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Блог пользователя adnan_toky

Codeforces Round #848 (Div. 2) Editorial

Автор adnan_toky, 23 месяца назад, По-английски

Tutorial

Code

#include <bits/stdc++.h>
using namespace std;
 
const int sz = 1e5 + 10;
int ara[sz];
 
int main()
{
    int t;
    scanf("%d", &t);
 
    while(t--) {
        int n;
        scanf("%d", &n);
 
        int sum = 0;
 
        for(int i = 1; i <= n; i++) {
            scanf("%d", &ara[i]);
            sum += ara[i];
        }
 
        int ans = -1e9;
 
        for(int i = 1; i < n; i++) {
            if(ara[i] == ara[i+1]) {
                if(ara[i] == 1) ans = max(ans, sum-4);
                else ans = max(ans, sum+4);
            }
            else
                ans = max(ans, sum);
        }
 
        printf("%d\n", ans);
    }
 
    return 0;
}

1778B - The Forbidden Permutation

Tutorial

Code

#include <bits/stdc++.h>
using namespace std;
 
const int sz = 1e5+10;
int p[sz], a[sz], pos[sz];
 
int main()
{
    int t;
    scanf("%d", &t);
 
    while(t--) {
        int n, m, d;
        scanf("%d %d %d", &n, &m, &d);
 
        for(int i = 1; i <= n; i++) {
            scanf("%d", &p[i]);
            pos[ p[i] ] = i;
        }
 
        for(int i = 1; i <= m; i++) {
            scanf("%d", &a[i]);
        }
 
        int ans = 1e9;
 
        for(int i = 1; i < m; i++) {
            if(pos[a[i+1]] <= pos[a[i]] || pos[a[i+1]]-pos[a[i]] > d) {
                ans = 0;
                break;
            }
 
            ans = min(ans, pos[a[i+1]] - pos[a[i]]);
 
            int dist = pos[a[i+1]]-pos[a[i]];
 
            int swapNeed = d-dist+1;
            int swapPossible = (pos[a[i]]-1) + (n - pos[a[i+1]]);
            if(swapPossible >= swapNeed) ans = min(ans, swapNeed);
        }
 
        printf("%d\n", ans);
    }
 
    return 0;
}

1778C - Flexible String

Tutorial

Code

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
#define EL '\n'
#define fastio std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
string a, b;
string char_list;
bool mark[26];
ll ans, k;
 
ll count_matching_pair()
{
    ll tot_pair = 0, match_count = 0;
 
    for(ll i = 0; i < a.size(); i++) {
        if(a[i] == b[i] || mark[ a[i]-'a' ])
            match_count++;
        else {
            tot_pair += match_count*(match_count+1)/2;
            match_count = 0;
        }
    }
    tot_pair += match_count*(match_count+1)/2;
 
    return tot_pair;
}
 
void solve(ll pos, ll cnt)
{
    if(cnt > k) return;
 
    if(pos == char_list.size()) {
        if(cnt == k) ans = max(ans, count_matching_pair());
        return;
    }
 
    solve(pos+1, cnt);
 
    mark[ char_list[pos]-'a' ] = 1;
    solve(pos+1, cnt+1);
    mark[ char_list[pos]-'a' ] = 0;
}
 
int main()
{
    fastio;
    ll t;
    cin >> t;
 
    while(t--) {
        ll n; cin >> n >> k;
        cin >> a >> b;
 
        unordered_set <char> unq;
        for(auto &ch : a) unq.insert(ch);
 
        char_list.clear();
        for(auto &x : unq) char_list.pb(x);
 
        k = min(k, (ll)unq.size());
        memset(mark, 0, sizeof mark);
        ans = 0;
        solve(0, 0);
 
        cout << ans << EL;
    }
 
    return 0;
}

1778D - Flexible String Revisit

Tutorial

1778D - Flexible String Revisit

Let $$$k$$$ be the number of indices where the characters between two strings are different and $$$f(x)$$$ be the expected number of moves to make two strings equal given that the strings have $$$x$$$ differences. We have to find the value of $$$f(k)$$$.

