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;
}