题目链接

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;

}

巴特西

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

题目链接

题面

题意

思路

代码实现如下

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