题目链接

传送门

题面

题意

给你一个\(a\)数组和一个\(k\)数组，进行\(q\)次操作，操作分为两种：

• 将\(a_i\)增加\(x\)，此时如果\(a_{i+1}<a_i+k_i\)，那么就将\(a_{i+1}\)变成\(a_i+k_i\)，如果\(a_{i+2}<a_i+k_i\)，则将\(a_{i+2}\)变成\(a_{i+1}+k_{i+1}\)，以此类推。
• 查询\(\sum\limits_{i=l}^{r}a_i\)。
  

  思路

  我们首先存下\(k\)数组的前缀和\(sum1\)，再存\(sum1\)的前缀和\(sum2\)。
  那么修改操作产生的影响我们就可以先通过二分出右端点\(r\)，然后进行区间覆盖，将\([l,r]\)覆盖成\(a_i\)，然后这一段区间的\(sum=a_i\times (r-l+1)+sum2[r]-sum2[l]-sum1[l]\times(r-l)\)。
  查询操作就是普通的区间求和。

  代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

char op[3];
int n, q, l, r;
LL sum1[maxn], sum2[maxn];
int k[maxn];

struct node {
    bool lazy;
    int l, r;
    LL val, sum;
}segtree[maxn<<2];

void push_up(int rt) {
    segtree[rt].sum = segtree[lson].sum + segtree[rson].sum;
}

void push_down(int rt) {
    if(!segtree[rt].lazy) return;
    segtree[rt].lazy = 0;
    segtree[lson].lazy = segtree[rson].lazy = 1;
    segtree[lson].val = segtree[rt].val;
    segtree[rson].val = sum1[segtree[rson].l] - sum1[segtree[rt].l] + segtree[rt].val;
    segtree[lson].sum = segtree[lson].val * (segtree[lson].r - segtree[lson].l + 1) + sum2[segtree[lson].r] - sum2[segtree[lson].l] - sum1[segtree[lson].l] * (segtree[lson].r - segtree[lson].l);
    segtree[rson].sum = segtree[rson].val * (segtree[rson].r - segtree[rson].l + 1) + sum2[segtree[rson].r] - sum2[segtree[rson].l] - sum1[segtree[rson].l] * (segtree[rson].r - segtree[rson].l);
}

void build(int rt, int l, int r) {
    segtree[rt].l = l, segtree[rt].r = r;
    segtree[rt].sum = segtree[rt].lazy = 0;
    if(l == r) {
        scanf("%lld", &segtree[rt].val);
        segtree[rt].sum = segtree[rt].val;
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    push_up(rt);
}

void update1(int rt, int pos, int x) {
    if(segtree[rt].l == segtree[rt].r) {
        segtree[rt].val += x;
        segtree[rt].sum += x;
        return;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(pos <= mid) update1(lson, pos, x);
    else update1(rson, pos, x);
    push_up(rt);
}

void update2(int rt, int l, int r, LL x) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        segtree[rt].val = x;
        segtree[rt].lazy = 1;
        segtree[rt].sum = x * (r - l + 1) + sum2[r] - sum2[l] - sum1[l] * (r - l);
        return;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) update2(lson, l, r, x);
    else if(l > mid) update2(rson, l, r, x);
    else {
        update2(lson, l, mid, x);
        update2(rson, mid + 1, r, x + sum1[mid + 1] - sum1[l]);
    }
    push_up(rt);
}

LL query1(int rt, int pos) {
    if(segtree[rt].l == segtree[rt].r) return segtree[rt].val;
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(pos <= mid) return query1(lson, pos);
    else return query1(rson, pos);
}

LL query2(int rt, int l, int r) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        return segtree[rt].sum;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) return query2(lson, l, r);
    else if(l > mid) return query2(rson, l, r);
    else return query2(lson, l, mid) + query2(rson, mid + 1, r);
}

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
    scanf("%d", &n);
    build(1, 1, n);
    for(int i = 1; i < n; ++i) {
        scanf("%d", &k[i]);
    }
    for(int i = 2; i <= n; ++i) {
        sum1[i] = sum1[i-1] + k[i-1];
    }
    for(int i = 2; i <= n; ++i) {
        sum2[i] = sum2[i-1] + sum1[i];
    }
    scanf("%d", &q);
    while(q--) {
        scanf("%s%d%d", op, &l, &r);
        if(op[0] == '+') {
            int ub = n, lb = l + 1, mid, ans = -1;
            update1(1, l, r);
            LL val = query1(1, l);
            while(ub >= lb) {
                mid = (ub + lb) >> 1;
                if(query1(1, mid) < val + sum1[mid] - sum1[l]) {
                    ans = mid;
                    lb = mid + 1;
                } else {
                    ub = mid - 1;
                }
            }
            if(ans != -1) update2(1, l, ans, val);
        } else {
            printf("%lld\n", query2(1, l, r));
        }
    }
    return 0;
}

Nastya Hasn't Written a Legend（Codeforces Round #546 (Div. 2)E+线段树）

原文地址：https://www.cnblogs.com/Dillonh/p/11181343.html