2022-2023 Southern And Volga Russian Regional

№	Пользователь	Рейтинг
1	Benq	3792
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3	jiangly	3631
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№	Пользователь	Вклад
1	Qingyu	162
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3	Um_nik	146
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8	YuukiS	135
9	chromate00	134
10	soullless	133

Idea: BledDest, preparation: awoo

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

int n, m, m2;
vector<vector<char>> g;

int T;
vector<int> mt;
vector<int> used;

bool try_kuhn(int v){
    if (used[v] == T)
        return false;
    used[v] = T;
    forn(u, m2) if (g[v][u] && (mt[u] == -1 || try_kuhn(mt[u]))){
        mt[u] = v;
        return true;
    }
    return false;
}

int main() {
    cin >> n >> m;
    vector<string> b(m, string(n, '0'));
    forn(i, n){
        string t;
        cin >> t;
        forn(j, m) b[j][i] = t[j];
    }
    
    vector<string> nw = b;
    sort(nw.begin(), nw.end());
    nw.resize(unique(nw.begin(), nw.end()) - nw.begin());
    m2 = nw.size();
    g.assign(m2, vector<char>(m2, 0));
    forn(i, m2) forn(j, m2) if (i != j){
        bool in = true;
        forn(k, n) in &= nw[i][k] >= nw[j][k];
        if (in) g[i][j] = 1;
    }
    
    mt.assign(m2, -1);
    used.assign(m2, -1);
    T = 0;
    int k = m2;
    forn(i, m2) if (try_kuhn(i)){
        ++T;
        --k;
    }
    
    vector<int> nxt(m2, -1);
    vector<char> st(m2, true);
    forn(i, m2) if (mt[i] != -1){
        nxt[mt[i]] = i;
        st[i] = false;
    }
    
    vector<int> gr(m2), req(m2);
    int t = 0;
    forn(i, m2) if (st[i]){
        int v = i;
        int pos = 2;
        while (v != -1){
            gr[v] = t;
            req[v] = pos;
            ++pos;
            v = nxt[v];
        }
        ++t;
    }
    assert(t == k);
    
    vector<int> num(m);
    forn(i, m) num[i] = lower_bound(nw.begin(), nw.end(), b[i]) - nw.begin();
    
    printf("%d\n", k);
    forn(i, m) printf("%d ", gr[num[i]] + 1);
    puts("");
    forn(i, m) printf("%d ", req[num[i]]);
    puts("");
    forn(i, n){
        vector<int> l(k, 1);
        forn(j, m2) if (nw[j][i] == '1')
            l[gr[j]] = max(l[gr[j]], req[j]);
        forn(j, k)
            printf("%d ", l[j]);
        puts("");
    }
    return 0;
}

1765B - Broken Keyboard

Idea: vovuh, preparation: vovuh

Tutorial

Solution (vovuh)

for _ in range(int(input())):
    n = int(input())
    s = input()
    print('YES' if n % 3 != 2 and not False in [s[i * 3 + 1] == s[i * 3 + 2] for i in range(n // 3)] else 'NO')

1765C - Card Guessing

Idea: DStepanenko, preparation: BledDest

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)
#define fore(i, l, r) for (int i = int(l); i < int(r); i++)

using namespace std;

const int MOD = 998244353;

int add(int a, int b){
    a += b;
    if (a >= MOD)
        a -= MOD;
    return a;
}

int mul(int a, int b){
    return a * 1ll * b % MOD;
}

int binpow(int a, int b){
    int res = 1;
    while (b){
        if (b & 1)
            res = mul(res, a);
        a = mul(a, a);
        b >>= 1;
    }
    return res;
}

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    
    vector<int> fact(4 * n + 1);
    fact[0] = 1;
    fore(i, 1, fact.size()) fact[i] = mul(fact[i - 1], i);
    vector<int> rfact(4 * n + 1);
    rfact.back() = binpow(fact.back(), MOD - 2);
    for (int i = int(fact.size()) - 2; i >= 0; --i)
        rfact[i] = mul(rfact[i + 1], i + 1);
    
    auto cnk = [&](int n, int k){
        return mul(fact[n], mul(rfact[k], rfact[n - k]));
    };
    
