题解 CF1942F【Farmer John's Favorite Function】
萌萌 F 题，上大分。
首先，如下定义 \(g(i)\)：
- \(g(1)=\lfloor\sqrt{a_1}\rfloor\)；
- 对于所有 \(i > 1\)，\(g(i)=\lfloor\sqrt{g(i-1)+a_i}\rfloor\)。
也就是将 \(f(i)\) 的每一步运算后都向下取整。注意到 \(\lfloor f(i)\rfloor=g(i)\) 恒成立，于是我们只需要转而求每次修改后 \(g(n)\) 的值，避免了浮点数运算。
将询问离线，对时间轴（操作序列）维护数据结构，按空间轴（序列 \(a_i\)）下标从小到大扫描线。每扫描到一个新的下标 \(i\)，需要做的操作是将每个时刻的 \(g\) 加上对应时刻修改后的 \(a_i\)，然后将每个时刻的 \(g\) 开根号下取整。因此，数据结构需要支持 \(n+q\) 次区间加操作和 \(n\) 次全局开根号下取整操作，并在最后输出每个位置的值。
可以使用势能线段树维护：每个节点维护区间最小值、最大值、区间加法标记。区间加操作是平凡的，全局开根号下取整操作直接暴力递归，直到 \(mx-mn=\lfloor\sqrt{mx}\rfloor-\lfloor\sqrt{mn}\rfloor\)，此时打区间加法标记即可。也可以使用分块等维护。
// Problem: F. Farmer John's Favorite Function
// Contest: Codeforces - CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!)
// URL: https://codeforces.com/contest/1942/problem/F
// Memory Limit: 256 MB
// Time Limit: 5000 ms
//
// Powered by CP Editor (https://cpeditor.org)
//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;
mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
uniform_int_distribution<int> dist(L, R);
return dist(rnd);
}
template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}
template<int mod>
inline unsigned int down(unsigned int x) {
return x >= mod ? x - mod : x;
}
template<int mod>
struct Modint {
unsigned int x;
Modint() = default;
Modint(unsigned int x) : x(x) {}
friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
friend Modint operator/(Modint a, Modint b) {return a * ~b;}
friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
friend Modint operator~(Modint a) {return a ^ (mod - 2);}
friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
friend Modint& operator++(Modint& a) {return a += 1;}
friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
friend Modint& operator--(Modint& a) {return a -= 1;}
friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};
const ll N = 2e5 + 5;
ll n, m, a[N], qk[N], qx[N];
vector<tuple<ll, ll>> qs[N];
inline ll mysqrt(ll x) {
ll k = sqrtl(x);
while(k * k > x) --k;
while((k + 1) * (k + 1) <= x) ++k;
return k;
}
struct SegTree {
ll mx[N << 2], mn[N << 2], tag[N << 2];
#define lc(u) (u << 1)
#define rc(u) (u << 1 | 1)
void pushup(ll u) {
mx[u] = max(mx[lc(u)], mx[rc(u)]);
mn[u] = min(mn[lc(u)], mn[rc(u)]);
}
void pushdown(ll u) {
tag[lc(u)] += tag[u];
tag[rc(u)] += tag[u];
mx[lc(u)] += tag[u];
mx[rc(u)] += tag[u];
mn[lc(u)] += tag[u];
mn[rc(u)] += tag[u];
tag[u] = 0;
}
void rangeadd(ll u, ll l, ll r, ll ql, ll qr, ll k) {
if(ql > qr) return;
if(ql <= l && r <= qr) {
tag[u] += k;
mx[u] += k;
mn[u] += k;
return;
}
pushdown(u);
ll mid = (l + r) >> 1;
if(ql <= mid) rangeadd(lc(u), l, mid, ql, qr, k);
if(qr > mid) rangeadd(rc(u), mid + 1, r, ql, qr, k);
pushup(u);
}
void rangesqrt(ll u, ll l, ll r, ll ql, ll qr) {
if(ql > qr) return;
if(l == r) {
mx[u] = mn[u] = mysqrt(mx[u]);
return;
}
if(ql <= l && r <= qr && mx[u] - mn[u] == mysqrt(mx[u]) - mysqrt(mn[u])) {
ll diff = mysqrt(mx[u]) - mx[u];
tag[u] += diff;
mx[u] += diff;
mn[u] += diff;
return;
}
pushdown(u);
ll mid = (l + r) >> 1;
if(ql <= mid) rangesqrt(lc(u), l, mid, ql, qr);
if(qr > mid) rangesqrt(rc(u), mid + 1, r, ql, qr);
pushup(u);
}
void print(ll u, ll l, ll r, char ENDPRINT = '\n') {
if(l == r) {
cout << mx[u] << ENDPRINT;
return;
}
pushdown(u);
ll mid = (l + r) >> 1;
print(lc(u), l, mid, ENDPRINT);
print(rc(u), mid + 1, r, ENDPRINT);
pushup(u);
}
#undef lc
#undef rc
}sgt;
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> n >> m;
rep(i, 1, n) cin >> a[i];
rep(i, 1, m) cin >> qk[i] >> qx[i];
rep(i, 1, m) qs[qk[i]].emplace_back(qx[i], i);
rep(i, 1, n) {
ll lstval = a[i], lstkey = 1;
for(auto [val, key] : qs[i]) {
sgt.rangeadd(1, 1, m, lstkey, key - 1, lstval);
lstval = val;
lstkey = key;
}
sgt.rangeadd(1, 1, m, lstkey, m, lstval);
// sgt.print(1, 1, m, ' '); cout << endl;
sgt.rangesqrt(1, 1, m, 1, m);
// sgt.print(1, 1, m, ' '); cout << endl;
}
sgt.print(1, 1, m);
return 0;
}
