WC模擬（1.14） T2 Everyone will meet some difficult

Everyone will meet some difficult

題目背景：

1.14 WC模擬T2  

分析：數學 + 數論

 

100分做法我不會，也不想去學，所以就說一下80分做法吧，首先，我們可以知道答案是

顯然，對於一個固定的k，組合數是一個m - n次多項式，並且對於任意k，這個多項式的係數顯然都是相同的，那麼我們定義，k的i次項係數爲a_k，那麼顯然：

顯然對於a_i，我們可以直接爆拆組合數在(m - n)²的時間搞定。

所以直接考慮後面部分，令：

直接用矩陣快速冪優化一下上面的遞推式就可以了。

時間複雜度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;
}