AtCoder Grand Contest 019 F Yes or No

洛谷传送门

AtCoder 传送门

首先容易发现最优策略是回答剩余多的选项。设 \(n\) 为剩余 Yes 的数量，\(m\) 为剩余 No 的数量，考虑将 \((n,m)\) 放到平面上，若 \(n > m\) 则回答 Yes 并向左走，\(n < m\) 则回答 No 并向下走，\(n=m\) 则随意。

如果按照这样的策略则至少能对 \(\max(n,m)\) 个，考虑计算 \(n=m\) 时能蒙对几个。

设当前在 \((i,i)\)，经过 \((i,i)\) 的路径有 \(\binom{2i}{i} \times \binom{n-i+m-i}{n-i}\) 条，总路径有 \(\binom{n+m}{n}\) 条，则这部分的贡献为：

\[\dfrac{1}{2} \times \sum\limits_{i=1}^{min(n,m)} \dfrac{\binom{2i}{i} \times \binom{n-i+m-i}{n-i}}{\binom{n+m}{n}} \]

加上 \(\max(n,m)\) 即为最终答案。

Code

// Problem: F - Yes or No
// Contest: AtCoder - AtCoder Grand Contest 019
// URL: https://atcoder.jp/contests/agc019/tasks/agc019_f
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100;
const int N = 1000000;
const ll mod = 998244353;
const ll inv2 = (mod + 1) / 2;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, m, fac[maxn], ifac[maxn];

void init() {
	fac[0] = 1;
	for (int i = 1; i <= N; ++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	ifac[N] = qpow(fac[N], mod - 2);
	for (int i = N - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
}

inline ll C(ll n, ll m) {
	if (n < m || n < 0 || m < 0) {
		return 0;
	} else {
		return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
	}
}

void solve() {
	scanf("%lld%lld", &n, &m);
	ll ans = max(n, m), val = qpow(C(n + m, n) * 2 % mod, mod - 2);
	for (int i = 1; i <= min(n, m); ++i) {
		ans = (ans + C(i * 2, i) * C(n + m - i * 2, n - i) % mod * val % mod) % mod;
	}
	printf("%lld\n", ans);
}

int main() {
	init();
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}