Submission #237658

#	Time	Username	Problem	Language	Result	Execution time	Memory
237658		zaneyu	Gap (APIO16_gap)	C++14	100 / 100	77 ms	3308 KiB

gap

/*input
2 4
2 3 6 8
*/
#include "gap.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;
 
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long 
#define ld long double
#define pii pair<double,double>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#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()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e12;
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%MOD)*(b%MOD))%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=2e5+5;
const ll maxlg=__lg(maxn)+2;
 
ll arr[maxn];
long long findGap(int T, int n){
	if(T==1){
		int l=0,r=n-1;
		MinMax(0,(ll)1e18,arr+l,arr+r);
		++l;
		--r;
		while(l<=r){
			MinMax(arr[l-1]+1,arr[r+1]-1,arr+l,arr+r);
			++l;
			--r;
		}
		ll ans=0;
		REP1(i,n-1) MXTO(ans,arr[i]-arr[i-1]);
		return ans;
	}
	else{
		//pigeon
		ll l,r;
		MinMax(0,(ll)1e18,&l,&r);
		ll sz=(r-l+n)/(n-1);
		vector<ll> v;
		REP(i,n-1){
			ll tl,tr;
			MinMax(l+i*sz,l+(i+1)*sz-1,&tl,&tr);
			if(tl!=-1){
				v.pb(tl);
				v.pb(tr);
			}
		}
		v.pb(r);
		ll ans=0;
		REP1(i,sz(v)-1){
			MXTO(ans,v[i]-v[i-1]);
		}
		return ans;
	}
}

• Subtask 1: 30 / 30
• Subtask 2: 70 / 70