【矩阵乘法加速递推】洛谷P1939 矩阵加速（数列）

版权声明：转载无所谓的。。。 https://blog.csdn.net/xuxiayang/article/details/82627961

链接

https://www.luogu.org/problemnew/show/P1939

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大意

$a[1]=a[2]=a[3]=1$
$a[x]=a[x-3]+a[x-1] (x>3)$
求$a$数列的第$n$项对$10^9+7$取余的值

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思路

矩阵乘法加速递推

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代码

#include<cstdio>
#include<cstring>
#define ymw 1000000007
#define LL long long
#define r(i,x,y) for(register int i=x;i<y;i++)
using namespace std; 
LL n,g,f[3],a[3][3]; 
inline void mulself()
{
    LL c[3][3]={{0,0,0},{0,0,0},{0,0,0}}; 
    r(i,0,3) r(j,0,3) r(k,0,3) c[i][j]=(c[i][j]+a[i][k]*a[k][j])%ymw; 
    memcpy(a,c,sizeof(c)); 
}
inline void mulans()
{
    LL c[3]={0,0,0};  
    r(j,0,3) r(k,0,3) c[j]=(c[j]+f[k]*a[k][j])%ymw; 
    memcpy(f,c,sizeof(c));
}
signed main()
{
    scanf("%lld",&g); 
    while (g--)
    {
        scanf("%lld",&n); 
        if(n<4) {printf("1\n");continue;}
        n-=3;  
        f[0]=f[1]=f[2]=1;
        a[0][0]=a[0][1]=a[1][1]=a[1][2]=a[2][0]=0;
        a[0][2]=a[1][0]=a[2][1]=a[2][2]=1;
        for(;n;mulself(),n>>=1) if (n&1) mulans();
        printf("%lld\n",f[2]%ymw);
    }
}