2020牛客暑期多校训练营（第九场）

Groundhog and 2-Power Representation

s = input()
a = []
for i in range(len(s)):
    if s[i] == '(':
        a.append(-1)
    elif s[i] == ')':
        a.append(-2)
    elif s[i] == '+':
        a.append(-3)
    else:
        a.append(int(s[i]))
stack = []
for i in range(len(a)):
    if a[i] != -2:
        stack.append(a[i])
    else:  # a[i]==)
        while len(stack) != 0:
            x = stack.pop()
            if len(stack) == 0:
                stack.append(x)
                break
            else:
                op = stack.pop()
                if op == -1:  # (
                    x = pow(stack.pop(), x)
                    stack.append(x)
                    break
                else:  # op==-3 +
                    y = stack.pop()
                    stack.append(x + y)
while len(stack) != 0:
    x = stack.pop()
    if len(stack) != 0:
        stack.pop()
        y = stack.pop()
        stack.append(x + y)
    else:
        stack.append(x)
        break
print(stack.pop())

万万没想到版本

print(eval(input().replace('(', '**(')))

---

Groundhog and Apple Tree
Groundhog and Gaming Time
Groundhog and Golden Apple

Groundhog Chasing Death

#include 
using namespace std;
typedef long long ll;
const ll mod = 998244353;

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

ll a, b, c, d, x, y;
map<int, int> mp;//记录x里质因数p的个数
ll res = 1;
ll phi = mod - 1;

// 统计答案
void statisticsRes(ll p, ll cntX, ll cntY) {
    ll a1 = a * cntX, b1 = b * cntX;
    ll c1 = c * cntY, d1 = d * cntY;

    ll sum = 0;
    for (ll j = a1; j <= b1; j += cntX) {
        // 只取< 等于再下面那个for里统计
        if (j < c1)
            sum += (d1 / cntY - (c1 - 1) / cntY) * j;
        else if (j < d1)
            sum += (d1 / cntY - j / cntY) * j;

        sum %= phi;
    }

    for (ll j = c1; j <= d1; j += cntY) {
        // <=
        if (j <= a1)
            sum += (b1 / cntX - (a1 - 1) / cntX) * j;
        else if (j <= b1)
            sum += (b1 / cntX - j / cntX + (((j - a1) % cntX) == 0)) * j;
        
        sum %= phi;
    }
    sum = sum % phi + phi;
    res = res * qpow(p, sum) % mod;
}

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

    cin >> a >> b >> c >> d >> x >> y;

    for (int i = 2; i * i <= x; i++) {
        if (x % i == 0) {
            int cnt = 0;
            while (x % i == 0) {
                cnt++;
                x /= i;
            }
            mp[i] = cnt;
        }
    }
    if (x > 1)  mp[x] = 1;
    
    for (int i = 2; i * i <= y; i++) {
        if (y % i == 0) {
            int cnt = 0;
            while (y % i == 0) {
                cnt++;
                y /= i;
            }
            if (mp.count(i)) 
                statisticsRes(i, mp[i], cnt);
        }
    }

    if (y > 1 && mp.count(y)) 
            statisticsRes(y, mp[y], 1);
        
    cout << res << endl;
    return 0;
}

Groundhog Looking Dowdy

#include 
using namespace std;
typedef pair<int, int> pii;
const int N = 1e6 + 10;
int n, m, k;

vector<pii> a;
int len;
int vis[N];

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

    cin >> n >> m;
    for (int i = 1, x; i <= n; i++) {
        cin >> k;

        for (int j = 1; j <= k; j++) {
            cin >> x;
            a.push_back({x, i});
        }
    }

    sort(a.begin(), a.end());
    len = a.size();

    int cnt = 0;
    int res = INT_MAX;
    for (int r = 0, l = 0; r < len; r++) {
        vis[a[r].second]++;
        if (vis[a[r].second] == 1) {
            cnt++;
        }

        while (l < r && cnt >= m) {
            res = min(res, a[r].first - a[l].first);
            vis[a[l].second]--;
            if (!vis[a[l].second]) cnt--;
            l++;
        }
    }

    cout << res << endl;
    return 0;
}

Groundhog Playing Scissors
Groundhog Speaking Groundhogish

The Crime-solving Plan of Groundhog

#include 
using namespace std;
typedef pair<int, int> pii;
const int N = 1e6 + 10;
int n;

vector<int> mul(vector<pii> a, int b) {
    vector<int> c;
    int t = 0;

    for (int i = a.size() - 1; i >= 0; i--) {
        int x = a[i].first;
        int cnt = a[i].second;

        for (int j = 0; j < cnt; j++) {
            t += x * b;
            c.push_back(t % 10);
            t /= 10;
        }
    }
    while (t)c.push_back(t % 10), t /= 10;
    return c;
}

void output(vector<int> a) {
    for (int i = a.size() - 1; i >= 0; i--)
        cout << a[i];//注意了 输出的是一串数 所以不能有空格
    cout << endl;
}

int a[N];
vector<pii> v;

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


    int T;
    cin >> T;
    while (T--) {
        for (int i = 0; i <= 9; i++) {
            a[i] = 0;
        }
        v.clear();

        cin >> n;
        for (int i = 1, x; i <= n; i++) {
            cin >> x;
            a[x]++;
        }

        int x;
        for (int i = 1; i <= 9; i++) {
            if (a[i]) {
                x = i;
                a[i]--;
                break;
            }
        }
        for (int i = 1; i <= 9; i++) {
            if (a[i]) {
                if (v.empty()) {
                    v.push_back({i, 1});
                    a[i]--;
                    if (a[0]) v.push_back({0, a[0]});
                    if (a[i]) v.push_back({i, a[i]});
                } else {
                    v.push_back({i, a[i]});
                }
            }
        }
        output(mul(v, x));
    }
    return 0;
}

The Escape Plan of Groundhog

#include 
using namespace std;
const int N = 1e6 + 10;
vector<int> e[N];
int n, m, t;

int fa[N];

void dfs1(int u, int f) {
    fa[u] = f;
    for (int i = 0, lim = e[u].size(); i < lim; i++) {
        int v = e[u][i];
        if (v != f) {
            dfs1(v, u);
        }
    }
}

vector<int> path;
int st = 0;
int res = 0;

void dfs2(int u, int f, int k, int x, int y) {

    if (u == n) return;

    int t1 = (y + 1) / 2;
    int t2 = x;

    if (t1 <= t2) { // 到达该点被抓
        res = max(t1, res);
        return;
    }

    int path_v = k ? path[k] : 0;//判断是否是 st-n 上的点
    int lim = e[u].size();
    for (int i = 0; i < lim; i++) {
        int v = e[u][i];
        if (v != f && v != path_v) {
            dfs2(v, u, 0, x + 1, y + 1);
        }
    }

    // 停在这个位置等待被抓
    if (lim == 1) {
        res = max(res, (y + 1) / 2);
    }
    if (k) {
        dfs2(path_v, u, k + 1, x + 1, max(y - 1, 0));
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    cin >> n >> t;
    for (int i = 1, u, v; i < n; i++) {
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }

    dfs1(1, 0);

    for (int i = n; i; i = fa[i]) {
        path.push_back(i);
    }

    m = path.size() - 1;
    if (m <= t) {
        cout << 0 << endl;
        return 0;
    }

    //翻转
    reverse(path.begin(), path.end());
    //起始点
    st = path[t];
    dfs2(st, 0, t + 1, 0, m - t);

    cout << res << endl;
    return 0;
}

The Flee Plan of Groundhog
The Shopping Plan of Groundhog