#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;
}
