제출 #916168

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
916168		GrindMachine	Gap (APIO16_gap)	C++17	100 / 100	43 ms	3628 KiB

gap

#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

read some solutions a long time ago, slightly remember some ideas
may have still solved it if not for those ideas

T = 1:
can restore the whole array
in 1st op, find a[1] and a[n]
in 2nd op, find a[2] and a[n-1] (set s = a[1]+1, t = a[n]-1)
in 3rd op, find a[3] and a[n-2] (set s = a[2]+1, t = a[n-1]-1)
...

after restoring the whole array, find the max gap

T = 2:
key idea:
ans >= ceil2(mx-mn,n-1) (lower bound on ans)

proof by contradiction
b = ceil2(mx-mn,n-1)
if max gap < b, then mn+(n-1)*b < mx, so max must be lesser than mx, which is a contradiction

how does this help?

split the range [mn,mx] into blocks of size b+1
note that 2 adj guys with gap > b occur in 2 adj blocks (a[i] = last guy of block x, a[i+1] = first guy of block x+1)
ans = max(max_gap_between_the_ends_of_adj_blocks,b)

cost:
n+1 to find (mn,mx)
n in total for all guys (+1 for each guy), because each guy occurs in exactly 1 block (just contribution of each guy, +1 for query not added)
<= n-1 for all queries (<= n-1 queries in total)

in total, cost <= 3n

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "gap.h"

long long findGap(int T, int n)
{
    if(T == 1){
        vector<ll> a(n+5);
        ll l = 0, r = inf2;
        ll mn,mx;

        rep1(i,ceil2(n,2)){
            MinMax(l,r,&mn,&mx);
            a[i] = mn;
            a[n-i+1] = mx;
            l = mn+1, r = mx-1;
        }

        ll ans = 0;
        rep1(i,n-1) amax(ans,a[i+1]-a[i]);

        return ans;
    }

    ll mn,mx;
    MinMax(0,inf2,&mn,&mx);
    ll b = ceil2(mx-mn,n-1);
    ll ptr = mn;
    ll prev_guy = -1;
    ll ans = b;

    while(ptr <= mx){
        ll curr_mn,curr_mx;
        MinMax(ptr,ptr+b,&curr_mn,&curr_mx);
        if(curr_mn != -1){
        	if(prev_guy != -1){
        		amax(ans,curr_mn-prev_guy);
        	}
        	prev_guy = curr_mx;
        }

    	ptr += b+1;
    }

    return ans;
}

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