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Max and Electrical Panel

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Max and Electrical Panel

 Solution 01:


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

int main()
{
    int n,c;
    cin >> n >>c;
    int lo = 0, hi = n, res = 0;
    while(hi-lo > 1){
        if(res)cout << 2<<endl<<flush;
        int mid = lo + (hi-lo-1)/8+1;
        cout << 1 << " "<<mid<<endl<<flush;
        cin >> res;
        if(res)hi = mid;
        else lo = mid;
    }
    cout << 3 << " "<<hi<<endl<<flush;
    return 0;
}

Solution 02:



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

int main()
{
    int n,c;
    cin >> n >>c;
    int lo = 1, hi = n+1, res = 0;
    while(hi-lo > 1){
        if(res)cout << 2<<endl<<flush;
        int mid = lo + (hi-lo)/60;
        cout << 1 << " "<<mid<<endl<<flush;
        cin >> res;
        if(res)hi = mid+1;
        else lo = mid+1;
    }
    cout << 3 << " "<<lo<<endl<<flush;
    return 0;
}

Solution 03:



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

bool solve(int x)
{
    cout << "1 "<<x<<endl<<flush;
    int a;
    cin >> a;
    return a;
}

int main()
{
    int n,c;
    cin >> n >>c;
    int ans = -1, lg = 0, le = -1;
    int range = sqrt(n);
    //cout << range<<endl;
    for(int i = 1; i <= n; i += range){
           // cout << i << endl;
        if(solve(i)){
            ans = i;
            break;
        }
        lg = i;
        le = min(i+range-1, n);
    }
    cout << 2<<endl<<flush;
    for(int i = lg; i <= le; i++){
        if(solve(i)){
            ans = i;
            break;
        }
    }
    cout << 3 << " "<<ans<<endl<<flush;
    return 0;
}

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315 - Network

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315 - Network

Topic : Graph Theory(AP)

Solution 01:



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

#define Max 100000
vector<int> graph[Max];
int parent[Max];
int low[Max];
int d[Max];
int visited[Max];
bool isArticulationPoint[Max];
int Time = 0;

void dfs(int u, int root)
{
    Time = Time + 1;
    visited[u] = Time;
    d[u] = low[u] = Time;
    int noOfChildren = 0;
    for(int i = 0; i <graph[u].size(); i++){
        int v = graph[u][i];
        if(v == parent[u])continue;
        parent[v] = u;
        if(visited[v]) low[u] = min(low[u], d[v]);
        else{
            noOfChildren = noOfChildren + 1;
            dfs(v, root);
            low[u] = min(low[u], low[v]);
            if(low[v] >= d[u] and u != root)isArticulationPoint[u] = true;
        }
    }
    if(u == root and noOfChildren > 1)isArticulationPoint[u] = true;
}

int main()
{
    int n,x,y;
    char c;
    while(cin >> n && n)
    {
        Time = 0;
        for(int t = 0; t <= n; t++){
            graph[t].clear();
            parent[t]= low[t] = d[t] = visited[t] = 0;
            isArticulationPoint[t] = false;
        }
        while(scanf("%d", &x) == 1 && x)
        {
            while(scanf("%d%c", &y, &c) == 2)
            {
                graph[x].push_back(y);
                    graph[y].push_back(x);
                if(c == '\n') break;
            }
        }
        dfs(1, 1);
        cout << count(isArticulationPoint+1, isArticulationPoint+n+1, true)<<endl;
    }
    return 0;
}

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Articulation Points and Bridges

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Resource:

বাইকানেক্টেড কম্পোনেন্ট , ব্রিজ, আরটিকুলেশন পয়েন্ট [ থিওরি ]

গ্রাফ থিওরিতে হাতেখড়ি ১৩: আর্টিকুলেশন পয়েন্ট এবং ব্রিজ

Visualization

CF blog 

Finding bridges in a graph in O(N+M)

Finding articulation points in a graph in O(N+M)

Articulation Points:01


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

#define Max 100000
vector<int> graph[Max];
int parent[Max];
int low[Max];
int d[Max];
int visited[Max];
bool isArticulationPoint[Max];
int Time = 0;

void dfs(int u, int root)
{
    Time = Time + 1;
    visited[u] = Time;
    d[u] = low[u] = Time;
    int noOfChildren = 0;
    for(int i = 0; i <graph[u].size(); i++){
        int v = graph[u][i];
        if(v == parent[u])continue;
        parent[v] = u;
        if(visited[v]) low[u] = min(low[u], d[v]);
        else{
            noOfChildren = noOfChildren + 1;
            dfs(v, root);
            low[u] = min(low[u], low[v]);
            if(low[v] >= d[u] and u != root)isArticulationPoint[u] = true;
        }
    }
    if(u == root and noOfChildren > 1)isArticulationPoint[u] = true;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  //  freopen("in.txt", "r", stdin);
    int n,e;
    cin >> n >> e;
    for(int i = 0; i < e; i++){
        int u,v;
        cin >> u >> v;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }
    dfs(1,1);
    for(int i = 1; i <= n; i++){
        if(isArticulationPoint[i])cout << i<<endl;
    }
    return 0;
}

