Link:斐波那契
输出描述:
在一行中输出斐波那契数列的前n项平方和模 1e9+7
示例1
输入
5
输出
40
说明
1^2+^2+2^2+3^2+5^2=40
Problem solving:
我们可以推出来一个公式,然后直接矩阵快速幂就可以了。
{% fb_img https://qn.cndrew.cn/20191221220951.png 公式 %}
但是这里我没用矩快,因为我了解到了一个更nb的算法——杜教BM
快速的解决解决线性递推求项
时间复杂度可以看这里:https://blog.csdn.net/Ike940067893/article/details/84781307
这个记着板子就好了
Code:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <string> #include <map> #include <set> #include <cassert> #include <iostream> 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 _; ll 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\n",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() { scanf("%lld", &n); long long a = linear_seq::gao(VI{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 }, n - 1); long long b = linear_seq::gao(VI{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 }, n); printf("%lld\n", (a * b) % mod); }
推导过程:
{% fb_img https://qn.cndrew.cn/20191222064532.jpg 过程 %}