题解 | #小红的连续段#

简单题

非常像寒假营第六场的计数题小 L 的彩球

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 1e9 + 7;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream &operator>>(istream &is, i128 &val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char &c: str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream &operator<<(ostream &os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

struct Hash {
    vector<int> h, p;
    int B = 131;
    Hash (const string &s) {
        int n = s.size();
        h.resize(n + 1, 0);
        p.resize(n + 1, 1);
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * B % MOD;
            h[i + 1] = (h[i] * B + s[i]) % MOD;
        }
    }

    LL get_hash(int l, int r) {
        LL v = h[r] - h[l - 1] * p[r - l + 1] % MOD;
        v = (v % MOD + MOD) % MOD;
        return v;
    }
};

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

LL fact[N], infact[N];

LL q_pow(LL a, LL b) {
    LL ans = 1;
    a %= MOD;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

void init() {
    fact[0] = 1, infact[0] = 1;
    for (LL i = 1; i < N; ++i) {
        fact[i] = fact[i - 1] * i % MOD;
        infact[i] = q_pow(fact[i], MOD - 2) % MOD;
    }
}

LL C(LL a, LL b) {
    if (a < b || b < 0) return 0;
    return fact[a] % MOD * infact[b] % MOD * infact[a - b] % MOD;
}

void solve() {
    int x, y;
    cin >> x >> y;
    for (int i = 1; i <= x + y; ++i) {
        int a = i / 2;
        int b = (i + 1) / 2;
        LL ans = C(x - 1, a - 1) * C(y - 1, b - 1) % MOD + C(y - 1, a - 1) * C(x - 1, b - 1) % MOD;
        ans %= MOD;
        cout << ans << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    init();

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}