Submission #446941

#	Time	Username	Problem	Language	Result	Execution time	Memory
446941		zaneyu	Semiexpress (JOI17_semiexpress)	C++14	18 / 100	1 ms	204 KiB

semiexpress

/*input
1000000000 2 3000
1000000000 1 2
1000000000
1
1000000000
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("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<ll,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=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*b%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;
}
const ll maxn=3e3+5;
const ll maxlg=__lg(maxn)+2; 
int arr[maxn];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,m,k;
    ll a,b,c;
    cin>>n>>m>>k>>a>>b>>c;
    ll t;
    cin>>t;
    REP(i,m) cin>>arr[i],--arr[i];
    int ans=0;
    if((n-1)*b<=t) ++ans;
    vector<ll> v;
    REP(i,m-1){
        int l=arr[i],r=arr[i+1];
        int p=l;
        int cnt=0;
        ll cur=l*b;
        while(p<r and cnt<=k-m){
            ll extra=t-cur;
            if(extra<0) break;
            extra/=a;
            ll nxt=p+extra+1;
            MNTO(nxt,(ll)r);
            if(p==l) ans+=nxt-p;
            else v.pb(nxt-p);
            cur+=(nxt-p)*c;
            p=nxt;
            ++cnt;
        }
    }
    sort(ALL(v));
    REP(i,min(sz(v),k-m)) ans+=v[i];
    cout<<(ans-1);
}

• Subtask 1: 18 / 18
• Subtask 2: 0 / 30
• Subtask 3: 0 / 52