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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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🔥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 : @srksvk

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📈 Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 046 subscribers, ranking 14 501 in the Education category and 28 425 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 14 046 subscribers.

According to the latest data from 12 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -103 over the last 30 days and by 0 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.71%. Within the first 24 hours after publication, content typically collects 1.85% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 522 views. Within the first day, a publication typically gains 260 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“🔥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...”

Thanks to the high frequency of updates (latest data received on 13 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

14 046
Subscribers
No data24 hours
-267 days
-10330 days
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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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