2 1 1 3 2 3
6 196

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod=10007;
struct matrix
{
long long ma[5][5];
};
matrix multi(matrix x,matrix y)//矩阵相乘
{
matrix ans;
memset(ans.ma,0,sizeof(ans.ma));
for(int i=1;i<=4;i++)
{
for(int j=1;j<=4;j++)
{
if(x.ma[i][j])//稀疏矩阵优化
for(int k=1;k<=4;k++)
{
ans.ma[i][k]=(ans.ma[i][k]+(x.ma[i][j]*y.ma[j][k])%mod)%mod;
}
}
}
return ans;
}
matrix pow(matrix a,long long m)
{
matrix ans;
for(int i=1;i<=4;i++)
{
for(int j=1;j<=4;j++)
{
if(i==j)
ans.ma[i][j]=1;
else
ans.ma[i][j]=0;
}
}
while(m)
{
if(m&1)
ans=multi(ans,a);
a=multi(a,a);
m=m>>1;
}
return ans;
}
int main()
{
long long x,y,n;
while(~scanf("%I64d%I64d%I64d",&n,&x,&y))
{
matrix a,b;
memset(a.ma,0,sizeof(a.ma));
memset(b.ma,0,sizeof(b.ma));
a.ma[1][1]=1;
a.ma[1][2]=1;
a.ma[2][2]=(x*x)%mod;
a.ma[2][3]=(y*y)%mod;
a.ma[2][4]=(2*x*y)%mod;
a.ma[3][2]=1;
a.ma[4][2]=x;
a.ma[4][4]=y;
b.ma[1][1]=1;
b.ma[2][1]=1;
b.ma[3][1]=1;
b.ma[4][1]=1;
a=pow(a,n);
a=multi(a,b);
printf("%I64d\n",a.ma[1][1]);
}
return 0;
}
hdu 3306 Another kind of Fibonacci(矩阵快速幂)
原文地址:http://blog.csdn.net/caduca/article/details/40680943