51nod1121 机器人排队

题面

题目链接

解题思路

考虑如何求解阶乘，跟这篇题解类似，是求阶乘的最低十八位。组合数能够写成阶乘的形式。

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

#pragma GCC optimize(2)

typedef long long ll;
typedef __int128 Int;

Int add(Int a, Int b, Int mod) { return a + b >= mod ? a + b - mod : a + b; }
Int mul(Int a, Int b, Int mod) { return 1LL * a * b % mod; }
Int qpow(Int x, Int n, Int mod) { Int r = 1; for (; n; n >>= 1, x = mul(x, x, mod)) if (n & 1) r = mul(x, r, mod); return r; }
void exgcd(Int a, Int b, Int &x, Int &y) { b == 0 ? (x = 1, y = 0) : (exgcd(b, a % b, y, x), y -= a / b * x); }
Int rev(Int a, Int p) { Int x, y; exgcd(a, p, x, y); return (x + p) % p; }

struct node {
	Int s2[110][110];

	void init(Int k, Int mod) {
		s2[0][0] = 1;
		for (int i = 1; i <= k + 10; i++) {
			for (int j = 1; j <= i; j++) {
				s2[i][j] = add(s2[i - 1][j - 1], mul(j, s2[i - 1][j], mod), mod);
			}
		}
	}

	Int get(Int n, Int k, Int mod) {
		if (n == 0) return 0;
		Int res = 0;
		for (Int i = 1; i <= k && i <= n; i++) {
			Int sum = s2[k][i], flag = true;
			for (Int j = n - i + 1; j <= n + 1; j++) {
				if (j % (i + 1) == 0 && flag) sum = mul(sum, j / (i + 1), mod), flag = false;
				else sum = mul(sum, j, mod);
			}
			res = add(res, sum, mod);
		}
		return res;
	}

	Int g(Int t, Int d, Int p, Int k, Int mod) {
		if (t <= k) {
			Int res = 1;
			for (Int i = 0; i <= t; i++) res = mul(res, add(mul(i, p, mod), d, mod), mod);
			return res;
		}
		else {
			static Int dp[110], cd[110], cc[110], cp[110], cnt[110];
			cd[k - 1] = qpow(d, t - k + 2, mod);
			for (Int i = k - 2; i >= 0; i--) cd[i] = mul(d, cd[i + 1], mod);
			cc[0] = 1;
			for (Int i = 1; i < k; i++) cc[i] = get(t, i, mod);
			cp[0] = 1;
			for (Int i = 1; i < k; i++) cp[i] = mul(cp[i - 1], p, mod);
			cnt[0] = 0;
			for (Int i = 1; i < k; i++) cnt[i] = 0;
			dp[0] = 1; Int ma = 0;
			for (Int i = 1; i < k; i++) {
				dp[i] = 0;
				for (int j = 1, f = 1; j <= i; j++, f = -f) {
					Int r = mul(cc[j], mul(dp[i - j], cp[ma - cnt[i - j]], mod), mod);
					if (f > 0) dp[i] = add(dp[i], r, mod);
					else dp[i] = add(dp[i], mod - r, mod);
				}
				Int x = i; cnt[i] = ma;
				while (x % p == 0) x /= p, cnt[i]++, ma++;
				dp[i] = mul(dp[i], rev(x, mod), mod);
			}
			Int res = 0;
			for (int i = 0; i < k; i++) res = add(res, mul(cp[i - cnt[i]], mul(cd[i], dp[i], mod), mod), mod);
			return res;
		}
	}

	Int f(Int n, Int p, Int k, Int mod) {
		if (n < p) {
			Int res = 1;
			for (int i = 2; i <= n; i++) res = mul(res, i, mod);
			return res;
		}
		else {
			Int res = f(n / p, p, k, mod);
			for (int i = 1; i < p; i++) res = mul(res, g(n / p + (n % p >= i ? 0 : -1), i, p, k, mod), mod);
			return res;
		}
	}

	Int ans, cnt;

	void Fac(Int n, Int p, Int k, Int mod, Int op) {
		if (op == 1) ans = mul(ans, f(n, p, k, mod), mod);
		else ans = mul(ans, rev(f(n, p, k, mod), mod), mod);
		while (n > 0) cnt += n / p * op, n /= p;
	}

	void solve(Int n, Int m, Int p, Int k, Int mod) {
		init(k, mod);
		ans = 1, cnt = 0;
		Fac(n + m, p, k, mod, 1);
		Fac(n + 1, p, k, mod, -1); Fac(m, p, k, mod, -1);
		Int x = n - m + 1;
		while (x % p == 0) cnt++, x /= p;
		ans = mul(ans, x, mod);
	}
}a2, a5;

ll n, m;


int main() {
	//freopen("0.txt", "r", stdin);
	scanf("%lld%lld", &m, &n);
	if (m == 0) puts("1");
	else if (m > n) puts("0");
	else {
		Int n2 = qpow(2, 18, 1e20), n5 = qpow(5, 18, 1e20), nn = n2 * n5;
		a2.solve(n, m, 2, 18, n2);
		a5.solve(n, m, 5, 18, n5);
		Int r2 = a2.ans, r5 = a5.ans;
		Int c2 = a2.cnt, c5 = a5.cnt;
		r5 = mul(r5, qpow(rev(2, n5), c2, n5), n5);
		r2 = mul(r2, qpow(rev(5, n2), c5, n2), n2);
		Int t5 = Int(n2) * rev(n2, n5) % nn * r5 % nn;
		Int t2 = Int(n5) * rev(n5, n2) % nn * r2 % nn;
		Int r = (t5 + t2) % nn;
		r = mul(r, qpow(2, c2, nn), nn);
		r = mul(r, qpow(5, c5, nn), nn);
		printf("%lld\n", ll(r));
	}
	return 0;
}