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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ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 083 obunachidan iborat bo'lib, Taʼlim toifasida 14 133-o'rinni va Hindiston mintaqasida 28 152-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 14 083 obunachiga ega bo‘ldi.
06 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -120 ga, so‘nggi 24 soatda esa -10 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.44% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.82% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 485 marta ko‘riladi; birinchi sutkada odatda 256 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 07 Oktabr, 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.
14 083
Obunachilar
-1024 soatlar
-337 kun
-12030 kun
Postlar arxiv
14 076
+2
Infosys sp exam help successfully done by remote access ✅✅
3/3 code fully passed ✅✅✅ 🔥🔥🔥
Contact for placement exam @srksvk
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#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(string s1, string s2) {
ll m = s1.length();
ll n = s2.length();
vector<vector<ll>>dp(m+1,vector<ll>(n + 1));
for (ll i=1;i<= m;i++) {
dp[i][0]=i;
}
for (ll j = 1; j <= n;j++) {
dp[0][j] = j;
}
for (ll i = 1; i <= m; ++i)
{
for (ll j = 1; j <= n; ++j)
{
if (s1[i - 1] == s2[j - 1])
{
dp[i][j] = dp[i - 1][j - 1];
}
else
{
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
if (i > 1 && j > 1 && s1[i - 1] == s2[j - 2])
{
dp[i][j] = min(dp[i][j], 1 + dp[i - 1][j - 2]);
}
if (i > 1 && j > 1 && s1[i - 2] == s2[j - 1])
{
dp[i][j] = min(dp[i][j], 1 + dp[i - 2][j - 1]);
}
}
}
}
return dp[m][n];
}
signed main()
{
string s1,s2; cin>>s1>>s2;
cout<<solve(s1,s2)<<endl;
return 0;
}
String operation.
Just paste it
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const int MOD 1000000007;
int subsetSumCount(const vector& A, int L, int R, int K) {
vector dp(K+1, 0); dp[0] = 1;
for (int i=L; i<=R; ++i) { for (int j = K; j >= A[1];
--j) {
dp[j] = (dp[j] + dp[j- A[i]]) % MOD; }
}
return dp[K];}
int findXOR(int n, int Q,
const vector& A, const vector>& B) {
int result = 0;
for (const auto& query: B) {
int L= query [0] - 1;
int R= query [1] -1;
int K = query[2];
int P=subsetSumCount(A,L,R,K);
result ^= P;
}
return result;
}
Subset sum
Fully passed ✅✅✅✅
Just paste it
14 076
Share with your Friends and in Big groups we will definitely post answers here
We will Try to share Some answers Here also in our Groups.
So must Join
1) @codeing_area
2) @codeing_area
3) @codeing_area
14 076
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll MOD=1e9+7;
struct FenwickTree {
vector<ll> bit;
ll n;
FenwickTree(ll size) {
n = size;
bit.assign(n + 1, 0);
}
void update(ll idx, ll val) {
for (; idx <= n; idx += idx & -idx)
bit[idx] = max(bit[idx], val);
}
ll query(ll idx) {
ll res = 0;
for (; idx > 0; idx -= idx & -idx)
res = max(res, bit[idx]);
return res;
}
};
ll solve(vector<ll>& a)
{
ll n=a.size();
