【洛谷】P1962

题目链接：https://www.luogu.org/problemnew/show/P1962

题意：求fib数列的第n项，很大。mod 1e9+7.

题解：BM直接推。

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b)
{
    ll res = 1; a %= mod;
    assert(b >= 0); 
    for (; b; b >>= 1)
    {
        if (b & 1)
            res = res * a%mod;
        a = a * a%mod;
    }
    return res;
}
// head

int _, n;
namespace linear_seq 
{
    const int N = 10010;
    ll res[N], base[N], _c[N], _md[N];
    vector<int> Md;
    void mul(ll *a, ll *b, int k) 
    {
        rep(i, 0, k + k) _c[i] = 0;
        rep(i, 0, k) 
            if (a[i]) 
                rep(j, 0, k) 
                    _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
        for (int i = k + k - 1; i >= k; i--) 
            if (_c[i])
                rep(j, 0, SZ(Md)) 
                    _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
        rep(i, 0, k) a[i] = _c[i];
    }
    int solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d
",SZ(b));
        ll ans = 0, pnt = 0;
        int k = SZ(a);
        assert(SZ(a) == SZ(b));
        rep(i, 0, k) 
            _md[k - 1 - i] = -a[i]; _md[k] = 1;
        Md.clear();
        rep(i, 0, k) 
            if (_md[i] != 0) Md.push_back(i);
        rep(i, 0, k) 
            res[i] = base[i] = 0;
        res[0] = 1;
        while ((1ll << pnt) <= n) pnt++;
        for (int p = pnt; p >= 0; p--) 
        {
            mul(res, res, k);
            if ((n >> p) & 1) 
            {
                for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;
                rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
            }
        }
        rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
        if (ans < 0) ans += mod;
        return ans;
    }
    VI BM(VI s) 
    {
        VI C(1, 1), B(1, 1);
        int L = 0, m = 1, b = 1;
        rep(n, 0, SZ(s)) 
        {
            ll d = 0;
            rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
            if (d == 0) ++m;
            else if (2 * L <= n) 
            {
                VI T = C;
                ll c = mod - d * powmod(b, mod - 2) % mod;
                while (SZ(C) < SZ(B) + m) C.pb(0);
                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                L = n + 1 - L; B = T; b = d; m = 1;
            }
            else 
            {
                ll c = mod - d * powmod(b, mod - 2) % mod;
                while (SZ(C) < SZ(B) + m) C.pb(0);
                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                ++m;
            }
        }
        return C;
    }
    int gao(VI a, ll n) 
    {
        VI c = BM(a);
        c.erase(c.begin());
        rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
        return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
    }
};

int main() {
    long long n;      
    vector<int>v;
    v.push_back(1);
    v.push_back(1);
    v.push_back(2);
    v.push_back(3);
    v.push_back(5);
    v.push_back(8);
    v.push_back(13);
    v.push_back(21);
    v.push_back(34);
    while(~scanf("%lld",&n)){
        printf("%lld
", linear_seq::gao(v, n - 1));
        //VI{1,2,4,7,13,24}
        //printf("%lld
", linear_seq::gao(v, n - 1));
        //printf("%d
",linear_seq::gao(VI{x1,x2,x3,x4},n-1));
    }
}

---

update:

矩阵快速幂

egin{pmatrix} 1 & 1\ 1 & 0 end{pmatrix}  *

 egin{pmatrix} f(n-1)\ f(n-2) end{pmatrix} =

egin{pmatrix} f(n)\ f(n-1) end{pmatrix}

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
const int maxn = 3;
const ll mod = 1e9+7;

ll n,p;

//矩阵结构体 
struct Matrix{
    ll a[maxn][maxn];
    void init(){    //初始化为单位矩阵 
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= maxn;i++){
            a[i][i] = 1;
        }
    }
};

//矩阵乘法 
Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i = 1;i <= 2;++i){
        for(int j = 1;j <= 2;++j){
            ans.a[i][j] = 0;
            for(int k = 1;k <= 2;++k){
                ans.a[i][j] = ans.a[i][j] % mod + a.a[i][k] * b.a[k][j] % mod;
            }
        }
    } 
    return ans;
}

//矩阵快速幂 
Matrix qpow(Matrix a,ll b){
    Matrix ans;
    ans.init();
    while(b){
        if(b & 1)
            ans = mul(ans,a);
        a = mul(a,a);
        b >>= 1;
    }
    return ans;
}

void print(Matrix a){
    for(int i = 1; i <= n;++i){
        for(int j = 1;j <= n;++j){
            cout << a.a[i][j]%mod<< " ";
        }
        cout << endl;
    }
}

int main(){
    Matrix base;
    Matrix ans;
    ans.a[1][1] = 1;
    ans.a[1][2] = 0;
    base.a[1][1] = 1;
    base.a[1][2] = 1;
    base.a[2][1] = 1;
    base.a[2][2] = 0;
    cin>>n;
    ans = mul(ans,qpow(base,n-1));
    cout<<ans.a[1][1]<<endl;
    return 0;
}