arc060 E - Tak and Hotels

E - Tak and Hotels

先找到每个点一步最远跳到哪个点，然后用个倍增即可

#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define sc second
#define pb push_back
#define ll long long
#define trav(v,x) for(auto v:x)
#define all(x) (x).begin(), (x).end()
#define VI vector<int>
#define VLL vector<ll>
#define pll pair<ll, ll>
#define double long double
//#define int long long
using namespace std;
const int N = 1e6 + 100;
const int inf = 1e9;
//const ll inf = 1e18;
const ll mod = 998244353;//1e9 + 7;#ifdef LOCAL
void debug_out(){cerr << endl;}
template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T)
{cerr << " " << to_string(H);debug_out(T...);
}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endifvoid sol()
{ll n, m;cin >> n;VLL a(n);for(int i = 0; i < n; i++)cin >> a[i];cin >> m;vector<VLL> jp(n, VLL(20, 0));for(int i = n - 1; i >= 0; i--){if(i == n - 1){for(int j = 0; j < 20; j++)jp[i][j] = n;continue;}int x = upper_bound(all(a), a[i] + m) - a.begin();--x;jp[i][0] = x;for(int j = 1; j < 20; j++)jp[i][j] = (jp[i][j - 1] == n ? n : jp[jp[i][j - 1]][j - 1]);}int qq;cin >> qq;while(qq--){int x, y;cin >> x >> y;--x, --y;if(x > y)swap(x, y);int ans = 0;for(int i = 19; i >= 0; i--){if(jp[x][i] <= y)x = jp[x][i], ans += 1 << i;}if(x < y)++ans;cout << ans << '\n';}
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);
//  int tt;
//  cin >> tt;
//  while(tt--)sol();
}

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