Submission #442972

#	Time	Username	Problem	Language	Result	Execution time	Memory
442972		zaneyu	Snowball (JOI21_ho_t2)	C++14	0 / 100	1 ms	332 KiB

Main

/*input
1 4
1000000000000
1000000000000
-1000000000000
-1000000000000
-1000000000000
*/
#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;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//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=1e18+1;
const int INF=0x3f3f3f3f;
const int MOD=1e9+9;
const ld PI=acos(-1);
const ld eps=1e-3;
#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>=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=2e5+5;
const int maxlg=__lg(maxn)+2;
ll arr[maxn];
ll L[maxn],R[maxn];
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,q;
    cin>>n>>q;
    REP(i,n){
        cin>>arr[i];
    }
    vector<pii> v;
    REP1(i,n-1) v.pb({arr[i]-arr[i-1],i});
    REP(i,n) L[i]=R[i]=-INF64;
    sort(ALL(v));
    ll l=0,r=0,cur=0;
    int p=0;
    while(q--){
        ll x;
        cin>>x;
        cur+=x;
        MNTO(l,cur);
        MXTO(r,cur);
        while(p<sz(v) and v[p].f<=r-l){
            int i=v[p].s;
            if(x<0){
                R[i-1]=r+arr[i-1];
                L[i]=R[i-1];
            }
            else{
                L[i]=l+arr[i];
                R[i-1]=L[i];
            }
            ++p;
        }
    }
    REP(i,n){
        if(L[i]==-INF64) L[i]=l+arr[i];
        if(R[i]==-INF64) R[i]=r+arr[i];
        cout<<R[i]-L[i]<<'\n';
    }
}  

• Subtask 1: 0 / 33
• Subtask 2: 0 / 67