Submission #321304

#	Time	Username	Problem	Language	Result	Execution time	Memory
321304		zaneyu	JJOOII 2 (JOI20_ho_t2)	C++14	100 / 100	20 ms	1924 KiB

ho_t2

/*input
9 1
IIIOOOJJJ
*/
//#include "supertrees.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;
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
typedef long long int ll;
#define ld double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll 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()
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
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)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    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;
}
vector<int> v[3];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,k;
    cin>>n>>k;
    REP(i,n){
        char x;
        cin>>x;
        if(x=='J') v[0].pb(i);
        else if(x=='O') v[1].pb(i);
        else v[2].pb(i);
    }
    int ans=INF;
    REP(i,sz(v[0])-k+1){
        int p=v[0][i+k-1];
        int nx=GET_POS(v[1],p)+k-1;
        if(nx>=sz(v[1])){
            break;
        }
        p=v[1][nx];
        nx=GET_POS(v[2],p)+k-1;
        if(nx>=sz(v[2])){
            break;
        }
        MNTO(ans,v[2][nx]-v[0][i]+1-3*k);
    }
    if(ans==INF) ans=-1;
    cout<<ans;
}

• Subtask 1: 1 / 1
• Subtask 2: 12 / 12
• Subtask 3: 87 / 87