【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。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
#define rg register
#define N 100
#define n 3
using namespace std;
const LL Mod=1e9+7;
LL T,m,tmp[N][N];
struct Mat{LL m[N][N];}t,ans;
inline LL read(){
    LL k=0; char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar();
    return k;
}
inline void mul(Mat &a,Mat b){
    for(rg int i=1;i<=n;i++)
    for(rg int j=1;j<=n;j++) 
    for(rg int k=1;k<=n;k++)
    tmp[i][j]+=a.m[i][k]*b.m[k][j],tmp[i][j]%=Mod;
    for(rg int i=1;i<=n;i++)
    for(rg int j=1;j<=n;j++) a.m[i][j]=tmp[i][j],tmp[i][j]=0;
}
int main(){
    T=read();
    while(T--){
        memset(t.m,0,sizeof(t.m));
        memset(ans.m,0,sizeof(ans.m));
        t.m[1][1]=1; t.m[1][3]=1; t.m[2][1]=1; t.m[3][2]=1;
        for(rg int i=1;i<=n;i++) ans.m[i][i]=1;
        m=read();
        if(m<=3){
            puts("1"); continue;
        }
        m-=3;
        while(m){
            if(m&1) mul(ans,t); mul(t,t); m>>=1;
        }
        printf("%lld
",(ans.m[1][1]+ans.m[1][2]+ans.m[1][3])%Mod);
    }
    return 0;
}