vector<pair<ll,ll>>ia(n);
for (ll i=0;i<n;i++)
{
ia[i] = make_pair(a[i], i);
}
sort(ia.begin(),ia.end(),[](const pair<ll,ll>&p1,const pair<ll,ll>&p2) {
return p1.first<p2.first;
});
FenwickTree fenwick(n);
ll maxBeauty=0;
for (ll i=0;i<n;i++)
{
ll val=ia[i].first;
ll og=ia[i].second;
ll beauty=0;
for (ll j=og-1;j>=0;j--)
{
if (__gcd(abs(a[j]),abs(a[og]))>1) {
beauty=max(beauty,fenwick.query(j+1)+(a[og]-a[j])*(a[og]-a[j]) % MOD);
}
}
fenwick.update(og + 1, beauty);
maxBeauty = max(maxBeauty, beauty);
}
return maxBeauty;
}
signed main()
{
ll n; cin>>n;
vector<ll>a(n);
for (ll i=0;i<n;i++) cin>>a[i];
cout<<solve(a)<<endl;
return 0;
}
subsequence beauty
Share group @srksvk
14 076
dp[i+1][0][0][(j + addv)%k][pv+1] += dp[i][1][0][j][pv];
dp[i+1][0][0][(j + addv)%k][pv+1] %= mod;
}
else{
dp[i+1][1][0][(j + addv)%k][pv+1] += dp[i][1][0][j][pv];
dp[i+1][1][0][(j + addv)%k][pv+1] %= mod;
}
// le = 0, re = 1
if(r[i] == '0'){
dp[i+1][0][1][j][pv-1] += dp[i][0][1][j][pv];
dp[i+1][0][1][j][pv-1] %= mod;
}
else{
dp[i+1][0][1][(j + addv)%k][pv+1] += dp[i][0][1][j][pv];
dp[i+1][0][1][(j + addv)%k][pv+1] %= mod;
dp[i+1][0][0][j][pv-1] += dp[i][0][1][j][pv];
dp[i+1][0][0][j][pv-1] %= mod;
}
// le = 0, re = 0
dp[i+1][0][0][j][pv-1] += dp[i][0][0][j][pv];
dp[i+1][0][0][j][pv-1] %= mod;
dp[i+1][0][0][(j + addv)%k][pv+1] += dp[i][0][0][j][pv];
dp[i+1][0][0][(j + addv)%k][pv+1] %= mod;
}
}
}
}
int ans = 0;
for(int i = 0; i <= n; i++){
for(int j = 0; j < 2; j++){
for(int z = 0; z < 2; z++){
ans += dp[n][j][z][0][i];
ans %= mod;
}
}
}
return ans;
}
int32_t main(){
cin.tie(0); cout.tie(0);
ios_base::sync_with_stdio(false);
int m, n, k;
cin>>m>>n>>k;
string l, r;
cin>>l>>r;
cout<
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import sys
MOD = 1000000007
def getCount(N, M, S):
dp = [[[0] * (M + 1) for _ in range(N + 1)] for _ in range(N + 1)]
dp[0][0][0] = 1
for i in range(N):
for j in range(N + 1):
for k in range(M + 1):
if dp[i][j][k] == 0:
continue
if j + 1 <= N:
if k < M and S[k] == '(':
dp[i + 1][j + 1][k + 1] = (dp[i + 1][j + 1][k + 1] + dp[i][j][k]) % MOD
else:
dp[i + 1][j + 1][k] = (dp[i + 1][j + 1][k] + dp[i][j][k]) % MOD
if j > 0:
if k < M and S[k] == ')':
dp[i + 1][j - 1][k + 1] = (dp[i + 1][j - 1][k + 1] + dp[i][j][k]) % MOD
else:
dp[i + 1][j - 1][k] = (dp[i + 1][j - 1][k] + dp[i][j][k]) % MOD
return dp[N][0][M]
def main():
N = int(sys.stdin.readline().strip())
M = int(sys.stdin.readline().strip())
S = sys.stdin.readline().strip()
result = getCount(N, M, S)
print(result)
if name == "main":
main()
Bracket sequence,✅✅✅✅
14 076
Pre bookings slot help successfully done ...
Now I will upload answer... guy's just share group fast ⏩⏩⏩
14 076
Share with your Friends and in Big groups we will definitely post answers here
We will Try to share Some answers Here also in our Groups.
So must Join
1) @codeing_area
2) @codeing_area
3) @codeing_area
14 076
Don't ask these code everyone...
Note:: This one prebooking 👍👍👍
So I am not share these code... because plug 👍👍