For all $$$x$$$ where $$$1 \leq x \leq n-1$$$,

$$$ \begin{align} f(x) = \frac{x}{n} \cdot [1 + f(x-1)] + \frac{n-x}{n} \cdot [1 + f(x+1)]\\ or,~~~ f(x) = 1 + \frac{x}{n}\cdot f(x-1) + \frac{n-x}{n} \cdot f(x+1) \end{align} $$$

Now, $$$f(0) = 0$$$ and $$$f(1) = 1 + \frac{n-1}{n}f(2)$$$.

We can represent any $$$f(i)$$$ in the form $$$f(i) = a_i + b_i \cdot f(i+1)$$$.

Let, $$$a_{1} = 1$$$ and $$$b_{1} = \frac{n-1}{n}$$$. So, we can write $$$f(1) = a_{1} + b_{1}\cdot f(2)$$$.

When $$$1 \lt i \lt n$$$, $$$f(i) = 1 + \frac{i}{n} \cdot f(i-1) + \frac{n-i}{n} \cdot f(i+1)$$$. We can substitute the value of $$$f(i-1)$$$ with $$$a_{i-1} + b_{i-1}\cdot f(i)$$$ and calculate the value of $$$f(i)$$$. Thus we can get the value of $$$a_i$$$ and $$$b_i$$$ using the value of $$$a_{i-1}$$$ and $$$b_{i-1}$$$ by considering $$$a_1$$$ and $$$b_1$$$ as the base case.

We get, $$$a_{i} = \frac{n+i\cdot a_{i-1}}{n-i\cdot b_{i-1}}$$$ and $$$b_{i} = \frac{n-i}{n-i \cdot b_{i-1}}$$$ for $$$2 \leq i \leq n$$$.

Calculation

Similarly, $$$f(n) = 1+f(n-1)$$$.

We can represent any $$$f(i)$$$ in the form $$$f(i) = c_i + d_i \cdot f(i-1)$$$.

Let, $$$c_{n} = 1$$$ and $$$d_{n} = 1$$$. So, we can write $$$f(n) = c_{n} + d_{n}\cdot f(n-1)$$$.

When $$$1 \lt i \lt n$$$, $$$f(i) = 1 + \frac{i}{n} \cdot f(i-1) + \frac{n-i}{n} \cdot f(i+1)$$$. We can substitute the value of $$$f(i+1)$$$ with $$$c_{i+1} + d_{i+1}\cdot f(i)$$$ and calculate the value of $$$f(i)$$$. Thus we can get the value of $$$c_i$$$ and $$$d_i$$$ using the value of $$$c_{i+1}$$$ and $$$d_{i+1}$$$ by considering $$$c_n$$$ and $$$d_n$$$ as the base case.

We get, $$$c_{i} = \frac{n+(n-i)\cdot c_{i+1}}{n-(n-i)\cdot d_{i+1}}$$$ and $$$d_{i} = \frac{i}{n-(n-i) \cdot d_{i+1}}$$$.

Calculation

Now, $$$f(i) = c_i + d_i \cdot f(i-1)$$$ and $$$f(i-1) = a_{i-1} + b_{i-1} \cdot f(i)$$$. By solving these two equations, we find that $$$f(i) = \frac{c_i+d_i \cdot a_{i-1}}{1-d_i \cdot b_{i-1}}$$$.

Time Complexity: $$$\mathcal{O}(n\cdot \log m)$$$.

