首页 > 编程语言> > 斐波那契和（“科大讯飞杯”第十七届同济大学程序设计预选赛暨高校网络友谊赛 J，矩阵快速幂）

斐波那契和（“科大讯飞杯”第十七届同济大学程序设计预选赛暨高校网络友谊赛 J，矩阵快速幂）

2020-05-13 18:41:39 作者：互联网

一.题目链接：

斐波那契和

二.题目大意：

求 $\sum _{i=1} ^{n} i^kfib(i) ,\;\; 1 \leq n \leq 10^{18}, 1 \leq k \leq 100$

三.分析：

比赛时套杜教 BM 一直wa，赛后才发现模数写错...(太蠢了

由于杜教 BM 直接套上模板改改模数就能 AC，这里只给出非杜教 BM 解法(杜教 BM 他不香吗？

首先做一下符号解释

$F_k(n) = \sum _{i=1}^n i^kfib(i)$

$G_k(n) = \sum _{i = 1} ^{n} (n - i)^k fib(i)$

符号化后，题目即求 $F_k(n)$

然后求 $F_k(n)$ 的递推式

$G_k(n) = \sum_{i = 1} ^n (n - i)^k fib(i)$

$G_k(n) = \sum_{i = 1} ^n ((-i) + n)^k fib(i)$

$G_k(n) = \sum_{i = 1} ^n \sum_{j = 0}^k \binom{k}{j} (-i)^j n^{k-j} fib(i)$

$G_k(n) = \sum_{i = 1} ^n \sum_{j = 0}^k \binom{k}{j} (-1)^ji^j n^{k-j} fib(i)$

$G_k(n) = \sum_{j = 0}^k \binom{k}{j} (-1)^j n^{k-j} \sum_{i = 1} ^ni^j fib(i)$

$G_k(n) = \sum_{j = 0}^k \binom{k}{j} (-1)^j n^{k-j} F_j(n)$

$G_k(n) = (-1)^kF_k(n) + \sum_{j = 0}^{k-1} \binom{k}{j} (-1)^j n^{k-j} F_j(n)$

$(-1)^kF_k(n) = G_k(n) - \sum_{j = 0}^{k-1} \binom{k}{j} (-1)^j n^{k-j} F_j(n)$

当计算 $F_k(n)$ 时，前面 $F_j(n)$ 的值均已计算完毕，因此我们只需求 $G_i(n),\;\; 0 \leq i \leq k$

下面求 $G_k(n)$ 的递推式

当 $k = 0$ 时

$G_0(n) = \sum_{i=1}^{n} fib(i) = fib(i + 2) - 1$

当 $k \geq 1$ 时

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n+1} (n +1- i)^kfib(i) - \sum _{i=1}^{n} (n - i)^kfib(i)$

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n} (n +1- i)^kfib(i) - \sum _{i=1}^{n} (n - i)^kfib(i)$

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n} ((n +1- i)^k - (n - i)^k)fib(i)$

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n} (((n - i) +1)^k - (n - i)^k)fib(i)$

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n} (\sum_{j = 0}^{k} \binom{k}{j}(n-i)^j - (n - i)^k)fib(i)$

$G_k(n+1) - G_k(n) = \sum _{i=1}^{n} \sum_{j = 0}^{k-1} \binom{k}{j}(n-i)^jfib(i)$

$G_k(n+1) - G_k(n) = \sum_{j = 0}^{k-1} \binom{k}{j} \sum _{i=1}^{n}(n-i)^jfib(i)$

$G_k(n+1) - G_k(n) = \sum_{j = 0}^{k-1} \binom{k}{j} G_j(n)$

$G_k(n+1) = G_k(n) + \sum_{j = 0}^{k-1} \binom{k}{j} G_j(n)$

$G_k(n+1) = \sum_{j = 0}^{k} \binom{k}{j} G_j(n)$

至此我们即可用矩阵快速幂求解 $G_i(n),\;\; 0 \leq i \leq k$

具体来讲

$G = \left[ \begin{matrix} G_k(n)& G_{k-1}(n) & \cdots & G_0(n) & fib(n + 2) & fib(n + 1) & 1\\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ \vdots &\vdots & \cdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \end{matrix} \right]_{k+4}$

