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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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📈 Análisis del canal de Telegram ACCENTURE EXAM SOLUTIONS

El canal ACCENTURE EXAM SOLUTIONS (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 14 052 suscriptores, ocupando la posición 14 335 en la categoría Educación y el puesto 28 336 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 14 052 suscriptores.

Según los últimos datos del 11 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -105, y en las últimas 24 horas de 0, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.68%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.83% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 518 visualizaciones. En el primer día suele acumular 257 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
  • Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 12 octubre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

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#include <bits/stdc++.h> using namespace std; void solve(vector<vector<int>> vv, int operation, int xx, int yy, int &res) {     for (int i = 1; i < 3; i++)     {         int sum1 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum1 += vv[j][i - 1];         }         int sum2 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum2 += vv[j][i];         }         if (sum1 == sum2)         {             res = min(res, operation);         }         return;     }     for (int i = 0; i < vv.size(); i++)     {         solve(vv, operation, xx, yy, res);         vector<int> p1 = vv[i];         reverse(p1.begin(), p1.end());         solve(vv, operation + yy, xx, yy, res);         vector<int> p2 = vv[i];         int temp1 = p2[0];         int temp2 = p2[1];         int temp3 = p2[2];         p2[2] = temp1;         p2[1] = temp3;         p2[0] = temp2;         solve(vv, operation + xx, xx, yy, res);         vector<int> p3 = vv[i];         temp1 = p2[0];         temp2 = p2[1];         temp3 = p2[2];         p2[2] = temp2;         p2[1] = temp1;         p2[0] = temp3;         solve(vv, operation + xx, xx, yy, res);     }     return; } int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     int t = 1;     while (t--)     {         int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, sum = 0, diff = 0, maxN = 0, minN = 0, count = 0, temp = 0;         bool flag = false;         cin >> n;         cin >> m;         int xx;         cin >> xx;         int yy;         cin >> yy;         vector<vector<int>> vv(n, vector<int>(m));         for (int i = 0; i < n; i++)         {             for (int j = 0; j < m; j++)             {                 cin >> vv[i][j];             }         }         if (n == 1)         {             cout << -1 << endl;             continue;         }         int res = INT_MAX;         solve(vv, 0, xx, yy, res);         cout << res << endl;     }     return 0; } Pay for gift Join @codeing_area Share group everyone ✅🎉🎉

int help(int i, int n, vector<int>&a, vector<vector<int>>& dp, int prev){    if(i == n)       return 0;    int ans = INT_MIN;    if(dp[i][prev+1] != -1)       return dp[i][prev+1];    for(int j=0;j<4;j++){       if(j != prev){          if(j==0){             ans = max(ans, a[i] + help(i+1, n, a, dp, j));          }          if(j == 1){             ans = max(ans, a[i]*2 + help(i+1, n, a, dp, j));          }          if(j==2){             ans = max(ans, a[i]/2 + help(i+1, n, a, dp, j));          }          if(j==3){             ans = max(ans, a[i]+2 + help(i+1, n, a, dp, j));          }       }    }    return dp[i][prev+1] = ans; } int solve(int n, vector<int>&a){     vector<vector<int>>dp(n,vector<int>(5,-1));     return help(0, n, a, dp, -1); } No do at same

int solve(vector&arr) { if(arr.size()==1) { if(arr[0]==0) return 0; else return 1; } int u=0; int s=0; int n=arr.size(); int r=0; for(int i=0;is) r+=i-s+1; } return 2*r; } Count subset

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int solve(vector<int>&a) { int n=a.size(); vector<int> dp(1e5+1),d(1e5+1); int ans = 0; for (int i = 0;i<n; i++) { dp[a[i]] = 1; for (int j = 2; j * j <= a[i]; j++) { if (a[i] % j == 0) { dp[a[i]] = max(dp[a[i]], dp[d[j]] + 1); dp[a[i]] = max(dp[a[i]], dp[d[a[i] / j]] + 1); d[j] = a[i]; d[a[i] / j] = a[i]; } } ans = max(ans, dp[a[i]]); d[a[i]] = a[i]; } return ans; } @codeing_area

int help(int i, int n, vector<int>&a, vector<vector<int>>& dp, int prev){    if(i == n)       return 0;    int ans = INT_MIN;    if(dp[i][prev+1] != -1)       return dp[i][prev+1];    for(int j=0;j<4;j++){       if(j != prev){          if(j==0){             ans = max(ans, a[i] + help(i+1, n, a, dp, j));          }          if(j == 1){             ans = max(ans, a[i]*2 + help(i+1, n, a, dp, j));          }          if(j==2){             ans = max(ans, a[i]/2 + help(i+1, n, a, dp, j));          }          if(j==3){             ans = max(ans, a[i]+2 + help(i+1, n, a, dp, j));          }       }    }    return dp[i][prev+1] = ans; } int solve(int n, vector<int>&a){     vector<vector<int>>dp(n,vector<int>(5,-1));     return help(0, n, a, dp, -1); } @codeing_area

#include <bits/stdc++.h> using namespace std; #define ll long long ll dp[10005][2]; const int mod = (1e9 + 7); ll solve(ll i, ll o, ll n, ll k, vector<ll> &v) { if (i > n) { return 0; } if (dp[i][o] != -1) return dp[i][o]; ll ans = 0; if (o == 0) { for (int j = i; j < min(n + 1, i + k); j++) { ans = max(ans, solve(j + 1, 1 - o, n, k, v)) % mod; } } else { ll ta = 0; ll mx = 0; for (int j = i; j < min(n + 1, i + k); j++) { if (ta + v[j] >= 0) { ta += v[j]; mx = max(mx, ta); } else { ta = 0; } ll c = solve(j + 1, 1 - o, n, k, v) % mod; ans = max(ans, ((j - i + 1) * mx) % mod + c % mod); } } return dp[i][o] = ans; } int main() { ll n, k; cin >> n >> k; vector<ll> v; ll neg = 0; for (int i = 0; i < n; i++) { ll x; cin >> x; v.push_back(x); if (x <= 0) neg++; } if (neg == v.size()) { cout << 0 << endl; return 1; } memset(dp, -1, sizeof(dp)); cout << max(solve(0, 0, v.size() - 1, k, v), solve(0, 1, v.size() - 1, k, v)) << endl; } // Array Segments @codeing_area

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