cf1136E. Nastya Hasn't Written a Legend(二分 线段树)

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 << '\n';}
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 << '\n'; 
        }
        c = 'g';
    }
    return 0;
}