$G' = \left[ \begin{matrix} G_k(n-1)& G_{k-1}(n-1) & \cdots & G_0(n-1) & fib(n + 1) & fib(n) & 1\\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ \vdots &\vdots & \cdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \end{matrix} \right]_{k+4}$

$B = \left[ \begin{matrix} \binom{k}{k} & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 & 0\\ \binom{k}{k-1} & \binom{k-1}{k-1} & 0 & 0 & \cdots & 0 & 0 & 0 & 0 & 0\\ \binom{k}{k-2} & \binom{k-1}{k-2} & \binom{k-2}{k-2} & 0 & \cdots & 0 & 0 & 0 &0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \binom{k}{1} & \binom{k-1}{1} & \binom{k-2}{1} & \binom{k-3}{1} & \cdots & \binom{1}{1} & 0 & 0 & 0 & 0 \\ \binom{k}{0} & \binom{k-1}{0} & \binom{k-2}{0} & \binom{k-3}{0} & \cdots & \binom{1}{0} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -1 & 0 & 0 & 1 \end{matrix} \right]_{k+4}$

可知 $G = G'B$

$A = \left[ \begin{matrix} 0 & 0 & \cdots & 1 & 2 & 1 & 1\\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ \vdots &\vdots & \cdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & 0 \end{matrix} \right]_{k+4}$

递归计算可知 $G = AB^{n-1}$

四.代码实现：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int M = (int)1e2 + 4;
const ll mod = (ll)998244353;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;

ll n; int k;
ll g[M + 5];
ll f[M + 5];

struct node
{
    ll D[M + 5][M + 5];
};

ll quick(ll a, ll b)
{
    ll sum = 1;
    while(b)
    {
        if(b & 1) sum = sum * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return (sum + mod) % mod;
}

node mul(node a, node b)
{
    node c; memset(c.D, 0, sizeof(c.D));
    for(int i = 0; i < M; ++i)
    for(int j = 0; j < M; ++j)
    for(int l = 0; l < M; ++l)
    c.D[i][j] = (c.D[i][j] + a.D[i][l] * b.D[l][j] % mod) % mod;
    return c;
}

node quick(node a, ll b)
{
    node sum; memset(sum.D, 0, sizeof(sum.D));
    for(int i = 0; i < M; ++i) sum.D[i][i] = 1;
    while(b)
    {
        if(b & 1) sum = mul(sum, a);
        a = mul(a, a);
        b >>= 1;
    }
    return sum;
}

int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    scanf("%lld %d", &n, &k);
    node A, B;
    memset(A.D, 0, sizeof(A.D));
    memset(B.D, 0, sizeof(B.D));
    A.D[0][k] = 1, A.D[0][k + 1] = 2, A.D[0][k + 2] = 1, A.D[0][k + 3] = 1;
    for(int j = k; j >= 0; --j)
    {
        B.D[k][j] = 1;
        for(int i = k - 1; i >= j; --i) B.D[i][j] = (B.D[i][j + 1] + B.D[i + 1][j + 1]) % mod;
    }
    B.D[k][k] = 0;
    B.D[k + 1][k] = B.D[k + 1][k + 1] = B.D[k + 1][k + 2] = 1;
    B.D[k + 2][k] = B.D[k + 2][k + 1] = 1;
    B.D[k + 3][k] = -1, B.D[k + 3][k + 3] = 1;
    node G = mul(A, quick(B, n - 1));
    for(int i = 0; i <= k; ++i) g[i] = G.D[0][k - i];
    f[0] = G.D[0][k + 1] - 1;
    for(int i = 1; i <= k; ++i)
    {
        f[i] = g[i];
        for(int j = 0; j <= i - 1; ++j) f[i] = (f[i] + (((j&1)<<1)-1) * B.D[k - j][k - i] * quick(n % mod, i - j) % mod * f[j] % mod) % mod;
        if(i & 1) f[i] *= -1;
        f[i] = (f[i] + mod) % mod;
    }
    printf("%lld\n", f[k]);
    return 0;
}

标签：node,int,BM,ll,斐波,飞杯,友谊赛,sum,mod
来源： https://blog.csdn.net/The___Flash/article/details/106066577

斐波那契和（“科大讯飞杯”第十七届同济大学程序设计预选赛暨高校网络友谊赛 J，矩阵快速幂）

一.题目链接：