Source

https://www.lydsy.com/JudgeOnline/problem.php?id=4870

Solution

首先，组合数有一个重要的性质：

C_{n}^{m} = C_{n - 1}^{m} + C_{n - 1}^{m - 1}

$C_n^m=C_{n-1}^m+C_{n-1}^{m-1}$
设：

f (m, r) = \sum_{i = 0}^{\infty} C_{m}^{i k + r}

$f(m,r)=\sum_{i=0}^{\infty}C_{m}^{ik+r}$
根据组合数的性质推出：

f (m, r) = \sum_{i = 0}^{\infty} (C_{m - 1}^{i k + r} + C_{m - 1}^{i k + (r + k - 1) mod k})

$f(m,r)=\sum_{i=0}^{\infty}(C_{m-1}^{ik+r}+C_{m-1}^{ik+(r+k-1)\bmod k})$

= \sum_{i = 0}^{\infty} C_{m - 1}^{i k + r} + \sum_{i = 0}^{\infty} C_{m - 1}^{i k + (r + k - 1) mod k}

$=\sum_{i=0}^{\infty}C_{m-1}^{ik+r}+\sum_{i=0}^{\infty}C_{m-1}^{ik+(r+k-1)\bmod k}$

= f (m - 1, r) + f (m - 1, (r + k - 1) mod k)

$=f(m-1,r)+f(m-1,(r+k-1)\bmod k)$
边界

f (0, 0) = 1

$f(0,0)=1$ 。
由于

k

$k$ 不大，因此用矩阵乘法就能解决这个问题。

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define For(i, a, b) for (i = a; i <= b; i++)
using namespace std; typedef long long ll;
const int N = 55;
int n, PYZ, k, r;
struct cyx {
    int n, m, a[N][N]; cyx() {} cyx(int _n, int _m) : n(_n), m(_m)
        {memset(a, 0, sizeof(a));}
    friend inline cyx operator * (const cyx &a, const cyx &b) {
        int i, j, k; cyx res = cyx(a.n, b.m);
        For (i, 1, res.n) For (j, 1, res.m) For (k, 1, a.m)
            res.a[i][j] = (res.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % PYZ) % PYZ;
        return res;
    }
    friend inline cyx operator ^ (cyx a, ll b) {
        int i; cyx res = cyx(a.n, a.m); For (i, 1, res.n) res.a[i][i] = 1;
        while (b) {
            if (b & 1) res = res * a; a = a * a; b >>= 1;
        }
        return res;
    }
} P, Y, Z;
int main() {
    int i; cin >> n >> PYZ >> k >> r; P = cyx(k, k);
    For (i, 0, k - 1) P.a[i + 1][i + 1]++, P.a[i + 1][(i + k - 1) % k + 1]++;
    Y = cyx(1, k); Y.a[1][1] = 1; Z = Y * (P ^ 1ll * n * k);
    cout << Z.a[1][r + 1] << endl; return 0;
}

[BZOJ4870][Shoi2017]组合数问题（矩阵乘法）

Source

Solution

Code

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