Alternative Solution

Code

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

#define mod 998244353 
#define N 1000005
 
template<int MOD>
struct ModInt {
  unsigned x;
  ModInt() : x(0) { }
  ModInt(signed sig) : x(sig) {  }
  ModInt(signed long long sig) : x(sig%MOD) { }
  int get() const { return (int)x; }
  ModInt pow(ll p) { ModInt res = 1, a = *this; while (p) { if (p & 1) res *= a; a *= a; p >>= 1; } return res; }
 
  ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
  ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
  ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
  ModInt &operator/=(ModInt that) { return (*this) *= that.pow(MOD - 2); }
 
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
  ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
  bool operator<(ModInt that) const { return x < that.x; }
  friend ostream& operator<<(ostream &os, ModInt a) { os << a.x; return os; }
};
typedef ModInt<998244353> mint;
 
mint a[N], b[N], c[N], d[N];
 
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
 
        string s1, s2;
        cin >> s1 >> s2;
 
        int diff = 0;
        for(int i = 0; i < n; i++){
            diff += s1[i]!=s2[i];
        }
 
        c[n] = 1, d[n] = 1, a[1] = 1, b[1] = ((mint) n-1)/n;
 
        for(int i = 2; i <= n; i++){
            a[i] = ((mint) n + a[i-1]*i)/((mint) n - b[i-1]*i);
            b[i] = ((mint) n-i)/((mint) n - b[i-1]*i);
        }
 
        for(int i = n-1; i >= 1; i--){
            c[i] = ((mint) n + c[i+1]*(n-i))/((mint) n - d[i+1]*(n-i));
            d[i] = (mint) i / ((mint) n - d[i+1]*(n-i));
        }
 
        mint ans = (c[diff]+d[diff]*a[diff-1])/((mint) 1 - d[diff]*b[diff-1]);
 
        cout << ans << endl;
    }
 
    return 0;
}

Alternative Code

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
 
template<int MOD>
struct ModInt {
  unsigned x;
  ModInt() : x(0) { }
  ModInt(signed sig) : x(sig) {  }
  ModInt(signed long long sig) : x(sig%MOD) { }
  int get() const { return (int)x; }
  ModInt pow(ll p) { ModInt res = 1, a = *this; while (p) { if (p & 1) res *= a; a *= a; p >>= 1; } return res; }
 
  ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
  ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
  ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
  ModInt &operator/=(ModInt that) { return (*this) *= that.pow(MOD - 2); }
 
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
  ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
  bool operator<(ModInt that) const { return x < that.x; }
  friend ostream& operator<<(ostream &os, ModInt a) { os << a.x; return os; }
};
typedef ModInt<998244353> mint;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
 
    mint two = 2;
 
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
 
        string a, b;
        cin >> a >> b;
 
        int cnt = 0;
        for(int i = 0; i < n; i++){
            if(a[i]!=b[i]) cnt++;
        }
 
        vector<mint> dp(n+2);
        dp[0] = two.pow(n);
        dp[1] = dp[0]-1;
        dp[0] = 0;
 
        for(long long i = 2; i < n; i++){
            mint x = i-1;
            x /= n;
            dp[i] = (dp[i-1]-x*dp[i-2]-1)*n/(n-i+1);
        }
 
        dp[n] = dp[n-1]+1;
 
        cout << dp[cnt] << '\n';
    }
 
    return 0;
}

1778E - The Tree Has Fallen!

Tutorial

Code

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
#define nn '\n'
#define fastio std::ios_base::sync_with_stdio(false); cin.tie(NULL);
 
const int sz = 2e5 + 10, d = 30;
vector <int> g[sz], Tree[sz];
int a[sz], discover_time[sz], finish_time[sz], nodeOf[sz], tim;
 
struct BASIS {
    int basis[d];
    int sz;
 
    void init() {
        for(int i = 0; i < d; i++) basis[i] = 0;
        sz = 0;
    }
 
    void insertVector(int mask) {
        for (int i = d-1; i >= 0; i--) {
            if (((mask>>i)&1) == 0) continue;
 
            if (!basis[i]) {
                basis[i] = mask;
                ++sz;
                return;
            }
            mask ^= basis[i];
        }
    }
 
