oj.uz

Submission #407444

# Submission time Handle Problem Language Result Execution time Memory
407444 2021-05-19T01:16:37 Z zaneyu The short shank; Redemption (BOI21_prison) C++14
0 / 100
 251 ms 6144 KB 
/*input
5 1 42
13 37 47 11 42
*/
 
 
#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<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=5e5+5;
int dp[maxn],pv[maxn];
int cst[maxn];
int arr[maxn],b[maxn];
int main(){
    int n,d,t;
    cin>>n>>d>>t;
    REP(i,n){
        cin>>arr[i];
    }
    {
        int cnt=0;
        int i=0;
        for(int j=i;j<n;j++){
            if(j!=i) b[j]=min(arr[j],b[j-1]+1);
            else b[j]=arr[j];
            if(b[j]<=t) cnt++;
            cst[j]=cnt;
        }
    }
    vector<int> v;
    int ans=n;
    for(int i=n-1;i>=0;i--){
        MNTO(ans,(n-i-sz(v)-1)+cst[i]);
        while(sz(v) and v.back()<=i-(t-arr[i])) v.pop_back();
        if(arr[i]>t) v.pb(i);
    }
    cout<<ans;
}

Subtask #1
0 / 15.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 1 ms 332 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #2
0 / 10.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 251 ms 6144 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #3
0 / 20.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 1 ms 332 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #4
0 / 10.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 34 ms 1096 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #5
0 / 25.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 1 ms 332 KB Output isn't correct
3 Halted 0 ms 0 KB -

Subtask #6
0 / 20.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 1 ms 332 KB Output isn't correct
3 Halted 0 ms 0 KB -