【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 = ;
 ll powmod(ll a, ll b)
 {
     ll res = ; a %= mod;
     assert(b >= );
     for (; b; b >>= )
     {
         if (b & )
             res = res * a%mod;
         a = a * a%mod;
     }
     return res;
 }
 // head

 int _, n;
 namespace linear_seq
 {
     const int N = ;
     ll res[N], base[N], _c[N], _md[N];
     vector<int> Md;
     void mul(ll *a, ll *b, int k)
     {
         rep(i, , k + k) _c[i] = ;
         rep(i, , k)
             if (a[i])
                 rep(j, , k)
                     _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
         for (int i = k + k - ; i >= k; i--)
             if (_c[i])
                 rep(j, , SZ(Md))
                     _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
         rep(i, , 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\n",SZ(b));
         ll ans = , pnt = ;
         int k = SZ(a);
         assert(SZ(a) == SZ(b));
         rep(i, , k)
             _md[k -  - i] = -a[i]; _md[k] = ;
         Md.clear();
         rep(i, , k)
             if (_md[i] != ) Md.push_back(i);
         rep(i, , k)
             res[i] = base[i] = ;
         res[] = ;
         while ((1ll << pnt) <= n) pnt++;
         for (int p = pnt; p >= ; p--)
         {
             mul(res, res, k);
             if ((n >> p) & )
             {
                 for (int i = k - ; i >= ; i--) res[i + ] = res[i]; res[] = ;
                 rep(j, , SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
             }
         }
         rep(i, , k) ans = (ans + res[i] * b[i]) % mod;
         if (ans < ) ans += mod;
         return ans;
     }
     VI BM(VI s)
     {
         VI C(, ), B(, );
         int L = , m = , b = ;
         rep(n, , SZ(s))
         {
             ll d = ;
             rep(i, , L + ) d = (d + (ll)C[i] * s[n - i]) % mod;
             if (d == ) ++m;
             else if ( * L <= n)
             {
                 VI T = C;
                 ll c = mod - d * powmod(b, mod - ) % mod;
                 while (SZ(C) < SZ(B) + m) C.pb();
                 rep(i, , SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                 L = n +  - L; B = T; b = d; m = ;
             }
             else
             {
                 ll c = mod - d * powmod(b, mod - ) % mod;
                 while (SZ(C) < SZ(B) + m) C.pb();
                 rep(i, , 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, , 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();
     v.push_back();
     v.push_back();
     v.push_back();
     v.push_back();
     v.push_back();
     v.push_back();
     v.push_back();
     v.push_back();
     while(~scanf("%lld",&n)){
         printf("%lld\n", linear_seq::gao(v, n - ));
         //VI{1,2,4,7,13,24}
         //printf("%lld\n", linear_seq::gao(v, n - 1));
         //printf("%d\n",linear_seq::gao(VI{x1,x2,x3,x4},n-1));
     }
 }

---

update:

矩阵快速幂

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

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

\begin{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 = ;
 const ll mod = 1e9+;

 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, , sizeof(a));
         for(int i = ; i <= maxn;i++){
             a[i][i] = ;
         }
     }
 };

 //矩阵乘法
 Matrix mul(Matrix a, Matrix b){
     Matrix ans;
     for(int i = ;i <= ;++i){
         for(int j = ;j <= ;++j){
             ans.a[i][j] = ;
             for(int k = ;k <= ;++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 & )
             ans = mul(ans,a);
         a = mul(a,a);
         b >>= ;
     }
     return ans;
 }

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

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