    void mergeBasis(const BASIS &from) {
        for(int i = d-1; i >= 0; i--) {
            if(!from.basis[i])
                continue;
 
            insertVector(from.basis[i]);
        }
    }
 
    int findMax()  {
        int ret = 0;
        for(int i = d-1; i >= 0; i--) {
            if(!basis[i] || (ret>>i & 1))
                continue;
 
            ret ^= basis[i];
        }
        return ret;
    }
} in[sz], out, pre[sz], suf[sz];
 
void in_dfs(int u, int p)
{
    in[u].insertVector(a[u]);
    discover_time[u] = ++tim;
    nodeOf[tim] = u;
 
    for(auto &v : g[u]) {
        if(v == p)
            continue;
 
        Tree[u].pb(v);
        in_dfs(v, u);
 
        in[u].mergeBasis(in[v]);
    }
    finish_time[u] = tim;
}
 
inline bool in_subtree(int sub_root, int v)
{
    return discover_time[sub_root] <= discover_time[v]
            && finish_time[sub_root] >= finish_time[v];
}
 
int findChildOnPath(int sub_root, int v)
{
    int lo = 0, hi = (int)Tree[sub_root].size()-1;
 
    while(lo <= hi) {
        int mid = lo+hi>>1, node = Tree[sub_root][mid];
 
        if(finish_time[node] < discover_time[v])
            lo = mid + 1;
        else if(discover_time[node] > discover_time[v])
            hi= mid - 1;
        else
            return node;
    }
}
 
void init(int n) {
 
    for(int i = 0; i <= n+5; i++) {
        g[i].clear(), Tree[i].clear();
        in[i].init();
        pre[i].init(), suf[i].init();
    }
    tim = 0;
}
 
int main()
{
    fastio;
 
    int t;
    cin >> t;
 
    while(t--) {
        int n; cin >> n;
 
        init(n);
 
        for(int i = 1; i <= n; i++) cin >> a[i];
 
        for(int i = 1; i < n; i++) {
            int u, v;
            cin >> u >> v;
            g[u].pb(v); g[v].pb(u);
        }
 
        in_dfs(1, -1);
 
        for(int i = 1; i <= n; i++) {
            pre[i].insertVector(a[ nodeOf[i] ]);
            pre[i].mergeBasis(pre[i-1]);
        }
 
        for(int i = n; i >= 1; i--) {
            suf[i].insertVector(a[ nodeOf[i] ]);
            suf[i].mergeBasis(suf[i+1]);
        }
 
        int q; cin >> q;
 
        while(q--) {
            int root, v;
            cin >> root >> v;
 
            if(root == v) {
                cout << in[1].findMax() << nn;
            }
            else if(in_subtree(v, root)) {
                int child = findChildOnPath(v, root);
 
                out.init();
                out.mergeBasis(pre[discover_time[child]-1]);
                out.mergeBasis(suf[finish_time[child]+1]);
                cout << out.findMax() << nn;
            }
            else
                cout << in[v].findMax() << nn;
        }
    }
 
    return 0;
}

1778F - Maximizing Root

Tutorial

Code

#include <bits/stdc++.h>
using namespace std;
 
const int N = 100005;
const int mod = 998244353;
 
int val[N];
vector<int> g[N];
vector<int> divisor[N];
int subtree_gcd[N], par[N];
 
int dp[N][1003];
int gcdd[1003][1003];
 
inline long long ___gcd(long long a, long long b){
    if(gcdd[a][b]) return gcdd[a][b];
 
    return gcdd[a][b] = __gcd(a, b);
}
 
inline long long lcm(long long a, long long b){
    return (a/___gcd(a, b))*b;
}
 
void dfs(int u, int p){
    par[u] = p;
    subtree_gcd[u] = val[u];
    for(int v: g[u]){
        if(v==p) continue;
        dfs(v, u);
        subtree_gcd[u] = ___gcd(subtree_gcd[u], subtree_gcd[v]);
    }
}
 
