ACCENTURE EXAM SOLUTIONS

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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

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#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

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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,✅✅✅✅

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String container Fully passed 🔥🔥🔥 Infosys 🔥🔥 All three code done 😎😎😎

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Count Travel routers 🔥🔥🔥🔥🔥 Fully passed ✅✅✅✅

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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[i]; --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; } //subarray subset sum

#include <bits/stdc++.h> #define int long long using namespace std; #define ll long long ll solve(ll n,ll k,vector<ll>&a) { k++; unordered_map<ll,ll>freq; map<ll,vector<ll>>mpp; for (ll i=0;i<n;i++) { freq[a[i]]++; mpp[freq[a[i]]].push_back(a[i]); } ll ans=0; for (auto it:mpp) ans+=it.second.size(); return ans; } signed main() { ll n,k; cin>>n>>k; vector<ll>a(n); for(ll i=0;i<n;i++) cin>>a[i]; cout<<solve(n,k,a); return 0; } sequence split (Infosys) 100💯✅ running

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