000000007const int maxn = 210;using namespace std;struct matrix{    LL f[10][10];};matrix mul(matrix a, matrix b, int n){    matrix c;    memset(c.f, 0, sizeof(c.f));    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j];            c.f[i][j] %= mod;        }    }    return c;}matrix pow_mod(matrix a, LL b, int n){    matrix s;    memset(s.f, 0 , sizeof(s.f));    for(int i = 0; i < n; i++) s.f[i][i] = 1LL;    while(b)    {        if(b&1) s = mul(s, a, n);        a = mul(a, a, n);        b >>= 1;    }    return s;}matrix Add(matrix a,matrix b, int n) {    matrix c;    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            c.f[i][j] = a.f[i][j]+b.f[i][j];            c.f[i][j] %= mod;        }    }    return c;}int main(){    LL n;    LL a, ax, ay;    LL b, bx, by;    while(~scanf("%I64d",&n))    {        scanf("%I64d %I64d %I64d",&a, &ax, &ay);        scanf("%I64d %I64d %I64d",&b, &bx, &by);        a %= mod;        ax %= mod;        ay %= mod;        b %= mod;        bx %= mod;        by %= mod;        LL ff = a*b%mod;        LL x = (a*ax+ay)%mod;        LL y = (b*bx+by)%mod;        LL pp = (x*y)%mod;        if(n == 0)        {            puts("0");            continue;        }        matrix c;        memset(c.f, 0 ,sizeof(c.f));        c.f[0][0] = ax*bx%mod;        c.f[0][1] = ax*by%mod;        c.f[0][2] = ay*bx%mod;        c.f[0][3] = ay*by%mod;        ///c.f[0][4] = 1LL;        c.f[1][1] = ax;        c.f[1][3] = ay;        c.f[2][2] = bx;        c.f[2][3] = by;        c.f[3][3] = 1LL;        c.f[4][0] = 1LL;        c.f[4][4] = 1LL;        matrix d = pow_mod(c, n-1LL, 5);        LL sum = 0LL;        sum += ((d.f[4][0]*pp%mod)+(d.f[4][4]*ff%mod))%mod;        sum += ((d.f[4][1]*x%mod) + (d.f[4][2]*y%mod) + d.f[4][3]%mod)%mod;        printf("%I64d\n",(sum+mod)%mod);    }    return 0;}