【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：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <string>
  7 #include <map>
  8 #include <set>
  9 #include <cassert>
 10 using namespace std;
 11 #define rep(i,a,n) for (int i=a;i<n;i++)
 12 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 13 #define pb push_back
 14 #define mp make_pair
 15 #define all(x) (x).begin(),(x).end()
 16 #define fi first
 17 #define se second
 18 #define SZ(x) ((int)(x).size())
 19 typedef vector<int> VI;
 20 typedef long long ll;
 21 typedef pair<int, int> PII;
 22 const ll mod = 1000000007;
 23 ll powmod(ll a, ll b)
 24 {
 25     ll res = 1; a %= mod;
 26     assert(b >= 0); 
 27     for (; b; b >>= 1)
 28     {
 29         if (b & 1)
 30             res = res * a%mod;
 31         a = a * a%mod;
 32     }
 33     return res;
 34 }
 35 // head
 36 
 37 int _, n;
 38 namespace linear_seq 
 39 {
 40     const int N = 10010;
 41     ll res[N], base[N], _c[N], _md[N];
 42     vector<int> Md;
 43     void mul(ll *a, ll *b, int k) 
 44     {
 45         rep(i, 0, k + k) _c[i] = 0;
 46         rep(i, 0, k) 
 47             if (a[i]) 
 48                 rep(j, 0, k) 
 49                     _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
 50         for (int i = k + k - 1; i >= k; i--) 
 51             if (_c[i])
 52                 rep(j, 0, SZ(Md)) 
 53                     _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
 54         rep(i, 0, k) a[i] = _c[i];
 55     }
 56     int solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
 57 //        printf("%d\n",SZ(b));
 58         ll ans = 0, pnt = 0;
 59         int k = SZ(a);
 60         assert(SZ(a) == SZ(b));
 61         rep(i, 0, k) 
 62             _md[k - 1 - i] = -a[i]; _md[k] = 1;
 63         Md.clear();
 64         rep(i, 0, k) 
 65             if (_md[i] != 0) Md.push_back(i);
 66         rep(i, 0, k) 
 67             res[i] = base[i] = 0;
 68         res[0] = 1;
 69         while ((1ll << pnt) <= n) pnt++;
 70         for (int p = pnt; p >= 0; p--) 
 71         {
 72             mul(res, res, k);
 73             if ((n >> p) & 1) 
 74             {
 75                 for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;
 76                 rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
 77             }
 78         }
 79         rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
 80         if (ans < 0) ans += mod;
 81         return ans;
 82     }
 83     VI BM(VI s) 
 84     {
 85         VI C(1, 1), B(1, 1);
 86         int L = 0, m = 1, b = 1;
 87         rep(n, 0, SZ(s)) 
 88         {
 89             ll d = 0;
 90             rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
 91             if (d == 0) ++m;
 92             else if (2 * L <= n) 
 93             {
 94                 VI T = C;
 95                 ll c = mod - d * powmod(b, mod - 2) % mod;
 96                 while (SZ(C) < SZ(B) + m) C.pb(0);
 97                 rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
 98                 L = n + 1 - L; B = T; b = d; m = 1;
 99             }
100             else 
101             {
102                 ll c = mod - d * powmod(b, mod - 2) % mod;
103                 while (SZ(C) < SZ(B) + m) C.pb(0);
104                 rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
105                 ++m;
106             }
107         }
108         return C;
109     }
110     int gao(VI a, ll n) 
111     {
112         VI c = BM(a);
113         c.erase(c.begin());
114         rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
115         return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
116     }
117 };
118 
119 int main() {
120     long long n;      
121     vector<int>v;
122     v.push_back(1);
123     v.push_back(1);
124     v.push_back(2);
125     v.push_back(3);
126     v.push_back(5);
127     v.push_back(8);
128     v.push_back(13);
129     v.push_back(21);
130     v.push_back(34);
131     while(~scanf("%lld",&n)){
132         printf("%lld\n", linear_seq::gao(v, n - 1));
133         //VI{1,2,4,7,13,24}
134         //printf("%lld\n", linear_seq::gao(v, n - 1));
135         //printf("%d\n",linear_seq::gao(VI{x1,x2,x3,x4},n-1));
136     }
137 }

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