乱搞做法重见天日。目前能过掉在数据范围内的所有已知数据。求更强的 hack，或证明该代码 hack 不掉。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using u64 = unsigned long long;
using u128 = __uint128_t;
static unsigned char p[50000005]; 
static unsigned char num[10000005];
inline u64 modmul(u64 a, u64 b, u64 mod) {
    return (u128)a * b % mod;
}
inline u64 modpow(u64 a, u64 d, u64 mod) {
    u64 r = 1;
    while (d) {
        if (d & 1) r = modmul(r, a, mod);
        a = modmul(a, a, mod);
        d >>= 1;
    }
    return r;
}
inline bool miller_rabin(u64 n) {
    if (n < 2) return false;
    for (u64 x : {2ull,3ull,5ull,7ull,11ull}) {
        if (n == x) return true;
        if (n % x == 0) return false;
    }
    u64 d = n - 1, s = 0;
    while ((d & 1) == 0) { d >>= 1; ++s; }
    auto check = [&](u64 a)->bool{
        if (a % n == 0) return true;
        u64 x = modpow(a, d, n);
        if (x == 1 || x == n-1) return true;
        for (u64 r = 1; r < s; ++r) {
            x = modmul(x, x, n);
            if (x == n-1) return true;
        }
        return false;
    };
    for (u64 a : {2ull,7ull,61ull}) {
        if (a >= n) break;
        if (!check(a)) return false;
    }
    return true;
}
inline bool isprime(int n) {
    if (p[n] != 0) return p[n] == 2;
    bool isp;
    if (n <= 1) isp = false;
    else if (n <= 3) isp = true;
    else if ((n & 1) == 0) isp = false;
    else isp = miller_rabin((u64)n);
    p[n] = isp ? 2 : 1;
    return isp;
}
int N, M;
int maxn = 0;
void dfs(int dep) {
    if (dep < 0) return;
    long long base = 1LL << dep;
    int limit = (base==0) ? M : (int)(M / base);
    if (limit < 3) {
        dfs(dep - 1);
        return;
    }
    for (int i = 3; i <= limit; i += 2) {
        if (isprime(i)) {
            long long prod = base * 1LL * i;
            if (prod >= N && prod <= M) {
                cout << dep + 1;
                return;
            } else if (prod > M) {
                dfs(dep - 1);
                return;
            }
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    const int threshold = 2097150;
    cin >> N >> M;
    int k = (int)floor(log2((double)M));
    if (((1 << k) >= N)) {
        cout << k;
        return 0;
    } else if ((M - N + 1) <= threshold) {
        if (M >= 1) num[1] = 0;
        if (M >= 2) {
            isprime(2);
            num[2] = 1;
        }
        for (int i = 3; i <= M; ++i) {
            if ((i & 1) == 0) {
                num[i] = num[2] + num[i >> 1];
                continue;
            }
            if (isprime(i)) {
                num[i] = 1;
                continue;
            }
            if (i % 3 == 0) {
                num[i] = num[3] + num[i / 3];
                continue;
            }
            int lim = (int)floor(sqrt((double)i));
            bool done = false;
            for (int j = 5; j <= lim; j += 6) {
                int a = j;
                if (i % a == 0) {
                    num[i] = num[a] + num[i / a];
                    done = true;
                    break;
                }
                int b = j + 2;
                if (b <= lim && i % b == 0) {
                    num[i] = num[b] + num[i / b];
                    done = true;
                    break;
                }
            }
            if (!done) {
                for (int j = 3; j * j <= i; j += 2) {
                    if (i % j == 0) {
                        num[i] = num[j] + num[i / j];
                        done = true;
                        break;
                    }
                }
                if (!done) {
                    num[i] = 1;
                }
            }
        }
        unsigned char best = 0;
        for (int i = max(2, N); i <= M; ++i) {
            if (num[i] > best) best = num[i];
        }
        cout << (int)best;
        return 0;
    } else {
        dfs(k - 1);
        return 0;
    }
}

请不要在意我的逆天码风。