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ACCENTURE EXAM HELP ! CISCO EXAM !

ACCENTURE EXAM HELP ! CISCO EXAM !

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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

Ko'proq ko'rsatish

📈 Telegram kanali ACCENTURE EXAM HELP ! CISCO EXAM ! analitikasi

ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 208 obunachidan iborat bo'lib, Taʼlim toifasida 15 294-o'rinni va Hindiston mintaqasida 31 490-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 208 obunachiga ega bo‘ldi.

30 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -144 ga, so‘nggi 24 soatda esa 6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 3.08% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.29% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 407 marta ko‘riladi; birinchi sutkada odatda 170 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
🔥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...

Yuqori yangilanish chastotasi (oxirgi ma’lumot 01 Iyul, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

13 208
Obunachilar
+624 soatlar
-127 kunlar
-14430 kunlar
Postlar arxiv
Ibm exam celerd ✅✅✅✅✅✅✅ Got next round mail ✅🎉🎉 Helped proof 👇👇👇👇 https://t.me/codeing_area/7198?single
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Ibm exam celerd ✅✅✅✅✅✅✅ Got next round mail ✅🎉🎉 Helped proof 👇👇👇👇 https://t.me/codeing_area/7198?single

Cognizant communication round exam successfully completed ✅✅✅✅ We are providing all answers with 💯% correct answer ✅✅ Contac
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UST global exam successfully done by remote access ✅✅✅ 6/6 code fully passed with all the test caes passed ✅✅ Contact for placement exam @srksvk

IBM exam successfully done by remote access ✅✅✅ Java - 2/2 code passed with all the test caes passed ✅ All MCQ done with remo
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IBM exam successfully done by remote access ✅✅✅ Java - 2/2 code passed with all the test caes passed ✅ All MCQ done with remote access ✅✅ Contact for placement exam @srksvk

Ibm exam successfully done by remote access ✅✅✅ Java - 2/2 code passed✅ All MCQ done with 💯 correct answer ✅✅ Contact for pl
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Ibm exam successfully done by remote access ✅✅✅ Java - 2/2 code passed✅ All MCQ done with 💯 correct answer ✅✅ Contact for placement exam @srksvk

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Jai shree Ram 🙏🙏🙏🙏🙏 🎉🎉

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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

Make it 7k everyone ✅🎉🎉 Share @codeing_area Shared ✅ share✅ share

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

Share everyone foe next code ✅✅ Share @codeing_area

#include<bits/stdc++.h> using namespace std; const int N=500005; typedef long long ll; int T,n,k,t,h[N]; ll dp[N][2]; struct edge{   int ver,net;   ll val; }f[N*2]; void add(int x,int y,int z){   f[++t].net=h[x];   h[x]=t,f[t].ver=y;   f[t].val=z; } void dfs(int x,int fa){   dp[x][0]=dp[x][1]=0;   priority_queue<ll> q;   for(int i=h[x];i;i=f[i].net){     int y=f[i].ver;     if(y==fa) continue;     dfs(y,x);     dp[x][0]+=dp[y][0];     q.push(dp[y][1]+f[i].val-dp[y][0]);   }   dp[x][1]=dp[x][0];   int p=k;   while(p--&&q.size()&&q.top()>0){     if(p) dp[x][1]+=q.top();     dp[x][0]+=q.top();     q.pop();   }   } int main(){   scanf("%d",&T);   while(T--){     int x,y,z;t=0;     scanf("%d%d",&n,&k);     for(int i=1;i<n;++i){       scanf("%d%d%d",&x,&y,&z);       add(x,y,z),add(y,x,z);     }     dfs(1,0);     printf("%lld\n",dp[1][0]);     for(int i=1;i<=n;++i) h[i]=0,dp[i][0]=dp[i][1]=0;   }   return 0; } Good edage

At least do 6.5k everyone Shared @codeing_area