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2020 牛客国庆集训派对 Day6

Contents

Contest Info

Practice Link

Solved A B C D E F G H I J K
6/11 O O - - O O - - O Ø -
  • O 在比赛中通过
  • Ø 赛后通过
  • ! 尝试了但是失败了
  • - 没有尝试

Solutions

Problem A. Fractions

Solved By Dup4 & lts. 01:38(+)

题意

给出 n，要求找出 k(1 \leq k \leq 10^5) 对分数 \displaystyle \frac{a_i}{b_i}，满足：

  • b_i \;|\; n, 并且有 1 \lt b_i \lt n。
  • 1 \leq a_i \lt b_i。
  • \displaystyle \sum\limits_{i = 1}^k \frac{a_i}{b_i} = 1 - \frac{1}{n}。

思路

  • 大力猜测两个分数就够了。
  • 然后枚举出两个分母，做 exgcd 即可。
Code 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n;

ll ex_gcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = ex_gcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;
    return d;
}

bool liEu(ll a, ll b, ll c, ll &x, ll &y) {
    ll d = ex_gcd(a, b, x, y);
    if (c % d != 0)
        return 0;
    ll k = c / d;
    x *= k;
    y *= k;
    return 1;
}

void run() {
    rd(n);
    vector<int> fac;
    for (int i = 1; 1ll * i * i <= n; ++i) {
        if (n % i == 0) {
            if (i != 1 && i != n)
                fac.push_back(i);
            if (n / i != 1 && n / i != n)
                fac.push_back(n / i);
        }
    }
    sort(all(fac));
    fac.erase(unique(all(fac)), fac.end());
    int m = SZ(fac);
    for (int i = 0; i < m; ++i) {
        if ((n - 1) % (n / fac[i]) == 0) {
            cout << "YES\n";
            pt(1);
            pt((n - 1) / (n / fac[i]), fac[i]);
            return;
        }
        for (int j = i + 1; j < m; ++j) {
            ll a = n / fac[i];
            ll b = n / fac[j];
            ll c = n - 1;
            ll x, y;
            ll g = __gcd(a, b);
            ll t;
            if (liEu(a, b, c, x, y)) {
                //	if (x < 1) {
                t = b / g;
                x = (x % t + t) % t;
                //	}
                //	if (y < 1) {
                t = a / g;
                y = (y % t + t) % t;
                //	}
                cout << "YES\n";
                pt(2);
                pt(x, fac[i]);
                pt(y, fac[j]);
                return;
            }
        }
    }
    cout << "NO\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(20);
    int _T = 1;
    // nextInt();
    while (_T--) run();
    //    for (int kase = 1; kase <= _T; ++kase) {
    //        cout << "Case #" << kase << ": ";
    //        run();
    //    }
    //	while (cin >> n) run();
    //	run();
    return 0;
}

Problem B. Guest Student

Solved By Dup4. 00:31(+1)

题意

给出一周七天，哪些天有课，然后给出 n，问上 n 天课需要的周期是多长。

思路

枚举第一周和最后一周的开始位置和结束位置。

Code 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n, a[10], f[10];

int ceil(int x, int y) {
    return (x + y - 1) / y;
}

void run() {
    rd(n);
    f[0] = 0;
    for (int i = 1; i <= 7; ++i) rd(a[i]), f[i] = f[i - 1] + a[i];
    int sum = f[7];
    int res = 7 * n;
    if (n <= sum) {
        for (int i = 1; i <= 7; ++i) {
            for (int j = i; j <= 7; ++j) {
                if (f[j] - f[i - 1] >= n) {
                    chmin(res, j - i + 1);
                }
            }
        }
    }
    for (int i = 1; i <= 7; ++i) {
        for (int j = 1; j <= 7; ++j) {
            int need = n - f[j] - (sum - f[i - 1]);
            int now = 7 - i + 1 + j;
            now += ceil(need, sum) * 7;
            //	dbg(i, j, need, now);
            chmin(res, now);
        }
    }
    pt(res);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(20);
    int _T = nextInt();
    while (_T--) run();
    //    for (int kase = 1; kase <= _T; ++kase) {
    //        cout << "Case #" << kase << ": ";
    //        run();
    //    }
    //	while (cin >> n) run();
    //	run();
    return 0;
}

Problem C. Harder Satisfiability

Upsolved By -.

