제출 #406339

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
406339		zaneyu	A Difficult(y) Choice (BOI21_books)	C++14	100 / 100	2 ms	296 KiB

books

/*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();    
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 10 / 10
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 20 / 20
• Subtask 7: 20 / 20