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IBM Oa Help | Oa Exam Helper

IBM Oa Help | Oa Exam Helper

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We are here to clear All types of Exams Admin : @Codercpp001 (aka) KMK ✅ INTERVIEW HELP AVAILABLE 1-Coding Round 2-Aptitude and Reasoning Round 3-Communication round 4-Resume building 🎉Job updates will be posted here.

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Good set

closest k origin

#include <bits/stdc++.h>
#define int long long
using namespace std;
#define ll long long
vector<vector<ll>> solve(ll n,ll k,vector<ll>&a,vector<ll>&b) 
{
    priority_queue<pair<double,pair<ll,ll>>> pq;
    for(int i=0;i<n;i++)
    {
        double x=a[i];
        double y=b[i];
        double dis=sqrt(x+y);
        pq.push({dis,{x,y}});
        if(pq.size()>k)  pq.pop();
    }
    vector<vector<ll>>ans;
    while(!pq.empty())
    {
        ans.push_back({pq.top().second.first,pq.top().second.second});
        pq.pop();
    }
    sort(begin(ans),end(ans));
    return ans;
}
signed main() 
{        
        ll n,k; cin>>n>>k;
        vector<ll>a(n),b(n);
        for(ll i=0;i<n;i++) cin>>a[i];
        for(ll i=0;i<n;i++) cin>>b[i];
        vector<vector<ll>>ans=solve(n,k,a,b);
        for(auto it:ans) cout<<it[0]<<" "<<it[1]<<endl;
            
    return 0;
}

#include <iostream> #include <vector> using namespace std; vector<bool> sieve(int max_val) {     vector<bool> is_prime(max_val + 1, true);     is_prime[0] = is_prime[1] = false;     for (int i = 2; i * i <= max_val; ++i) {         if (is_prime[i]) {             for (int j = i * i; j <= max_val; j += i) {                 is_prime[j] = false;             }         }     }     return is_prime; } int main() {     int N;     cin >> N;     vector<int> A(N);     int max_val = 0;     for (int i = 0; i < N; ++i) {         cin >> A[i];         if (A[i] > max_val) {             max_val = A[i];         }     }     vector<bool> is_prime = sieve(max_val);     int prime_count = 0, composite_count = 0;     for (int i = 0; i < N; ++i) {         if (is_prime[A[i]]) {             prime_count++;         } else {             composite_count++;         }     }     int good_pairs = prime_count * composite_count;     cout << good_pairs << endl;     return 0; }.   find good pairs in array Infosys ✅

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IF yes then 200 subscribers

NEED AMPHIBIAN ESCAPE?
Anonymous voting

RED array

#include <bits/stdc++.h> using namespace std; #define int long long int32_t main() { // your code goes here int t; cin>>t; while(t--){ int a; cin>>a; int ans=INT_MAX; vector<int>v(a); unordered_map<int,int>mp; for(int i=0;i<a;i++){ cin>>v[i]; mp[v[i]]++; } for(auto x:mp){ int r=a-x.second; int l=x.first; l*=r; ans=min(ans,l); } ans=min(ans,a); cout<<ans<<endl; } }

void dfs(int node, int parent, const vector>>& graph,          const vector& minActivity, vector& activity, vector& vulnerable) {     activity[node] = 0; // initial activity is 0     for (const auto& edge : graph[node]) {         int next_node = edge.first;         int weight = edge.second;                 if (next_node != parent) {             dfs(next_node, node, graph, minActivity, activity, vulnerable);             activity[node] += activity[next_node] + weight;         }     }     if (activity[node] > minActivity[node]) {         vulnerable[node] = true;     } } int getMinServers(int server_nodes, int server_edges, const vector& server_from,                   const vector& server_to, const vector& server_weight,                   int minActivity_count, const vector& minActivity) {     vector>> graph(server_nodes + 1);     for (int i = 0; i < server_edges; ++i) {         graph[server_from[i]].emplace_back(server_to[i], server_weight[i]);         graph[server_to[i]].emplace_back(server_from[i], -server_weight[i]);     }     vector activity(server_nodes + 1, 0);     vector vulnerable(server_nodes + 1, false);     dfs(1, -1, graph, minActivity, activity, vulnerable);     int vulnerableCount = count(vulnerable.begin(), vulnerable.end(), true);     return vulnerableCount; }

b

Wants lucky matrix
Anonymous voting

Candle bunch

Cross number puzzle
Cross number puzzle

Resource power
Resource power

def Resource(A, B, C):     same_type_systems = (A // 3) + (B // 3) + (C // 3)     remaining_A = A % 3     remaining_B = B % 3     remaining_C = C % 3     different_type_systems = min(A, B, C)     remaining_A -= different_type_systems     remaining_B -= different_type_systems     remaining_C -= different_type_systems     remaining_A = max(0, remaining_A)     remaining_B = max(0, remaining_B)     remaining_C = max(0, remaining_C)     total_systems = same_type_systems + different_type_systems     leftover_resources = remaining_A + remaining_B + remaining_C     total_systems += leftover_resources // 3         return total_systems Resource power ✅