Submission #568857

#	Time	Username	Problem	Language	Result	Execution time	Memory
568857		zaneyu	Fish 2 (JOI22_fish2)	C++14	5 / 100	4062 ms	700 KiB

fish2

/*input
5
6 4 2 2 6
6
2 1 5
2 1 3
1 3 1
2 2 5
2 1 5
2 2 4
*/
#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("Ofast")
//#pragma GCC target("avx2")
//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 maxn=5e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#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)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    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;
}
int arr[maxn];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,q;
    cin>>n;
    REP1(i,n) cin>>arr[i];
    cin>>q;
    while(q--){
        int t,a,b;
        cin>>t>>a>>b;
        if(t==1){
            arr[a]=b;
        }
        else{
            int ans=0;
            for(int i=a;i<=b;i++){
                int l=i-1,r=i+1;
                ll sum=arr[i];
                while((l>=a and sum>=arr[l]) or (r<=b and sum>=arr[r])){
                    if((l>=a and sum>=arr[l])) sum+=arr[l--];
                    else sum+=arr[r++];
                }
                if(l<a and r>b) ++ans;
            }
            cout<<ans<<'\n';
        }
    }
} 

• Subtask 1: 5 / 5
• Subtask 2: 0 / 8
• Subtask 3: 0 / 12
• Subtask 4: 0 / 23
• Subtask 5: 0 / 35
• Subtask 6: 0 / 17