    vector<vector<int>> sv(n + 1, vector<int>(5));
    forn(i, n + 1) forn(t, 5) sv[i][t] = mul(binpow(cnk(n, i), t), rfact[t]);
    
    vector<vector<vector<int>>> dp(2, vector<vector<int>>(4 * n + 1, vector<int>(5)));
    forn(ii, n + 1){
        int i = ii & 1;
        int ni = i ^ 1;
        dp[ni] = vector<vector<int>>(4 * n + 1, vector<int>(5));
        for (int t = 1; t <= 4; ++t)
            dp[ni][ii * t][t] = mul(n - ii, sv[ii][t]);
        forn(j, k + 1) for (int p = 1; p <= 4; ++p) if (dp[i][j][p]){
            dp[ni][j][p] = add(dp[ni][j][p], dp[i][j][p]);
            for (int t = 1; p + t <= 4; ++t)
                dp[ni][j + ii * t][p + t] = add(dp[ni][j + ii * t][p + t], mul(dp[i][j][p], sv[ii][t]));
        }
    }
    
    int ans = 0;
    forn(sum, k + 1){
        ans = add(ans, mul(mul(mul(
            sum < k ? 1 : 4 * n - k, 
            dp[(n & 1) ^ 1][sum][4]), 
            mul(binpow(4 * n - sum, MOD - 2), mul(rfact[4 * n], fact[4 * n - sum]))), 
            mul(fact[4], fact[sum]))
        );
    }
    
    printf("%d\n", ans);
    return 0;
}

1765D - Watch the Videos

Idea: BledDest, preparation: DmitryKlenov

Tutorial

Solution (DmitryKlenov)

#include <iostream>
#include <algorithm>
using namespace std;

#define N 200000

int a[N], n, s;

bool can(int x) {
    int l = x - 1, r = n - 1;
    while (l < r) {
        if (a[r] > s - a[l]) return false;
        ++l;
        if (l < r) {
            if (a[l] > s - a[r]) return false;
            --r;
        }
    }
    return true;
}


int main() {
    long long sum = 0;
    scanf("%d %d\n", &n, &s);
    for(int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
        sum += a[i];
    }

    sort(a, a + n, greater<int>());

    int l = 1, r = n;
    while (l < r) {
        int m = (l + r) >> 1;
        if (can(m)) r = m;
        else l = m + 1;
    }

    long long ans = sum + r;
    cout << ans << endl;

    return 0;
}

1765E - Exchange

Idea: BledDest, preparation: BledDest

Tutorial

Solution (BledDest)

#include <bits/stdc++.h>
 
using namespace std;

void solve() 
{
    int n, a, b;
    cin >> n >> a >> b;
    int x = (n + a - 1) / a;
    if(a > b) x = 1;
    cout << x << endl;
}

int main()
{
    int t;
    cin >> t;
    for(int i = 0; i < t; i++)
        solve();
}

1765F - Chemistry Lab

Idea: awoo, preparation: awoo

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

typedef long long li;

const li INF64 = 1e18;

struct base{
    int x, w, c;
};

li area(const base &a, const base &b){
    return (a.c + b.c) * li(abs(a.x - b.x));
}

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    vector<base> a(n);
    forn(i, n) scanf("%d%d%d", &a[i].x, &a[i].w, &a[i].c);
    sort(a.begin(), a.end(), [](const base &a, const base &b){ return a.x > b.x; });
    vector<li> dp(n, -INF64);
    li ans = 0;
    forn(i, n){
        dp[i] = max(dp[i], -a[i].w * 200ll);
        for (int j = i + 1; j < n; ++j)
            dp[j] = max(dp[j], dp[i] + area(a[i], a[j]) * k - a[j].w * 200ll);
        ans = max(ans, dp[i]);
    }
    printf("%.15Lf\n", ans / (long double)(200));
    return 0;
}

1765G - Guess the String

Idea: BledDest, preparation: BledDest

Tutorial

Solution (BledDest)

