WC模拟（1.14） T2 Everyone will meet some difficult

最新推荐文章于 2024-07-06 07:30:00 发布

scarlyw

最新推荐文章于 2024-07-06 07:30:00 发布

阅读量306

点赞数

CC 4.0 BY-SA版权

分类专栏： NOIP解题报告矩阵乘法容斥原理组合数文章标签： NOI 数学数论

本文链接：https://blog.youkuaiyun.com/scar_lyw/article/details/80212403

NOIP解题报告同时被 3 个专栏收录

195 篇文章

订阅专栏

矩阵乘法

5 篇文章

订阅专栏

容斥原理

3 篇文章

订阅专栏

本文介绍了一种使用矩阵快速幂优化递推式的方法来解决特定的组合数问题，该问题出现在1.14WC模拟赛T2题中，涉及数学与数论知识。文中提供了两种解决方案，一种适用于较小的数据范围，另一种适用于更大数据范围的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Everyone will meet some difficult

题目背景：

1.14 WC模拟T2

分析：数学 + 数论

100分做法我不会，也不想去学，所以就说一下80分做法吧，首先，我们可以知道答案是

显然，对于一个固定的k，组合数是一个m - n次多项式，并且对于任意k，这个多项式的系数显然都是相同的，那么我们定义，k的i次项系数为a_k，那么显然：

显然对于a_i，我们可以直接爆拆组合数在(m - n)²的时间搞定。

所以直接考虑后面部分，令：

直接用矩阵快速幂优化一下上面的递推式就可以了。

时间复杂度O((m - n)³log(m- n))。

Source:

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

/*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}

template<class T>
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

///*
template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
		if (c == '-') iosig = true;	
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int MAXN = 1000000 + 10;
const int mod = 1000000000 + 7;

long long fac[MAXN], inv_fac[MAXN];
long long s, t, n, m;

inline long long mod_pow(long long a, long long b) {
	int ans = 1;
	for (; b; b >>= 1, a = a * a % mod)
		if (b & 1) ans = ans * a % mod;
	return ans;
}

inline void get_c() {
	fac[0] = 1;
	for (int i = 1; i < MAXN; ++i) fac[i] = fac[i - 1] * i % mod;
	inv_fac[MAXN - 1] = mod_pow(fac[MAXN - 1], mod - 2);
	for (int i = MAXN - 2; i >= 0; --i) 
		inv_fac[i] = inv_fac[i + 1] * (i + 1) % mod;
}

inline long long c(int n, int m) {
	if (n < m) return 0;
	return fac[n] * inv_fac[m] % mod * inv_fac[n - m] % mod; 
}

inline void solve_1() {
	get_c();
	long long ans = 0;
	for (int i = 0, sign = 1; i <= n; ++i, sign = -sign) {
		ans = (((ans + (long long)sign * c(n, i) * c(s - i * t, m)) 
			% mod) + mod) % mod;
	}
	std::cout << ans;
}

const int MAXD = 100 + 10;

struct matrix {
	int n;
	long long a[MAXD][MAXD];
	matrix(int n = 0) : n(n) {
		for (int i = 0; i <= n; ++i)
			for (int j = 0; j <= n; ++j)
				a[i][j] = 0;
	}
	
	inline matrix operator * (const matrix &c) const {
		matrix ret(n);
		for (int i = 0; i <= n; ++i)
			for (int k = 0; k <= n; ++k)
				for (int j = 0; j <= n; ++j)
					ret.a[i][j] += a[i][k] * c.a[k][j] % mod;
		for (int i = 0; i <= n; ++i)
			for (int j = 0; j <= n; ++j)
				ret.a[i][j] %= mod;
		return ret;
	}
	
	inline matrix operator ^ (int b) const {
		matrix ans(n), a = *this;
		for (int i = 0; i <= n; ++i) ans.a[i][i] = 1;
		for (; b; b >>= 1, a = a * a)
			if (b & 1) ans = ans * a;
		return ans;
	}
} ;

long long last[MAXD], cur[MAXD], sum[MAXD], mul[MAXN];
inline void solve_2() {
	long long ans = 0;
	last[0] = 1, get_c(), s %= mod;
	for (int i = 0; i < m - n; ++i) {
		for (int j = 0; j <= i + 1; ++j) cur[j] = 0;
		for (int j = 0; j <= i; ++j) cur[j + 1] = last[j];
		for (int j = 0; j <= i; ++j) 
			cur[j] = (cur[j] + last[j] * (s - i) % mod) % mod;
		for (int j = 0; j <= i + 1; ++j) last[j] = cur[j];
	}
	long long ret = 1;
	for (int i = 1; i <= m - n; ++i) ret = ret * i % mod;
	ret = mod_pow(ret, mod - 2);
	for (int i = 0; i <= m - n; ++i) cur[i] = cur[i] * ret % mod;
	for (int i = 1; i <= t; ++i) mul[i] = 1;
	sum[0] = t;
	for (int i = 1; i <= m - n; ++i) {
		for (int j = 1; j <= t; ++j)
			mul[j] = mul[j] * j % mod, sum[i] += mul[j];
		sum[i] %= mod;
	}
	matrix move(m - n);
	for (int i = 0; i <= m - n; ++i)
		for (int j = 0; j <= m - n; ++j)
			move.a[i][j] = c(j, i) * sum[j - i] % mod;
	move = (move ^ n);
//	for (int i = 0; i <= m - n; ++i, std::cout << '\n')
//		for (int j = 0; j <= m - n; ++j)
//			std::cout << move.a[i][j] << " ";
	for (int i = 0, sign = 1; i <= m - n; ++i, sign = -sign)
		ans += (long long)sign * cur[i] * move.a[0][i] % mod;
	ans = (ans % mod + mod) % mod;
	std::cout << ans;
}


int main() {
	freopen("success.in", "r", stdin);
	freopen("success.out", "w", stdout);
	R(s), R(t), R(n), R(m);
	if (s < MAXN && m < MAXN) solve_1();
	else solve_2();
	return 0;
}