【luogu P1939 【模板】矩阵加速（数列）】 题解

题目链接：https://www.luogu.org/problemnew/show/P1939

对于矩阵推序列的式子：

由题意知：

f[x+1] =1f[x] + 0f[x-1] + 1f[x-2]
f[x] = 1f[x] + 0f[x-1] + 0f[x-2]
f[x-1] = 0f[x] + 1f[x-1] + 0*f[x-2]

所以矩阵初项的系数：
1 1 0
0 0 1
1 0 0

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn = 110;
const int mod = 1000000007;
struct Matrix{
    ll m[maxn][maxn];
}A, E, Ans;
ll n, q[maxn];
Matrix mul(Matrix A, Matrix B)
{
    Matrix C;
    for(int i = 1; i <= 3; i++)
        for(int j = 1; j <= 3; j++)
        {
            C.m[i][j] = 0;
            for(int k = 1; k <= 3; k++)
                C.m[i][j] = (C.m[i][j] + A.m[i][k] * B.m[k][j] % mod) % mod;
        }
    return C;
}
Matrix qpow(Matrix A, ll k)
{
    Matrix S = E;
    while(k)
    {
        if(k & 1) S = mul(S, A);
        A = mul(A, A);
        k = k >> 1;
    }
    return S;
}
int main()
{
    cin>>n;
    E.m[2][2] = 1;
    E.m[1][1] = 1;
    E.m[3][3] = 1;
    A.m[1][1] = 1;
    A.m[1][2] = 1;
    A.m[2][3] = 1;
    A.m[3][1] = 1;

    for(int i = 1; i <= n; i++)
    {
        ll k;
        cin>>k;
        Ans = qpow(A, k);
        cout<<Ans.m[1][2]<<endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/MisakaAzusa/p/9648991.html