#include <bits/stdc++.h>     

using namespace std;

mt19937 rnd(42);
uniform_int_distribution<int> d(1, 2);

int ask(int t, int i)
{
    cout << t << " " << i + 1 << endl;
    int x;
    cin >> x;
    return x;
}

void giveAnswer(const string& s)
{
    cout << 0 << " " << s << endl;
    int x;
    cin >> x;
    assert(x == 1);
}

void guessOne(string& s, int i)
{
    int res = ask(1, i);
    if(res == 0)
        s[i] = '1';
    else
        s[i] = s[res - 1];
}

char inv(char c)
{
    if(c == '0') return '1';
    return '0';
}

void guessTwo(string& s, int i)
{
    if(s[1] == '0')
    {
        if(d(rnd) == 1)
        {
            int res = ask(1, i);
            if(res >= 2)
            {
                s[i] = s[res - 1];
                s[i - 1] = s[res - 2];
            }
            else if(res == 1)
            {
                s[i] = '0';
                s[i - 1] = '1';
            }
            else
            {
                s[i] = '1';
                guessOne(s, i - 1);
            }
        }
        else
        {
            int res = ask(2, i);
            if(res >= 2)
            {
                s[i] = inv(s[res - 1]);
                s[i - 1] = inv(s[res - 2]);
            }
            else if(res == 1)
            {
                s[i] = '1';
                s[i - 1] = '0';
            }
            else
            {
                s[i] = '0';
                guessOne(s, i - 1);
            }
        }
    }
    else
    {
        if(d(rnd) == 1)
        {
            int res = ask(1, i);
            if(res >= 2)
            {
                s[i] = s[res - 1];
                s[i - 1] = s[res - 2];
            }
            else if(res == 1)
            {
                s[i] = '0';
                guessOne(s, i - 1);
            }
            else
            {
                s[i] = '1';
                s[i - 1] = '1';
            }
        }
        else
        {
            int res = ask(2, i);
            if(res >= 2)
            {
                s[i] = inv(s[res - 1]);
                s[i - 1] = inv(s[res - 2]);
            }
            else if(res == 1)
            {
                s[i] = '1';
                guessOne(s, i - 1);
            }
            else
            {
                s[i] = '0';
                s[i - 1] = '0';
            }
        }
    }
}

int main()
{
    int t;
    cin >> t;
    for(int i = 0; i < t; i++)
    {
        int n;
        cin >> n;
        string s(n, '0');
        for(int j = 1; j < n; j += 2)
        {
            if(j == 1) guessOne(s, j);
            else guessTwo(s, j);
        }
        if(n % 2 == 1) guessOne(s, n - 1);
        giveAnswer(s);
    }
}

1765H - Hospital Queue

Idea: Neon, preparation: Neon

Tutorial

Solution (Neon)

#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (auto &x : a) cin >> x;
  vector<vector<int>> g(n);
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    g[y - 1].push_back(x - 1);
  }
    
  vector<int> ans(n, -1);
  for (int s = 0; s < n; ++s) {
    vector<int> deg(n);
    for (int i = 0; i < n; ++i) 
      for (int j : g[i]) ++deg[j];
    priority_queue<pair<int, int>> q;
    for (int v = 0; v < n; ++v)
      if (deg[v] == 0 && v != s)
        q.push({a[v], v});
    for (int i = n; i > 0; --i) {
      if (q.empty() || q.top().first < i) {
        if (deg[s] == 0 && i <= a[s])
          ans[s] = i;
        break;
      }
      int v = q.top().second;
      q.pop();
      for (int u : g[v]) {
        --deg[u];
        if (deg[u] == 0 && u != s)
          q.push({a[u], u});
      }
    }
  }
  
  for (int &x : ans) cout << x << ' ';
}

1765I - Infinite Chess

Idea: DmitryKlenov, preparation: dmitryme

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

const string al = "KQRBN";

const int INF = 1e9;

struct mv{ int dx, dy; };
struct piece{ int x, y, t; };
struct seg{ int l, r; };
struct pos{ int x, y; };

bool operator <(const pos &a, const pos &b){
    if (a.x != b.x) return a.x < b.x;
    return a.y < b.y;
}

vector<vector<mv>> mvs({
    {{-1, -1}, {-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}},
    {{-1, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, -1}, {1, -1}, {1, 1}, {-1, 1}},
    {{-1, 0}, {0, 1}, {1, 0}, {0, -1}},
    {{-1, -1}, {1, -1}, {1, 1}, {-1, 1}},
    {{-2, -1}, {-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}}
});

int main() {
    int sx, sy, fx, fy;
    cin >> sx >> sy >> fx >> fy;
    swap(sx, sy), swap(fx, fy);
    --sx, --fx;
    
