[SHOI 2017] 组合数问题

[题目链接]

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

[算法]

       回顾组合数的定义 :

       C(N , M)表示将N个小球放入M个盒子里的方案数

       我们发现题目要求的其实就是将nk个小球放入模k意义下于r个盒子中的方案数

       不妨设Fi , j表示放了i个小球 ， j个盒子(模k意义下)的方案数

       有 : Fi , j = Fi - 1 , j - 1 + Fi - 1 , j

       矩阵乘法即可

       时间复杂度 : O(K ^ 3logNlogK)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int N = 1e9 + 10;
const int K = 55;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int n , p , k , r;
int mat[K][K];

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void multipy(int a[K][K] , int b[K][K])
{
    static int res[K][K];
    for (int i = 0; i < k; ++i)
    {
        for (int j = 0; j < k; ++j)
        {
            res[i][j] = 0;
        }
    }
    for (int x = 0; x < k; ++x)
    {
        for (int i = 0; i < k; ++i)
        {
            for (int j = 0; j < k; ++j)
            {
                res[i][j] = (res[i][j] + 1ll * a[i][x] * b[x][j] % p) % p;
            }
        }
    }
    for (int i = 0; i < k; ++i)
    {
        for (int j = 0; j < k; ++j)
        {
            a[i][j] = res[i][j];    
        }    
    }
}
inline void exp_mod(int mat[K][K] , ll n)
{
    static int b[K][K];
    for (int i = 0; i < k; ++i)
    {
        for (int j = 0; j < k; ++j)
        {
            b[i][j] = (i == j);
        }
    }
    while (n > 0)
    {
        if (n & 1) multipy(b , mat);
        multipy(mat , mat);
        n >>= 1;
    }
    for (int i = 0; i < k; i++)
    {
        for (int j = 0; j < k; j++)
        {
            mat[i][j] = b[i][j];
        }
    }
}

int main()
{
    
    read(n); read(p); read(k); read(r);
    for (int i = 0; i < k; ++i) 
    {
        ++mat[i][i]; 
        ++mat[i][((i - 1) % k + k) % k];
    }
    exp_mod(mat , (ll)n * k);
    printf("%d\n" , mat[r][0]);
    
    return 0;
}