牛客小白月赛20 A 斐波那契 - BM

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 %}