字符串匹配

int n;
string DD = "DFS", dd = "dfs";

void solve() {
    int D = 0, d = 0;
    string s; cin >> n >> s;
    for (int i = 0; i < n; i ++)
        if (s[i] == DD[D]) D ++;
        else if (s[i] == dd[d]) d ++;
    cout << (D >= 3) << ' '  << (d >= 3) << endl;
}

有以下四种基本图形，排列组合即可

int n;
pii a[N];

int solve() {
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i].bb >> a[i].aa;
    sort(a, a + n);
    int res = 3;
    bool down = false, L = false, R = false, left = false, right = false;
    bool ll = false, rr = false;
    for (int i = 0; i < n; i ++) {
        if (a[i].aa == 0 && a[i].bb == 2) down = true;
        if (a[i].aa == -1 && a[i].bb == 1) left = true;
        if (a[i].aa == 1 && a[i].bb == 1) right = true;
        if (a[i].aa < 0) ll = true;
        if (a[i].aa > 0) rr = true;
        if (i) {
            if (a[i].aa == a[i - 1].aa) {
                if (a[i].aa < 0) L = true;
                else R = true;
            } else if (a[i].aa - 1 == a[i - 1].aa && a[i].bb + a[i - 1].bb == 3) {
                if (a[i - 1].aa < 0) L = true;
                else if (a[i].aa > 0) R = true;
            }
        }
    }
    if (L && R || down && left && right) return 0;
    if (L && right && down || R && left && down) return 0;
    if (down && left || down && right || left && right) return 1;
    if (L && right || R && left || L && down || R && down) return 1;
    if (L && rr || R && ll) return 1;
    if (left && down && rr || right && down && ll) return 1;
    if (L || R || down || left || right || ll && rr) return 2;
    return 3;
}

排序后，前缀和人等待的时间，二分查找鸡插队的时间点

int n, m;
ll t, q;
ll a[N];

bool check(ll x) {
    int l = upper_bound(a + 1, a + n + 1, x) - a;
    ll res = (n - l + 1) * t;
    return res <= q;
}

ll solve() {
    ll l = 0, r = a[n];
    while (l < r) {
        ll mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    return l + t;
}

int main() {
    FastIO
    cin >> n >> m >> t;
    for (int i = 1; i <= n; i ++) cin >> a[i];
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= n; i ++) a[i] += a[i - 1];
    while (m --) {
        cin >> q;
        cout << solve() << endl;
    }
    return 0;
}

参照验题人代码

int n, m;
int a[N];
set<int> st;

bool check(int x) {
    ll res = 1;
    for (int i = 0; i < n; i ++) {
        res *= a[i] + x;
        if (abs(res) > 1e9) return false;
    }
    st.insert(res);
    return true;
}

string solve() {
    int x; cin >> x;
    return st.count(x) ? yes : no;
}

int main() {
    FastIO
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> a[i];
    sort(a, a + n);
    st.insert(0);
    for (int i = 0; i < n; i ++) {
        if (i && a[i] == a[i - 1]) continue;
        for (int l = -a[i] - 1; check(l); l --);
        for (int r = -a[i] + 1; check(r); r ++);
    }
    while (m --) cout << solve() << endl;
    return 0;
}

题为贪心误导人，暴力枚举情况，复杂度为 $O(T\times 3^n)\le 6\times 10^6$

int n, m, res;
int a[N], c[N];

void dfs(int d = 0) {
    if (d == m) {
        for (int i = 1; i <= n; i ++) c[i] = a[i];
        sort(c + 1, c + n + 1);
        int t = upper_bound(c + 1, c + n + 1, a[1]) - c - 1;
        t = n + 1 - t;
        res = min(res, t);
        return;
    }
    int u = b[d].aa, v = b[d].bb;
    a[u] += 3;
    dfs(d + 1);
    a[u] -= 3;
    a[v] += 3;
    dfs(d + 1);
    a[v] -= 3;
    a[u] ++, a[v] ++;
    dfs(d + 1);
    a[u] --, a[v] --;
}

int solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i ++) cin >> a[i];
    for (int i = 0; i < m; i ++) cin >> b[i].aa >> b[i].bb;
    res = n;
    dfs();
    return res;
}

第二类斯特林数

$S(n,m)$ 表示的是把 $n$ 个不同的小球放在 $m$ 个相同的盒子里方案数，也记作 $\left\{\begin{array}{c}n\\ m\end{array}\right\}$

求法

1. 递推 $S(n,m)=S(n-1,m-1)+mS(n-1,m)$
2. 容斥 $S(n,m)=\frac{1}{m!}{\sum }_{k=0}^m(-1{)}^{k}C(m,k)(m-k{)}^n$

性质

$n^k={\sum }_{i=0}^{k}S(k,i)\times i!\times C(n,i)$

ll n, m;

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

int inv(int a) { return qpow(a, mod - 2); }

int solve() {
    cin >> n >> m;
    if (n < m) return 0;
    ll res = 0;
    vector<ll> fac(n + 1, 1), infac(n + 1, 1);
    for (int i = 1; i <= n; i ++)
        infac[i] = inv(fac[i] = fac[i - 1] * i % mod);
    for (int i = 2; i <= m; i ++) res = res * inv(i) % mod;
    for (int i = 0; i <= m; i ++) {
        ll t = qpow(m - i, n) * infac[i] % mod * infac[m - i] % mod;
        res = (res + mod + (i & 1 ? -1 : 1) * t) % mod;
    }
    return res;
}

前缀和即可

int n, m;
pii a[N];

ll solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> a[i].aa >> a[i].bb;
    sort(a, a + n);
    for (int i = 1; i < n; i ++) a[i].bb += a[i - 1].bb;
    for (int i = n - 1; i >= 0; i --)
        if (a[i].aa - m <= a[i].bb) return a[i].bb + m;
    return m;
}

参照官方题解：

记所选物品重量或起来是 $c$，枚举 $m$ 里是 $1$ 的位，让 $c$ 里该位为 $0$，则该位将 $m$ 分成了两部分：

• 对于更高位，要求所选的物品这些位必须是 $m$ 的子集（即 $m$ 对应位是 $1$ 才能选）
• 对于更低位，可以全为 $1$，没有任何限制

因此，枚举 $m$ 每一位作为这个分界，每个物品就变成了要么能选要么不能选，所以把能选的都选上就好

int n, m;
int v[N], w[N];

ll solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i ++) cin >> v[i] >> w[i];
    ll res = 0;
    auto check = [&](int x) {
        ll ans = 0;
        for (int i = 0; i < n; i ++)
            if ((x & w[i]) == w[i]) ans += v[i];
        res = max(res, ans);
    };
    for (int i = 29; i >= 0; i --)
        if (m & 1 << i) check((m ^ 1 << i) | (1 << i) - 1);
    check(m);
    return res;
}

概率分析，第一种方法是圆心概率一定 第二种是半径概率一定。因此第一个人的圆心位置概率更平均，答案不唯一，我大概意思一下，一发过了

int n, a, b;

string solve() {
    cin >> n;
    a = b = 0;
    while (n --) {
        int x, y, r; cin >> x >> y >> r;
        x = max(abs(x), abs(y));
        if (x <= 70) a ++;
        else b ++;
    }
    if (1.0 * a / b > 1.2) return bb;
    return aa;
}

计算阴影部分面积即可

double solve() {
    double c, d, h, w; cin >> c >> d >> h >> w;
    return c * w * 3;
}

int solve() {
    int n; cin >> n;
    int res = n / 6;
    return n % 6 ? res * 2 : res;
}