Bridge 01:



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

#define Max 100000
vector<int> graph[Max];
vector<pair<int, int> > Bridge;
int parent[Max];
int low[Max];
int d[Max];
int visited[Max];
bool isArticulationPoint[Max];
int Time = 0;

void dfs(int u, int root)
{
    int v;
    Time = Time + 1;
    visited[u] = Time;
    d[u] = low[u] = Time;
    int noOfChildren = 0;
    for(int i = 0; i <graph[u].size(); i++){
         v = graph[u][i];
        if(v == parent[u])continue;
        parent[v] = u;
        if(visited[v]) low[u] = min(low[u], d[v]);
        else{
            noOfChildren = noOfChildren + 1;
            dfs(v, root);
            low[u] = min(low[u], low[v]);
            if(low[v] > d[u])
            {
                isArticulationPoint[u] = true;
             //    cout << u << "-------"<<v<<endl;
                 Bridge.push_back({u,v});
            }
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("in.txt", "r", stdin);
    int n,e;
    cin >> n >> e;
    for(int i = 0; i < e; i++){
        int u,v;
        cin >> u >> v;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }
    dfs(1,1);
    for(int i = 0; i < Bridge.size(); i++){
        cout << Bridge[i].first << " "<<Bridge[i].second<<endl;
    }
    return 0;
}

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11456 - Trainsorting

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11456 - Trainsorting

Solution:01



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#include<bits/stdc++.h>
using namespace std;
#define Max 2005

int main()
{
    int a[Max], x[Max], y[Max];
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        for(int i = 0; i < n; i++){
            cin >> a[i];
        }
        int ans = 0;
        for(int i = n-1; i >=0 ; i--){
            x[i] = 1;
            y[i] = 1;
            for(int k = i+1; k < n; k++){
                if(a[i] < a[k]){
                    x[i] = max(x[k]+1, x[i]);
                }
                if(a[i] > a[k]){
                    y[i] = max(y[k]+1, y[i]);
                }
            }
            ans = max(ans, x[i]+y[i]-1);
        }
        cout << ans << endl;
    }
    return 0;
}

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HTML CODE FOR COPY TEXT

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CODEBOX
        

<div class="main">
<div class="container">
<div class="codebox">
<div class="ipsCode_citation">
<div style="float: right;">
<button onclick="copyText()">//copy</button>
        </div>
//DELETE THIS LINE AND PASTE HERE WHICH CODE YOU WANT TO COPY ONE CLICK
</div>
</div>
</div>
</div>
<br />
<div id="output">
</div>
<script>
  function copyText(){
    var outputText = "";
    var targets = document.getElementsByClassName('codebox');
    for (var i = 0; i < targets.length; i++) {
      outputText += targets[i].innerText;
    }
    var output = document.getElementById('output');
    output.innerText = outputText;
    var range = document.createRange();
    range.selectNodeContents(output);
    var selection = window.getSelection();
    selection.removeAllRanges();
    selection.addRange(range);
    document.execCommand('copy');
    output.style.display = 'none';
  }
</script>

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RESOURCE

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1225 - Palindromic Numbers (II)

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1225 - Palindromic Numbers (II)

Solution:

 


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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll Max = 100005;

void palindrome(ll n, int cs)
{
  ll test = n;
  ll ans = 0;
  while(n)
  {
     ans = ans*10 + n%10;
    n /= 10;
  }
  cout << "Case "<<cs<<": ";
  if(test == ans)cout << "Yes\n";
  else cout << "No\n";
}

int main()
{
    ///*
  #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  #endif
    //*/
    int t;
    cin >> t;
    for(int cs = 1; cs <= t; cs++){
      ll n;
    cin >> n;
    palindrome(n, cs);
    }

  //double end_time = clock();
  //printf( "Time = %lf ms\n", ( (end_time - start_time) / CLOCKS_PER_SEC)*1000);

    return 0;
}
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E. Divisor Count of Odd Parity

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E. Divisor Count of Odd Parity

Topic: Divisor + parity

Solution: 

all possible solution at a time.



