Submission #404322

#TimeUsernameProblemLanguageResultExecution timeMemory
404322AmineWeslatiCigle (COI21_cigle)C++14
0 / 100
2 ms1228 KiB
//Never stop trying #include "bits/stdc++.h" using namespace std; #define boost ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0) typedef long long ll; typedef string str; typedef long double ld; typedef pair<int, int> pi; #define fi first #define se second typedef vector<int> vi; typedef vector<pi> vpi; #define pb push_back #define eb emplace_back #define sz(x) (int)x.size() #define all(x) begin(x), end(x) #define rall(x) rbegin(x), rend(x) #define endl "\n" #define FOR(i,a,b) for (int i = (a); i < (b); ++i) #define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i) const int MOD = 1e9 + 7; //998244353 const ll INF = 1e18; const int MX = 2e5 + 10; const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //right left down up template<class T> using V = vector<T>; template<class T> bool ckmin(T& a, const T& b) { return a > b ? a = b, 1 : 0; } template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; } ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up //constexpr int log2(int x) { return 31 - __builtin_clz(x); } // floor(log2(x)) mt19937 rng(chrono::system_clock::now().time_since_epoch().count()); //mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count()); ll random(ll a, ll b){ return a + rng() % (b - a + 1); } #ifndef LOCAL #define cerr if(false) cerr #endif #define dbg(x) cerr << #x << " : " << x << endl; #define dbgs(x,y) cerr << #x << " : " << x << " / " << #y << " : " << y << endl; #define dbgv(v) cerr << #v << " : " << "[ "; for(auto it : v) cerr << it << ' '; cerr << ']' << endl; #define here() cerr << "here" << endl; void IO() { #ifdef LOCAL freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif } /////////////////////////ONLY CLEAN CODES ALLOWED///////////////////////// int main() { boost; IO(); int N; cin>>N; vi a(N); FOR(i,0,N) cin>>a[i]; vector<vi>dp(N,vi(N,0)); FOR(l,1,N){ int l2=l,cur=0,prev=0,v=0; FOR(r,l,N){ cur+=a[r]; while(l2 && prev<cur){ l2--; if(prev && prev==cur-a[r]) v++; prev+=a[l2]; } dp[l][r]=dp[l2][r-1]+v; ckmax(dp[l][r],dp[l][r-1]); } } int ans=0; FOR(i,0,N) ckmax(ans,dp[i][N-1]); cout << ans << endl; return 0; } //Change your approach
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