【模板】矩阵加速（数列）
洛谷 1939
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。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 #define rg register 6 #define N 100 7 #define n 3 8 using namespace std; 9 const LL Mod=1e9+7; 10 LL T,m,tmp[N][N]; 11 struct Mat{LL m[N][N];}t,ans; 12 inline LL read(){ 13 LL k=0; char c=getchar(); 14 while(c<'0'||c>'9')c=getchar(); 15 while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar(); 16 return k; 17 } 18 inline void mul(Mat &a,Mat b){ 19 for(rg int i=1;i<=n;i++) 20 for(rg int j=1;j<=n;j++) 21 for(rg int k=1;k<=n;k++) 22 tmp[i][j]+=a.m[i][k]*b.m[k][j],tmp[i][j]%=Mod; 23 for(rg int i=1;i<=n;i++) 24 for(rg int j=1;j<=n;j++) a.m[i][j]=tmp[i][j],tmp[i][j]=0; 25 } 26 int main(){ 27 T=read(); 28 while(T--){ 29 memset(t.m,0,sizeof(t.m)); 30 memset(ans.m,0,sizeof(ans.m)); 31 t.m[1][1]=1; t.m[1][3]=1; t.m[2][1]=1; t.m[3][2]=1; 32 for(rg int i=1;i<=n;i++) ans.m[i][i]=1; 33 m=read(); 34 if(m<=3){ 35 puts("1"); continue; 36 } 37 m-=3; 38 while(m){ 39 if(m&1) mul(ans,t); mul(t,t); m>>=1; 40 } 41 printf("%lld\n",(ans.m[1][1]+ans.m[1][2]+ans.m[1][3])%Mod); 42 } 43 return 0; 44 }
