题目链接

题解

\[ans = \sum\limits_{i = 0}^{\infty}{nk \choose ik + r} \pmod p
\]

发现实际是求

\[ans = \sum\limits_{i = 0}^{\infty}{nk \choose i}[i \mod k = r] \pmod p
\]

设\(f[i][j]\)表示\(i\)个数选出\(x \mod k = j\)个数的方案数

利用组合数递推 + 矩乘转移即可

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<cmath>

#include<map>

#define LL long long int

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define cls(s,v) memset(s,v,sizeof(s))

#define mp(a,b) make_pair<int,int>(a,b)

#define cp pair<int,int>

using namespace std;

const int maxn = 55,maxm = 100005,INF = 0x3f3f3f3f;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}

	return flag ? out : -out;

}

int n,r,K,P;

struct Matrix{

	int s[maxn][maxn],n,m;

	Matrix(){cls(s,0);n = m = 0;}

}A,F0,F;

inline Matrix operator *(const Matrix& a,const Matrix& b){

	Matrix c;

	if (a.m != b.n) return c;

	c.n = a.n; c.m = b.m;

	for (int i = 0; i < c.n; i++)

		for (int j = 0; j < c.m; j++)

			for (int k = 0; k < a.m; k++)

				c.s[i][j] = (c.s[i][j] + 1ll * a.s[i][k] * b.s[k][j] % P) % P;

	return c;

}

inline Matrix qpow(Matrix a,LL b){

	Matrix re; re.n = re.m = a.n;

	for (int i = 0; i < re.n; i++) re.s[i][i] = 1;

	for (; b; b >>= 1,a = a * a)

		if (b & 1) re = re * a;

	return re;

}

int main(){

	n = read(); P = read(); K = read(); r = read();

	F0.n = K; F0.m = 1; F0.s[0][0] = 1;

	A.n = A.m = K;

	for (int j = 0; j < K; j++){

		A.s[j][j]++;

		A.s[j][(j - 1 + K) % K]++;

	}

	F = qpow(A,1ll * n * K) * F0;

	printf("%d\n",F.s[r][0]);

	return 0;

}

巴特西

BZOJ4870 [Shoi2017]组合数问题【组合数 + 矩乘】

题目链接

题解

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巴特西

BZOJ4870 [Shoi2017]组合数问题 【组合数 + 矩乘】

题目链接

题解

最新文章

热门文章

BZOJ4870 [Shoi2017]组合数问题【组合数 + 矩乘】