 
int solve(int u, int d, int p){
    if(subtree_gcd[u]%d==0) return 0;
    if((val[u]*val[u])%d) return (1<<30);
 
 
    if(dp[u][d]!=-1) return dp[u][d];
    
    long long req = d/___gcd(d, subtree_gcd[u]);
 
    long long res = (1<<30);
    for(int div: divisor[val[u]]){
        if((val[u]*div)%d==0 && d%div==0){
            long long r = 1;
            for(int v: g[u]){
                if(v==p) continue;
                r += solve(v, lcm(d/div, div), u);
            }
            res = min(res, r);
        }
    }
 
    return dp[u][d] = min(res, (1LL<<30));
}
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    for(int i = 2; i < 1001; i++){
        for(int j = i; j < 1001; j+=i){
            divisor[j].push_back(i);
        }
    }
 
    int t;
    cin >> t;
 
    while(t--){
        int n, k;
        cin >> n >> k;
 
        for(int i = 0; i <= n; i++){
            g[i].clear();
        }
 
        for(int i = 1; i <= n; i++){
            cin >> val[i];
        }
        
        for(int i = 0; i <= n; i++){
            for(int d: divisor[val[1]]){
                dp[i][d] = -1;
            }
        }
 
        for(int i = 0; i < n-1; i++){
            int u, v;
            cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
 
        dfs(1, 0);
 
        int ans = val[1];
 
        for(int d: divisor[val[1]]){
            int req = 0;
            int f = 1;
            for(int v: g[1]){
                int x = solve(v, d, par[v]);
                if(x>n) f = 0;
                req += x;
            }
 
            if(!f) continue;
 
            req++;
            if(req<=k){
                ans = max(ans, val[1]*d);
            }
        }
 
        cout << ans << "\n";
    }
    
    return 0;
}

Полный текст и комментарии »

Разбор задач Codeforces Round 848 (Div. 2)

+129

adnan_toky
23 месяца назад
144

Codeforces plagiarism checker should be smarter!

Автор adnan_toky, 3 года назад, По-английски

I am very frustrated after getting another unexpected warning from codeforces after the last educational round. I have already described the issue previously in this blog Unexpected warnings from CodeForces. It's disturbing to write over the same issue repeatedly. But I must write again this time because I did not receive any response from codeforces after writing the previous blog and commenting multiple times. It's very disappointing for me as a member of this community.

My submissions of this round skipped getting an unexpected warning that says my submission to problem C (1555C - Coin Rows) 124303153 coincides with the submission 124301236 by NZB.

The solution to problem C was very simple. We are given a grid of two rows as input and we have to calculate the answer from the sum of the suffix of the first row and the prefix of the second row. My solution was like this:

int n;
cin >> n;
 
ll x[2][n+1];
 
memset(x, 0, sizeof x);

// input for the first row
for (int i = 1; i <= n; i++){ 
    cin >> x[0][i];
    x[0][i] += x[0][i - 1]; // calculating prefix sum
}

// input for the second row
for (int i = 1; i <= n; i++){
    cin >> x[1][i];
    x[1][i] += x[1][i - 1]; // calculating prefix sum
}

ll mn = LL_INF;

// calculating the answer
for (int i = 1; i <= n; i++){
    mn = min(mn, max(x[0][n] - x[0][i], x[1][i - 1]));
}

Isn't it the simplest approach to solve the problem? More than 6000 people solved this problem during the contest. So there is a high possibility to coincide two submissions for a problem having this kind of simplest approach. Although the submission that coincides with my submission is different from many aspects except the approach of implementation. Codeforces plagiarism checker should be smarter to judge this kind of sensitive situation.

Here I'm attaching the submissions.