题意

思路

Problem D. Interval-Free Permutations

Upsolved By -.

题意

思路

Problem E. King Kog's Reception

Solved By Dup4. 03:20(+)

题意

给出 q 个事件，事件类型有如下三种：

  • + t d, 表示在第 t 时刻有个骑士需要参见国王，参见所需事件为 d。
  • - i，表示取消第 i 个事件。
  • ? t，表示询问公主如果在第 t 时刻来见国王，需要等多久。如果同一时刻，也有骑士在等，那么骑士优先。

思路

我们用 f_i 表示前 i 个时刻所有参见事件全都结束需要的时间，那么有：

\begin{eqnarray*} f_i = \max(f_{i - 1}, i) + d_i \end{eqnarray*}

然后发现这个东西可以用线段树维护。

Code 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e6 + 10;
int q, t[N], d[N];

struct SEG {
    struct node {
        ll sum, Max;
        node() {
            sum = Max = 0;
        }
        node(ll sum, ll Max) : sum(sum), Max(Max) {}
        node operator+(const node &other) const {
            node res = node();
            res.sum = sum + other.sum;
            res.Max = max(other.Max, Max + other.sum);
            return res;
        }
    } t[N << 2], res;
    void build(int id, int l, int r) {
        if (l == r) {
            t[id] = node(0, l);
            return;
        }
        int mid = (l + r) >> 1;
        build(id << 1, l, mid);
        build(id << 1 | 1, mid + 1, r);
        t[id] = t[id << 1] + t[id << 1 | 1];
    }
    void update(int id, int l, int r, int pos, ll v) {
        if (l == r) {
            t[id].sum += v;
            t[id].Max += v;
            return;
        }
        int mid = (l + r) >> 1;
        if (pos <= mid)
            update(id << 1, l, mid, pos, v);
        else
            update(id << 1 | 1, mid + 1, r, pos, v);
        t[id] = t[id << 1] + t[id << 1 | 1];
    }
    void query(int id, int l, int r, int ql, int qr) {
        if (l >= ql && r <= qr) {
            res = res + t[id];
            return;
        }
        int mid = (l + r) >> 1;
        if (ql <= mid)
            query(id << 1, l, mid, ql, qr);
        if (qr > mid)
            query(id << 1 | 1, mid + 1, r, ql, qr);
    }
} seg;

void run() {
    rd(q);
    char op[5];
    int m = 1e6;
    seg.build(1, 1, m);
    for (int i = 1, _t, x; i <= q; ++i) {
        cin >> op;
        if (*op == '+') {
            rd(t[i], d[i]);
            seg.update(1, 1, m, t[i], d[i]);
        } else if (*op == '-') {
            rd(x);
            seg.update(1, 1, m, t[x], -d[x]);
        } else {
            rd(_t);
            seg.res = SEG::node();
            seg.query(1, 1, m, 1, _t);
            pt(seg.res.Max - _t);
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(20);
    int _T = 1;
    // nextInt();
    while (_T--) run();
    //    for (int kase = 1; kase <= _T; ++kase) {
    //        cout << "Case #" << kase << ": ";
    //        run();
    //    }
    //	while (cin >> n) run();
    //	run();
    return 0;
}

Problem F. Lazyland

Solved By Dup4. 00:38(+)

题意

给出 n 个人以及 m 个任务，并且第 i 个人想要做第 a_i 个任务，要想劝说第 i 个人改变他的意志需要花费 b_i 的代价，现在要求每个任务至少有一个人需要做，最少需要花费多少代价。