    int k;
    cin >> k;
    vector<piece> a(k);
    forn(i, k){
        string t;
        cin >> t >> a[i].x >> a[i].y;
        swap(a[i].x, a[i].y);
        --a[i].x;
        a[i].t = al.find(t[0]);
    }
    
    sort(a.begin(), a.end(), [](const piece &a, const piece &b){
        if (a.x != b.x) return a.x < b.x;
        return a.y < b.y;
    });
    
    vector<int> base({sy, fy});
    forn(i, k) base.push_back(a[i].y);
    vector<int> ys;
    for (int y : base) for (int i = y - 16; i <= y + 16; ++i)
        ys.push_back(i);
    sort(ys.begin(), ys.end());
    ys.resize(unique(ys.begin(), ys.end()) - ys.begin());
    
    vector<vector<char>> tk(8, vector<char>(ys.size()));
    forn(i, k){
        a[i].y = lower_bound(ys.begin(), ys.end(), a[i].y) - ys.begin();
        tk[a[i].x][a[i].y] = true;
    }
    sy = lower_bound(ys.begin(), ys.end(), sy) - ys.begin();
    fy = lower_bound(ys.begin(), ys.end(), fy) - ys.begin();
    
    vector<vector<char>> bad = tk;
    
    forn(i, k){
        int x = a[i].x, y = a[i].y;
        if (a[i].t == 0 || a[i].t == 4){
            for (auto it : mvs[a[i].t]){
                int nx = x + it.dx;
                int ny = y + it.dy;
                if (0 <= nx && nx < 8)
                    bad[nx][ny] = true;
            }
        }
        else{
            for (auto it : mvs[a[i].t]){
                for (int nx = x + it.dx, ny = y + it.dy; ; nx += it.dx, ny += it.dy){
                    if (nx < 0 || nx >= 8) break;
                    if (ny < 0 || ny >= int(ys.size())) break;
                    if (tk[nx][ny]) break;
                    bad[nx][ny] = true;
                }
            }
        }
    }
    
    vector<vector<int>> d(8, vector<int>(ys.size(), INF));
    vector<vector<pos>> p(8, vector<pos>(ys.size()));
    set<pair<int, pos>> q;
    d[sx][sy] = 0;
    q.insert({0, {sx, sy}});
    while (!q.empty()){
        int x = q.begin()->second.x;
        int y = q.begin()->second.y;
        q.erase(q.begin());
        if (x == fx && y == fy){
            cout << d[x][y] << endl;
            return 0;
        }
        for (int ny : {y - 1, y, y + 1}){
            if (ny < 0 || ny >= int(ys.size())) continue;
            int dy = max(1, abs(ys[y] - ys[ny]));
            for (int nx = max(0, x - 1); nx <= min(7, x + 1); ++nx) if (!bad[nx][ny]){
                int nd = d[x][y] + dy;
                if (d[nx][ny] > nd){
                    q.erase({d[nx][ny], {nx, ny}});
                    d[nx][ny] = nd;
                    p[nx][ny] = {x, y};
                    q.insert({d[nx][ny], {nx, ny}});
                }
            }
        }
    }
    
    cout << -1 << endl;
    return 0;
}

1765J - Hero to Zero

Idea: BledDest, preparation: BledDest

Tutorial

Solution (BledDest)

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    vector<int> a(n), b(n);
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < n; i++) scanf("%d", &b[i]);
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    long long total = 0;
    for(int i = 0; i < n; i++)
    {
        total += n * 1ll * (a[i] + b[i]);
        total -= a[i] * 2ll * (b.end() - lower_bound(b.begin(), b.end(), a[i]));
        total -= b[i] * 2ll * (a.end() - upper_bound(a.begin(), a.end(), b[i]));
    }
    for(int i = 0; i < n; i++)
        total -= (n - 1) * 1ll * abs(a[i] - b[i]);
    cout << total << endl;
}

1765K - Torus Path

Idea: adedalic, preparation: adedalic

Tutorial

Solution (adedalic)