My Submission

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

#define ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pb push_back
#define ff first
#define ss second
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define nn "\n"
#define sci(x) scanf("%d", &x)
#define LL_INF (1LL << 62)
#define INF (1 << 30)
#define SetBit(x, k) (x |= (1LL << k))
#define ClearBit(x, k) (x &= ~(1LL << k))
#define CheckBit(x, k) (x & (1LL << k))
#define mod 1000000007
#define N 300005

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

    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;

        ll x[2][n+1];

        memset(x, 0, sizeof x);

        for(int i = 1; i <= n; i++){
            cin >> x[0][i];
            x[0][i] += x[0][i-1];
        }

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

        ll mn = LL_INF;

        for(int i = 1; i <= n; i++){
            mn = min(mn, max(x[0][n]-x[0][i], x[1][i-1]));
        }

        cout << mn << nn;
    
    }
    
    return 0;
}

NZB's Submission

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>

typedef long double ld;
typedef long long ll;
const int inf = 1e9 + 7;
const int mod = 1e9 + 7;
const int MAX_N = 2e5 + 7;

using namespace std;

int sum[2][MAX_N];

int main()
{
    int T;
	cin>>T;
	while(T--) {
		int N;
		cin>>N;
		for(int i=1;i<=N;i++) {
			cin>>sum[0][i];
			sum[0][i] += sum[0][i-1];
		}
		for(int i=1;i<=N;i++) {
			cin>>sum[1][i];
			sum[1][i] += sum[1][i-1];
		}
		int ans = inf;
		for(int i=1;i<=N;i++) {
			ans = min(ans, max(sum[0][N]-sum[0][i], sum[1][i-1]));
		}
		cout<<ans<<endl;
	}
}

If you want to downvote the post, please leave a comment before downvoting explaining how should we approach simple implementations like this and what would be your feeling after being convicted as a cheater twice just for the simplicity of your implementation? Isn't it heartbreaking and unexpected from such a reputed platform like codeforces?

I hope MikeMirzayanov will handle the situation gently this time and roll back the situation.

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+126

adnan_toky
3 года назад
6

List of Problems Having Short Statements

Автор adnan_toky, история, 4 года назад, По-английски

Who doesn't like short statements? (or maybe just me!)

Sometimes I get bored reading long statements while practicing and search for the problems with short statements. So I filtered such problems and ordered them by their statement size. I thought of sharing the list with everyone. Hope it may help the lazy programmers like me!

Problems (Categorized by Difficulty):

1700

1800

1900

2000

2100

2200

2300

2400

Remember that a short statement doesn't mean a simple and clear statement.

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+289

adnan_toky
4 года назад
32

Unexpected warnings from CodeForces

Автор adnan_toky, 4 года назад, По-английски

In the last round, I have received two unexpected warnings from CodeForces plagiarism checker. The first one says, my solution 105930169 for the problem 1476D - Journey significantly coincides with solutions bleh0.5/105914875. After sometimes I got another warning that my solution to the same problem (1476D) significantly coincides with solutions bleh0.5/105914875, tejas10p/105917332.

My approach for problem D was quite straight forward. Just three operations in three different loops. Whoever solved the problem using this idea, it is quite possible to match their solutions. I have explored the submissions of the first 100 people according to the common standings of this round. Among these 100 submissions, I found 7 submissions that are also quite similar to my solution. It does not mean that we all shared our code among us, it is just that our approaches were same. I'm mentioning these 7 submissions here.

My submission:

105930169

The two submissions that matched with my submission:

Similar submissions:

105883915 — Here only difference with my solution is I used array and he used vector.
105878527 — He used vector and wrote the solution in a separate function.

Other similar submissions:

I already commented on the post but nothing changed. So I am here annoying you again. From the day (16 Feb, 2020) I did my first contest here, i never missed a single round regardless of how busy i was. I love my cf profile more than anything and just can't let a scar stay on my profile without any wrongdoing. -_-

I am requesting MikeMirzayanov & awoo to look into these unexpected warnings.

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others

adnan_toky
4 года назад
7