hdu1588 Fibonacci and 矩阵连乘

题目描述：实际上是求f(b)+f(k+b)+f(2*k+b)+......+f((n-1)*k+b)

0 1          f(b-1)
f(b)

1 1        *     f(b)
= f(b+1)

故令A=0 1  则f(b)=A^b; f(k+b)=A^(k+b);

     1 1

所以题目改为求A^b+A^(k+b)+A^(2*k+b)+......+A^((n-1)*k+b)=A^b(A^k+a^2*k+......+A^(n-1)*k)

这样就可以利用矩阵连乘来求解。。。。

#include <iostream>
#include <cstdio>
using namespace std;
#define MAX 4
int mod;
typedef struct
{
long long m[MAX][MAX];
}Matrix;
Matrix p,I1;
Matrix I2={1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1};
Matrix q={0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
Matrix c;
int i,j,k;
for(i=0;i<MAX;i++)
for(j=0;j<MAX;j++)
{
c.m[i][j]=0;
for(k=0;k<MAX;k++)
c.m[i][j]+=((a.m[i][k]%mod)*(b.m[k][j]%mod))%mod;
c.m[i][j]=c.m[i][j]%mod;
}
return c;
}
Matrix quickpow(long long n,Matrix P,Matrix I)
{
Matrix m=P,b=I;
while(n>=1)
{
if(n&1)
b=matrixmul(b,m);
n=n>>1;
m=matrixmul(m,m);

}
return b;
}
int main()
{
int k,b,n;
long long ans;
Matrix ans1,ans2,ans3;
while(scanf("%d%d%d%d",&k,&b,&n,&mod)!=EOF)
{
p.m[0][0]=0;p.m[0][1]=1;
p.m[1][0]=1;p.m[1][1]=1;
I1.m[0][0]=1;I1.m[0][1]=0;
I1.m[1][0]=0;I1.m[1][1]=1;
ans1=quickpow(b,p,I1);
ans2=quickpow(k,p,I1);
q.m[0][0]=ans2.m[0][0];q.m[0][1]=ans2.m[0][1];
q.m[1][0]=ans2.m[1][0];q.m[1][1]=ans2.m[1][1];
ans3=quickpow(n,q,I2);
ans=((ans1.m[0][0]*ans3.m[0][3])%mod+(ans1.m[0][1]*ans3.m[1][3])%mod)%mod;
ans=(ans+mod)%mod;
cout<<ans<<endl;

}
return 0;
}