#include<bits/stdc++.h>

using namespace std;

int n;
vector< vector<int> > a;

bool read() {
    if (!(cin >> n))
        return false;
    a.resize(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cin >> a[i][j];
    }
    return true;
}

void solve() {
    long long sum = 0;
    for (int i = 0; i < n; i++)
        sum += accumulate(a[i].begin(), a[i].end(), 0LL);
    
    int mn = a[0][n - 1];
    for (int i = 0; i < n; i++)
        mn = min(mn, a[i][n - 1 - i]);
    
    cout << sum - mn << endl;
}

int main() {
#ifdef _DEBUG
    freopen("input.txt", "r", stdin);
    int tt = clock();
#endif
    
    if(read()) {
        solve();
        
#ifdef _DEBUG
        cerr << "TIME = " << clock() - tt << endl;
        tt = clock();
#endif
    }
    return 0;
}

1765L - Project Manager

Idea: BledDest, preparation: awoo

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

const string days[] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};

int main() {
    cin.tie(0);
    iostream::sync_with_stdio(false);
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<char>> ds(n, vector<char>(7));
    forn(i, n){
        int t;
        cin >> t;
        forn(_, t){
            string s;
            cin >> s;
            ds[i][find(days, days + 7, s) - days] = true;
        }
    }
    vector<int> h(m);
    forn(i, m) cin >> h[i];
    vector<vector<int>> a(k);
    forn(i, k){
        int p;
        cin >> p;
        a[i].resize(p);
        forn(j, p){
            cin >> a[i][j];
            --a[i][j];
        }
    }
    int j = 0;
    vector<int> ans(k, -1), lst(k);
    int done = 0;
    vector<map<int, int>> cur(7);
    vector<set<int>> wk(n);
    forn(i, k) forn(j, 7) if (ds[a[i][0]][j])
        ++cur[j][a[i][0]];
    forn(i, k)
        wk[a[i][0]].insert(i);
    for (int d = 1;; ++d){
        if (j < m && h[j] == d){
            ++j;
            continue;
        }
        int wd = (d - 1) % 7;
        vector<int> now, sv;
        for (auto it : cur[wd]) now.push_back(it.first);
        for (int x : now){
            forn(i, 7){
                auto it = cur[i].find(x);
                if (it != cur[i].end()){
                    if (it->second == 1)
                        cur[i].erase(it);
                    else
                        --it->second;
                }
            }
            int y = *wk[x].begin();
            sv.push_back(y);
            wk[x].erase(wk[x].begin());
        }
        forn(i, now.size()){
            int y = sv[i];
            ++lst[y];
            if (lst[y] == int(a[y].size())){
                ans[y] = d;
                ++done;
                continue;
            }
            wk[a[y][lst[y]]].insert(y);
            forn(j, 7) if (ds[a[y][lst[y]]][j])
                ++cur[j][a[y][lst[y]]];
        }
        if (done == k) break;
    }
    forn(i, k) cout << ans[i] << " ";
    cout << endl;
}

1765M - Minimum LCM

Idea: BledDest, preparation: Neon

Tutorial

Solution (Neon)

#include <bits/stdc++.h>

using namespace std;

int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a = 1;
    for (int g = 2; g * g <= n; ++g) {
      if (n % g == 0) {
        a = n / g;
        break;
      }
    }
    cout << a << ' ' << n - a << '\n';
  }
}

1765N - Number Reduction

Idea: Neon, preparation: Neon

Tutorial

Solution (Neon)

#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  
  int t;
  cin >> t;
  while (t--) {
    string x;
    cin >> x;
    int k;
    cin >> k;
    int n = x.size();
    vector<vector<int>> pos(10);
    for (int i = 0; i < n; ++i)
      pos[x[i] - '0'].push_back(i);
    for (int i = 0; i < 10; ++i)
      reverse(pos[i].begin(), pos[i].end());
    string ans;
    int lst = 0, len = n - k;
    for (int i = 0; i < len; ++i) {
      for (int d = (i == 0); d <= 9; ++d) {
        while (!pos[d].empty() && pos[d].back() < lst)
          pos[d].pop_back();
        if (!pos[d].empty() && pos[d].back() - lst <= k) {
          ans += d + '0';
          k -= pos[d].back() - lst;
          lst = pos[d].back() + 1;
          break;
        }
      }
    }
    cout << ans << '\n';
  }
}

Комментарии (27)

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BledDest

3 года назад, скрыть # |

Unfortunately, the editorials for two problems are not ready yet. They will appear here as soon as they're written.

