Submission #406339

#TimeUsernameProblemLanguageResultExecution timeMemory
406339zaneyuA Difficult(y) Choice (BOI21_books)C++14
100 / 100
2 ms296 KiB
/*input 3 5 1 4 4 3 4 1 4 1 4 2 1 4 4 4 3 */ #include<bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include "books.h" using namespace std; using namespace __gnu_pbds; typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set; //#pragma GCC optimize("O2","unroll-loops","no-stack-protector") //order_of_key #of elements less than x // find_by_order kth element using ll=long long; using ld=long double; using pii=pair<int,int>; #define f first #define s second #define pb push_back #define REP(i,n) for(int i=0;i<n;i++) #define REP1(i,n) for(int i=1;i<=n;i++) #define FILL(n,x) memset(n,x,sizeof(n)) #define ALL(_a) _a.begin(),_a.end() #define sz(x)(int)x.size() #define SORT_UNIQUE(c)(sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end())))) const ll INF64=4e18; const int INF=0x3f3f3f3f; const ll MOD=998244353; const ld PI=acos(-1); const ld eps=1e-9; #define lowb(x) x&(-x) #define MNTO(x,y) x=min(x,(__typeof__(x))y) #define MXTO(x,y) x=max(x,(__typeof__(x))y) inline ll mult(ll a,ll b){ if(a<0) a+=MOD; if(b<0) b+=MOD; if(a>=MOD) a%=MOD; if(b>=MOD) b%=MOD; return(a*b)%MOD; } inline ll mypow(ll a,ll b){ if(b<=0) return 1; ll res=1LL; while(b){ if(b&1) res=mult(res,a); a=mult(a,a); b>>=1; } return res; } const int maxn=1e6+5; map<int,ll> mp; ll query(int x){ if(x<=0) return 0; if(mp.count(x)) return mp[x]; return mp[x]=skim(x); } void solve(int n, int k, ll A, int s) { ll ans=0; vector<int> v; REP1(i,k){ ans+=query(i); v.pb(i); } if(ans>=A and ans<=2*A){ answer(v); return; } if(ans>2*A){ impossible(); return; } int l=1,r=n; while(l<r){ int mid=(l+r)/2; if(query(mid)>=A) r=mid; else l=mid+1; } v[k-1]=l; ll asd=ans+query(l)-query(k); if(asd>=A and asd<=2*A){ answer(v); return; } if(query(l)<A) n=l; else n=l-1; ans=0; REP1(i,k) ans+=query(i); REP(i,k){ ans-=query(k-i); ans+=query(n-i); v[k-i-1]=n-i; if(ans>=A and ans<=2*A){ answer(v); return; } } impossible(); }
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