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 #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
map<ll, ll>M;

bool findParity(ll n)
{
    ll cnt = 0;
    while(n){
        n = n & (n - 1);
        cnt++;
    }
    if(cnt & 1)return true;
    else return false;
}

bool _findParity(ll x)
{
    ll y = x ^ (x >> 1);
    y = y ^ (y >> 2);
    y = y ^ (y >> 4);
    y = y ^ (y >> 8);
    y = y ^ (y >> 16);
    y = y ^ (y >> 32);
 
    // Rightmost bit of y holds the parity value
    // if (y&1) is 1 then parity is odd else even
    if (y & 1)
        return 1;
    return 0;
}

bool findParity_loop(ll n)
{
    ll cnt = __builtin_popcountll(n);
    if(cnt & 1)return true;
    else return false;
}



  ll Divisors(ll n)/// while loop diye
{
    ll cnt = 0;
    for (ll i=1; i<=sqrt(n); i++)
    {
        if (n%i==0)
        {
            if (n/i == i)
            {
               // if(__builtin_parity(i))
                    if(findParity(i))
                {
                    cnt++;
                    M[i]++;
                    //printf("%d ", i);
                }
            }
            else
            {
                //if(__builtin_parity(i))
                    if(findParity(i))
                {
                    cnt++;
                    M[i]++;
                    //rintf("%d ", i);
                }
                //if(__builtin_parity(n/i))
                if(findParity(n/i))
                {
                    cnt++;
                    M[n/i]++;
                    //printf("%d ", n/i);
                }
            }
        }
    }
    return cnt;
}
   ll divisors(ll n)/// builtin function diye
{
    ll cnt = 0;
    for (ll i=1; i<=sqrt(n); i++)
    {
        if (n%i==0)
        {
            if (n/i == i)
            {
               if(__builtin_parityll(i))
               //     if(findParity(i))
                {
                    cnt++;
                    M[i]++;
                    //printf("%d ", i);
                }
            }
            else
            {
                if(__builtin_parityll(i))
                    //if(findParity(i))
                {
                    cnt++;
                    M[i]++;
                   //printf("%d ", i);
                }
                if(__builtin_parityll(n/i))
               // if(findParity(n/i))
                {
                    cnt++;
                    M[n/i]++;
                   // printf("%d ", n/i);
                }
            }
        }
    }
    return cnt;
}

  ll _Divisors(ll n)
{
    ll cnt = 0;
    for (ll i=1; i<=sqrt(n); i++)
    {
        if (n%i==0)
        {
            if (n/i == i)
            {
               //if(__builtin_parity(i))
                    if(_findParity(i))
                {
                    cnt++;
                    M[i]++;
                    //printf("%d ", i);
                }
            }
            else
            {
                //if(__builtin_parity(i))
                    if(_findParity(i))
                {
                    cnt++;
                    M[i]++;
                   //printf("%d ", i);
                }
               // if(__builtin_parity(n/i))
                if(_findParity(n/i))
                {
                    cnt++;
                    M[n/i]++;
                   // printf("%d ", n/i);
                }
            }
        }
    }
    return cnt;
}
 ll Divisors_loop(ll n)
{
    ll cnt = 0;
    for (ll i=1; i<=sqrt(n); i++)
    {
        if (n%i==0)
        {
            if (n/i == i)
            {
               // if(__builtin_parity(i))
                    if(findParity_loop(i))
                {
                    cnt++;
                    M[i]++;
                    //printf("%d ", i);
                }
            }
            else
            {
                //if(__builtin_parity(i))
                    if(findParity_loop(i))
                {
                    cnt++;
                    M[i]++;
                    //rintf("%d ", i);
                }
                //if(__builtin_parity(n/i))
                if(findParity_loop(n/i))
                {
                    cnt++;
                    M[n/i]++;
                    //printf("%d ", n/i);
                }
            }
        }
    }
    return cnt;
}



int  main()
{
   #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  #endif
    ll n;
    scanf("%lld", &n);
   /* int res = 0;
    for(int i = 0; i < (2*n+1); i++)
    {
        int x;
        scanf("%d",&x);
        res ^= x;
    }*/
    cout << "While diye--> "<< Divisors(n)<<endl;
    cout << "Builtin diye--> "<< divisors(n)<<endl;
    cout << "Xor diye--> " << _Divisors(n)<<endl;
    cout << "pOpcount  diye--> " << Divisors_loop(n)<<endl;
    for(auto x: M){
        if(x.second == 4)continue;
        else cout << x.first<<" ";
    }
    fprintf(stdout,"\nTIME: %.4lf sec\n", (double)clock()/(CLOCKS_PER_SEC));
    return 0;
}
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