思路

将每个任务不用劝说的人中，除了代价最高的那个人，其他人全都丢进最小堆中，然后贪心取出来去做其他任务。

Code 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n, k, a[N], b[N], c[N];

void run() {
    rd(n, k);
    for (auto &arr : {a, b}) {
        for (int i = 1; i <= n; ++i) {
            rd(arr[i]);
        }
    }
    memset(c, 0, sizeof c);
    vector<pII> vec;
    for (int i = 1; i <= n; ++i) {
        ++c[a[i]];
        vec.push_back(pII(b[i], a[i]));
    }
    ll sum = 0;
    int need = 0;
    for (int i = 1; i <= k; ++i) need += c[i] == 0;
    sort(all(vec), [&](pII a, pII b) {
        return a.fi > b.fi;
    });
    while (need) {
        pII back = vec.back();
        vec.pop_back();
        if (c[back.se] == 1)
            continue;
        else {
            --need;
            --c[back.se];
            sum += back.fi;
        }
    }
    pt(sum);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(20);
    int _T = 1;
    // nextInt();
    while (_T--) run();
    //    for (int kase = 1; kase <= _T; ++kase) {
    //        cout << "Case #" << kase << ": ";
    //        run();
    //    }
    //	while (cin >> n) run();
    //	run();
    return 0;
}

Problem G. Minegraphed

Upsolved By -.

题意

思路

Problem H. Archery Tournament

Upsolved By -.

题意

思路

Problem I. Box

Solved By groggy_. 01:09(+1)

题意

对于给出宽为 w，高为 h 的纸，需要剪出 a \cdot b \cdot c 的长方体，问能否裁出

思路

长方体的构成方式可以看作两种，用不同边长尝试出各种宽与高的可能性，判断即可

Code 
#include <algorithm>
#include <iostream>
#include <vector>

#define maxn 100005
using namespace std;

int w, h;
bool solve(int W, int H) {
    return W <= w && H <= h;
}

int main() {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    scanf("%d%d", &w, &h);

    int flag = 0;
    flag = flag | solve((a + b) * 2, c + 2 * min(a, b));
    flag = flag | solve((a + c) * 2, b + 2 * min(a, c));
    flag = flag | solve((b + c) * 2, a + 2 * min(b, c));

    flag = flag | solve(3 * b + a + c, c + a);
    flag = flag | solve(3 * a + b + c, b + c);
    flag = flag | solve(3 * c + a + b, a + b);

    flag = flag | solve(c + 2 * min(a, b), (a + b) * 2);
    flag = flag | solve(b + 2 * min(a, c), (a + c) * 2);
    flag = flag | solve(a + 2 * min(b, c), (b + c) * 2);

    flag = flag | solve(c + a, 3 * b + a + c);
    flag = flag | solve(b + c, 3 * a + b + c);
    flag = flag | solve(a + b, 3 * c + a + b);

    if (flag)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

Problem J. Connections

Upsolved By Dup4.

题意

给出 n 个点 m \gt 2 \cdot n 条边的有向强连通图，现在要求只保留 2n 条边，使得图仍然是强连通的。

思路

随便找个点作为根，跑出一个 dfs 树，然后在反图上，还是以这个点为根，跑出一个 dfs 树即可。

显然，这两棵树所用的边不会超过 2(n - 1)。

Code 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n, m, vis[N], f[N];
pII e[N];

vector<vector<pII>> G, T;

void dfs(int u) {
    for (auto &it : G[u]) {
        int v = it.fi, id = it.se;
        if (!vis[v]) {
            f[id] = 1;
            vis[v] = 1;
            dfs(v);
        }
    }
}

void run() {
    rd(n, m);
    G.clear();
    G.resize(n + 1);
    T.clear();
    T.resize(n + 1);
    for (int i = 1, u, v; i <= m; ++i) {
        rd(u, v);
        e[i] = pII(u, v);
        G[u].push_back(pII(v, i));
        T[v].push_back(pII(u, i));
        f[i] = 0;
    }
    memset(vis, 0, sizeof(vis[0]) * (n + 5));
    dfs(1);
    memset(vis, 0, sizeof(vis[0]) * (n + 5));
    G = T;
    dfs(1);
    int tot = m - n * 2;
    for (int i = 1; i <= m; ++i) {
        if (f[i] == 0 && tot) {
            --tot;
            pt(e[i].fi, e[i].se);
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(20);
    int _T = nextInt();
    while (_T--) run();
    //    for (int kase = 1; kase <= _T; ++kase) {
    //        cout << "Case #" << kase << ": ";
    //        run();
    //    }
    //	while (cin >> n) run();
    //	run();
    return 0;
}

Problem K. The Final Level

Upsolved By -.

题意

思路

Last update: March 23, 2022
Created: March 23, 2022
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