Submission #405978

#TimeUsernameProblemLanguageResultExecution timeMemory
405978zaneyuA Difficult(y) Choice (BOI21_books)C++14
20 / 100
330 ms560 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;
ll pref[maxn];
void solve(int n, int k, ll A, int s) {
    REP1(i,n){
        pref[i]=pref[i-1]+skim(i);
    }
    for(int i=k;i<=n;i++){
        if(pref[i]-pref[i-k]>=A){
            if(pref[i]-pref[i-k]<=2*A){
                vector<int> v;
                for(int j=i-k+1;j<=i;j++) v.pb(j);
                answer(v);
            }
            break;
        }
    }
    REP1(i,n){
        if(pref[i]-pref[i-1]>A){
            if(i<k){
                impossible();
                return;
            }
            ll ans=pref[i]-pref[i-1];
            REP1(z,k-1) ans+=pref[z]-pref[z-1];
            if(ans<=2*A){
                vector<int> v={i};
                REP1(z,k-1) v.pb(z);
                answer(v);
            }
            else{
                impossible();
            }
            return;
        }
    }
}
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