Description：
a[1]=a[2]=a[3]=1
a[x]=a[x-3]+a[x-1] (x>3)
求a数列的第n项对1000000007（10^9+7）取余的值。
Analysis：
[left[ egin{matrix} a_x \ a_{x-1}\a_{x-2} end{matrix}
ight] = left[ egin{matrix} 1 & 0 & 1\ 1 & 0 & 0\0 & 1 & 0 end{matrix}
ight] cdot left[ egin{matrix} a_{x-1} \ a_{x-2}\a_{x-3} end{matrix}
ight]
]
Code
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define ll long long
const int mod = 1000000007;
using namespace std;
int n,T;
struct Matrix{
ll data[10][10];
int r,c;
Matrix(int R,int C)
{
r = R;
c = C;
memset(data,0,sizeof(data));
}
ll* operator [](int pos)
{
return data[pos];
}
Matrix operator *(Matrix B)
{
Matrix C(r,B.c);
for(int i = 1;i <= r;++i){
for(int j = 1;j <= B.c;++j){
for(int k = 1;k <= c;++k){
C[i][j] = (C[i][j]%mod + (data[i][k]%mod)*(B[k][j]%mod)%mod)%mod;
}
}
}
return C;
}
Matrix operator *(int b){
Matrix B(R,C);
for(int i = 1;i <= R;++i){
for(int j = 1;j <= C;++j){
B[i][j] = data[i][j]*b;
}
}
return B;
}
Matrix T(){
Matrix B(C,R);
for(int i = 1;i <= C;++i){
for(int j = 1;j <= R;++j){
B[i][j] = data[j][i];
}
}
return B;
}
void print(){
for(int i = 1;i <= r;++i)
{
for(int j = 1;j <= c;++j)
{
printf("%lld ",data[i][j]);
}
puts("");
}
}
};
Matrix Pow(Matrix A,int b){
Matrix res(A.r,A.c);
for(int i = 1;i <= A.r;++i) res[i][i] = 1;//单位矩阵，相当于整数乘法中的1
for(;b;b >>= 1){
if(b&1) res = res*A;
A = A*A;
}
return res;
}
int main()
{
scanf("%d",&T);
for(int i = 1;i <= T;++i){
scanf("%d",&n);
Matrix A(3,3);
A[1][1] = A[1][3] = A[2][1] = A[3][2] = 1;
if(n <= 3){
printf("1
");
} else {
A = Pow(A,n-3);
printf("%lld
",(A[1][1]%mod + A[1][2]%mod + A[1][3]%mod)%mod);
}
}
return 0;
}