제출 #237867

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
237867		zaneyu	선물상자 (IOI15_boxes)	C++14	100 / 100	638 ms	149020 KiB

boxes

/*input
3 2 8
1 2 5
*/
#include "boxes.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long 
#define ld long double
#define pii pair<double,double>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#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()))))
#define MP make_pair
const ll INF64=4e12;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
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)
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1){
            res=(res*a)%MOD;
        }
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
long long delivery(int n, int k, int l, int arr[]) {
    vector<int> v1,v2;
    REP(i,n){
        if(!arr[i] or arr[i]==l){
            continue;
        }
        if(arr[i]<(l-arr[i])){
            v1.pb(arr[i]);
        }
        else{
            v2.pb(l-arr[i]);
        }
    }
    reverse(ALL(v2));
    ll ans=0;
    for(int i=sz(v1)-1;i>=0;i-=k) ans+=v1[i]*2;
    for(int i=sz(v2)-1;i>=0;i-=k) ans+=v2[i]*2;
    REP(circ,k+1){
        ll cur=0;
        for(int i=sz(v1)-1-circ;i>=0;i-=k) cur+=v1[i]*2;
        for(int i=sz(v2)-1-(k-circ);i>=0;i-=k) cur+=v2[i]*2;    
        MNTO(ans,cur+l);
    }
    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 15 / 15
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 30 / 30