hdu-4686 Arc of Dream

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http://acm.hdu.edu.cn/showproblem.php?pid=4686

题意：已知a0,ax,ay a[i] = ax * a[i-1] + ay;

b0,bx,by b[i] = bx * b[i-1] + by;

求 ai*bi(0 < i <(n-1)) 的和；

ac代码：

/***设前n想和为sum[n],则：*         a[n]*b[n] = a[n-1]*b[n-1]*ax*bx + a[n-1]*ax*by + b[n-1]*ay*bx + ay * by;**                  *                                                                                  {   1     0     0   0   0 }*                                                                                  { ax*bx  ax*bx  0   0   0 }*{sum[n],a[n-1]*b[n-1],a[n-1],b[n-1],1} = {sum[n-1],a[n-2]*b[n-2],a[n-1],b[n-2],1}*{ ax*by  ax*by  ax  0   0 }*                                                                                  { ay*bx  ay*bx  0   bx  0 }*                                                                                  { ay*by  ay*by  ay  by  1 }*****/#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;typedef __int64 LL;#define mod 1000000007struct Z{    LL m[5][5];    Z(){        memset(m,0,sizeof(m));    }    void init(){        for(int i = 0;i < 5;i++)            m[i][i] = 1;    }};Z operator * (Z a, Z b){    Z c;    for(int i = 0;i < 5;i++)        for(int k = 0;k < 5;k++)            for(int j = 0;j < 5;j++)                c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j])%mod;    return c;}Z Pow(Z a,LL x){    Z ret;    ret.init();    while(x){        if(x & 1) ret = ret * a;        a = a * a;        x >>= 1;    }    return ret;}int main(){    LL n,a0,ax,ay,b0,bx,by;    while(cin >> n)    {        cin >> a0 >> ax >> ay;        cin >> b0 >> bx >> by;        LL k = a0 * b0 % mod;        if(n == 0){cout << 0 <<endl;continue;}        Z s;        s.m[0][0] = 1;        s.m[1][0] = ax*bx%mod;s.m[1][1] = ax*bx%mod;        s.m[2][0] = ax*by%mod;s.m[2][1] = ax*by%mod;s.m[2][2] = ax%mod;        s.m[3][0] = ay*bx%mod;s.m[3][1] = ay*bx%mod;s.m[3][3] = bx%mod;        s.m[4][0] = ay*by%mod;s.m[4][1] = ay*by%mod;s.m[4][2] = ay%mod;        s.m[4][3] = by%mod;s.m[4][4] = 1;        s = Pow(s,n-1);        LL ans = (s.m[0][0]*k + s.m[1][0]*k + s.m[2][0]*a0 + s.m[3][0]*b0 + s.m[4][0])%mod;        cout << ans << endl;    }}

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