HDU 4686 矩阵快速幂 Arc of Dream

由式子的性质发现都是线性的，考虑构造矩阵，先有式子，a[i] = ax * a[i-1] + ay; b[i] = bx*b[i-1] +by;

a[i]*b[i] = ax*bx*a[i-1]*b[i-1] + ax*by*a[i-1] + bx*ay*b[i-1]+ay*by;

s[i] = s[i-1] + a[i-1]*b[i-1];

由此得到递推式 :设矩阵A=

ax	0	0	0	ay
0	bx	0	0	by
ax*by	bx*ay	ax*bx	0	ay*by
0	0	1	1	0
0	0	0	0	1

矩阵B[i]=（a[i-1],b[i-1],a[i-1]*b[i-1],s[i-1],1）' (转置）,B[i] =（a[i],b[i],a[i]*b[i],s[i],1）' （转置）,则有B[i] = A*B[i-1]

令s0 = 0,则有B[0] = （a0,b0,a0*b0,s0,1）',B[n] = A^n*B[0],矩阵乘法是服从结合律的，所以先用矩阵快速幂算出A^n,再算出B[n],那么B[n][4]即为所求。

贴代码：

 #include<cstdio>
 #include<cstring>
 typedef long long int ll;
 const int p = ;
 ll ax,ay,bx,by,a0,b0;
 struct matrix
 {
     ll m[][];
 } A;
 inline void init()
 {
     memset(A.m,,sizeof(A.m));
     A.m[][] =ax;
     A.m[][] = ay;
     A.m[][] = bx;
     A.m[][] = by;
     A.m[][] = ax*by%p;
     A.m[][] = ay*bx%p;
     A.m[][] = ax*bx%p;
     A.m[][] = ay*by%p;
     A.m[][] = A.m[][] = A.m[][] = ;
 }
 inline matrix mul(ll a[][],ll b[][])
 {
     matrix ans;
     memset(ans.m,,sizeof(ans.m));
     for(int i=; i<=; ++i)
         for(int j=; j<=; ++j)
             for(int k=; k<=; ++k)
                 ans.m[i][j] = (ans.m[i][j] + a[i][k]*b[k][j]%p)%p;
     return ans;
 }
 inline matrix qPow(ll x)
 {
     matrix ans;
     memset(ans.m,,sizeof(ans.m));
     for(int i=; i<=; ++i)
         ans.m[i][i] =;
     init();
     while(x)
     {
         if(x&) ans = mul(ans.m,A.m);
         A = mul(A.m,A.m);
         x >>= ;
     }
     return ans;
 }
 int main()
 {
 //    freopen("in.txt","r",stdin);
     ll n;
     while(~scanf("%I64d",&n))
     {
         scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a0,&ax,&ay,&b0,&bx,&by);
         matrix ans = qPow(n);
         ll res=;
         res = (res + ans.m[][]*a0%p)%p;
         res = (res + ans.m[][]*b0%p)%p;
         res = (res + ans.m[][]*((a0*b0)%p)%p)%p;
         res = (res + ans.m[][])%p;
         printf("%I64d\n",res);
     }
     return ;
 }