→ Ответить

zerefx

5 месяцев назад, скрыть # ^ |

guess what! it's been more than 3 years and the editorials are still coming soon.

i hope the editorials will be ready by 2030 ;)

bakaasta

NEVER DOUBT BledDest

Ibsan

2 месяца назад, скрыть # ^ |

it been 3 long years

awoo

To all the commenters above, the editorials have existed for a while, and you can see them if you refresh the page enough times. Idk what's causing that, so I attached a pdf of the editorial for all problems to the contest as well.

utility

In problem Torus Path wording "Note that you can't visit all vertices on the antidiagonal (vertices .... at the same time." — is quite confusing.

Jackyjuman

can problem N can be solved by using stack data structures ?

Cypher70

3 года назад, скрыть # ^ |

Yes indeed! You need to check for all 1 <= i <= n-1, if s[i] > s[i+1] then we can delete s[i] to obtain a smaller number. You will continue doing this till k is 0. One special case, when stack's first element is greater than 0 and all the remaining values are 0, when placing a value x > 0 and x < stack.top, if k >= stack size, delete all the elements from the stack and push the current one . P.S — You need vector to do this stack operation. Check my submission

xioachou

nice solution!

vivekananda1001

I liked your implementation! This code indeed requires a good sense of implementation! How did you come up with this?

Have you done similar problems before?

The problem K is easily solvable using DP. But the greedy method is quite tricky, at least for me.

2e5

Can you please explain your solution.. And how do you store which cells you have already visited? Thanks in advance

RANJAN888

In your code,using 'for loop', you have have handled the case of going from last to first within a row.But you have not done anything in case of going from last column to first column,whenever lvl==n if prev!=n-1 , you are just returning INT_MIN.why have you ignored that case??

jonnnnnsinaga

2 года назад, скрыть # ^ |

Do you have some kind of proof why the answer would be moving y-axis wise and not x-axis wise. In ur code, the incrementing lvl parameter is a sifn of u are moving yaxis wise

Taha_adeel

The problem D can be done in O(N) using just two pointers.

AC submission: https://mirror.codeforces.com/contest/1765/submission/186659367

KanishkGoel

Sorting does require O(NlogN) operations, you probably meant O(N) operations after sorting.

spkumar9952

Your solution seems ease to understand. Thank you for sharing!. But TC is O(Nlog(N)) because of sorting.

berrkannt_

Is there anyone to fail my code ? imo my logic is true and working every test cases that i used. Any help appreciated. 191443868

cantaloupe1

Try:

1
906
2

ivan100sic

Thank you BledDest for one of the most beautiful problems I've recently seen (J).

Killer250104

18 месяцев назад, скрыть # ^ |

Can you help me explain why this solution is opimized base on hungarian solution

In problem G, using the exact same idea from the editorial, and going backwards from the end instead of forwards, you end up with $$$1000/1.5 \approx 667$$$ steps on average. Even better, when you get a reply $$$x \gt 2$$$, you can find $$$x$$$ values in one move.

lucius_fox

2 года назад, скрыть # |

"and at least j values that are not less than aj (let's call them Group B)" in the editorial for D, I think instead of $$$a_j$$$ it should $$$a_{n-j+1}$$$

shrimpballs

12 месяцев назад, скрыть # |

the editorial for Problem C, the probability to guess the suit correctly is n-min(a,b,c,d)/4n−(a+b+c+d) or min(a,b,c,d)/4n−(a+b+c+d)?1765C - Card Guessing

vladgrin

5 месяцев назад, скрыть # |

thoughts on problem M — Minimum LCM

$$$LCM(a, b)$$$ reaches its minimum when $$$b$$$ is a multiple of $$$a$$$ (assuming $$$a ≤ b$$$), and then $$$LCM(a, b) = max(a, b) = b$$$.

So if we find the biggest divisor of $$$n$$$ and assign it to $$$a$$$, then $$$b = n - a$$$ will also be divisible by $$$a$$$ and will be the smallest multiple of $$$a$$$ such that $$$a + b = n$$$.

Yeah, I really liked this approach, wasn't able to think in this direction. It was a good problem.

Thanks for the solution <3.

wtf_utkarsh

11 дней назад, скрыть # ^ |

what about prime numbers. How will you check for prime numbers because it is causing tle

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