Submission #574080

#	Time	Username	Problem	Language	Result	Execution time	Memory
574080		MohammadAghil	K blocks (IZhO14_blocks)	C++14	0 / 100	1 ms	732 KiB

blocks

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
    using namespace std;
  typedef long long ll;
   typedef pair<int, int> pp;
    #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
      #define per(i,r,l) for(int i = (r); i >= (l); i--)
        #define rep(i,l,r) for(int i = (l); i < (r); i++)
           #define all(x) x.begin(), x.end()
              #define sz(x) (int)(x).size()
                  #define pb push_back
                      #define ss second
                           #define ff first
                                   void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e5 + 1, maxk = 101, lg = 21, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}

int dp[maxn], nw[maxn], a[maxn][lg], lgg[maxn];

int get(int l, int r){
     int g = lgg[r-l+1];
     return max(a[l][g], a[r-(1<<g)+1][g]);
}

void slv(int l, int r, int lx, int rx){
     int mid = (l + r)>>1;
     int cr = get(min(rx,mid), mid);
     pp sl = {inf, -1};
     per(i,min(rx,mid),max(lx,1)) cr = max(cr, a[i][0]), sl = min(sl, {dp[i-1] + cr, i});
     nw[mid] = sl.ff;
     if(l < mid) slv(l, mid-1, lx, sl.ss);
     if(r > mid) slv(mid+1, r, sl.ss, rx);
}

int main(){
     cin.tie(0) -> sync_with_stdio(0); 
     rep(i,2,maxn) lgg[i] = lgg[i>>1] + 1;
     int n, k; cin >> n >> k;
     rep(i,0,n) cin >> a[i][0];
     rep(j,1,lg) rep(i,0,n-(1<<j)+1) a[i][j] = max(a[i][j-1], a[i+(1<<(j-1))][j-1]);
     rep(i,0,n) dp[i] = max(i?dp[i-1]:0, a[i][0]);
     rep(i,2,k+1){
          slv(0, n-1, 0, n-1);
          rep(j,0,n) dp[j] = nw[j];
     }
     cout << dp[n-1] << '\n';
     return 0; 
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 18
• Subtask 3: 0 / 21
• Subtask 4: 0 / 47