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;
}