题意
Sol
yy出了一个暴躁线段树的做法。
因为题目保证了
(a_i + k_i <= a_{i+1})
那么我们每次修改时只需要考虑取max就行了。
显然从一个位置开始能影响到的位置是单调的，而且这些位置的每个改变量都是((a_i + x) + sum_{t=i}^{j-1} k_t)
那么可以建两棵线段树分别维护这两部分的值
每次修改的时候二分出要修改的位置。
打cf一定要记得开数据结构题啊qwq
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, a[MAXN], K[MAXN], S[MAXN], val[MAXN];
//第一棵线段树支持区间赋值 / 区间求和
//第二棵线段树支持区间赋值 / 区间求和 .....
struct Seg {
#define ls k << 1
#define rs k << 1 | 1
int sum[MAXN], tag[MAXN], siz[MAXN];
void update(int k) {
sum[k] = sum[ls] + sum[rs];
}
void ps(int k, int f) {
sum[k] = siz[k] * f;
tag[k] = f;
}
void pushdown(int k) {
if(tag[k] == -INF) return ;
ps(ls, tag[k]); ps(rs, tag[k]);
tag[k] = -INF;
}
void Build(int k, int l, int r) {
siz[k] = r - l + 1; tag[k] = -INF;
if(l == r) {sum[k] = val[l]; return ;}
int mid = l + r >> 1;
Build(ls, l, mid); Build(rs, mid + 1, r);
update(k);
}
void IntMem(int k, int l, int r, int ql, int qr, int v) {
if(ql <= l && r <= qr) {ps(k, v); return ;}
int mid = l + r >> 1;
pushdown(k);
if(ql <= mid) IntMem(ls, l, mid, ql, qr, v);
if(qr > mid) IntMem(rs, mid + 1, r, ql, qr, v);
update(k);
}
int IntQuery(int k, int l, int r, int ql, int qr) {
if(ql <= l && r <= qr) return sum[k];
int mid = l + r >> 1;
pushdown(k);
if(ql > mid) return IntQuery(rs, mid + 1, r, ql, qr);
else if(qr <= mid) return IntQuery(ls, l, mid, ql, qr);
else return IntQuery(ls, l, mid, ql, qr) + IntQuery(rs, mid + 1, r, ql, qr);
}
}T[2];
int GetA(int x) {
return T[0].IntQuery(1, 1, N, x, x) + K[x] - T[1].IntQuery(1, 1, N, x, x);
}
void Modify(int x, int v) {
int l = x, r = N, ans = l, tmp = GetA(x);
while(l <= r) {
int mid = l + r >> 1;
if(tmp + v + K[mid] - K[x] >= GetA(mid)) l = mid + 1, ans = mid;
else r = mid - 1;
}
T[0].IntMem(1, 1, N, x, ans, tmp + v);
T[1].IntMem(1, 1, N, x, ans, K[x]);
}
signed main() {
N = read();
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = 2; i <= N; i++)
K[i] = read() + K[i - 1];
memcpy(val, a, sizeof(a));
T[0].Build(1, 1, N);
memcpy(val, K, sizeof(K));
T[1].Build(1, 1, N);
for(int i = 1; i <= N; i++) S[i] = S[i - 1] + K[i];
int Q = read();
while(Q--) {
char c = 'g';
while(c != 's' && c != '+') c = getchar();
if(c == '+') {
int x = read(), v = read();
Modify(x, v);
}
else {
int l = read(), r = read();
int pre = T[0].IntQuery(1, 1, N, l, r);
int nxt = S[r] - S[l - 1] - T[1].IntQuery(1, 1, N, l, r);
cout << pre + nxt << '
';
}
c = 'g';
}
return 0;
}