Problem Description
In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1.add l r: add one for $$$a_l, a_{l+1}...a_r$$$
2.query l r: query $$$\sum_{i=l}^{r}\lfloor a_i/b_i\rfloor$$$
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1.add l r: add one for $$$a_l, a_{l+1}...a_r$$$
2.query l r: query $$$\sum_{i=l}^{r}\lfloor a_i/b_i\rfloor$$$
Input
There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
$$$1\le n,q\le 100000, 1\le l\le r\le n$$$, there're no more than 5 test cases.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
$$$1\le n,q\le 100000, 1\le l\le r\le n$$$, there're no more than 5 test cases.
Output
Output the answer for each 'query', each one line.
Sample Input
5 12
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5
Sample Output
1
1
2
4
4
6
1
2
4
4
6
题意
$$$a_1$$$~$$$a_n$$$最开始都是0，已知$$$b_1$$$~$$$b_n$$$，一共q次操作，add操作把$$$a_l$$$~$$$a_r$$$都加1，query操作求$$$\sum_{i=l}^{r}\lfloor a_i/b_i\rfloor$$$
思路
看到add l r和query l r，就觉得这是一道线段树，可是$$$b_i$$$的存在又破坏了区间的统一性。
根据线段树的特点，为了把add和query的复杂度降低到log(n)，结点应该记录这个区间被add了几次，以及区间的和$$$\sum_{i=l}^{r}\lfloor a_i/b_i\rfloor$$$，核心问题在于怎样更新每个结点的和。
由于复杂度的原因，不能每次都重新算一遍所有结点的和，但因为每次add操作都只加1，很多结点的值都不会立刻变化；实际上区间的和的变化有这样的特点：在若干次add之前，区间和都不会改变，只有当某一项$$$a_i$$$增加到$$$b_i$$$的倍数时，区间和才会相应的+1，也就是说，在整个区间被加若干次之前，都不用更新它，只把它总共被加的次数+1；一旦某一项变成$$$a_i$$$增加到$$$b_i$$$的倍数时，区间和+1，可以认为$$$a_i$$$又恢复为0，接下来整个区间又需要若干次才会改变。如果在区间和未改变时省掉更新操作，那么复杂度将一定程度上降低。
PS：题解视频里也讲啦，总更新次数很少的
考虑这样一个标签，它记录的是，整个区间至少被加几次以后，才会让其中一项发生变化。最开始的时候，$$$a_i$$$的标签都是$$$a_i$$$，父亲结点则记录两个儿子的标签较小的那个，当标签的值变为0的时候，更新区间和，并把对应的结点的标签重新设为$$$a_i$$$，并更新路径上的所有标签；而在标签大于0的时候，则不需要对区间进行更新。
具体的来说，假设用sum[]记录区间被完整的加了几次，low[]记录区间再被加几次就需要更新，ans[]记录区间和，每次add操作，找到对应的区间i，把sum[i]++，low[i]--，如果low[i]等于0了，就向下更新，把子区间的sum加上sum[i]，low减去sum[i]，把所有low变为小于等于0的区间继续向下更新，直到把发生变化的那个$$$a_i$$$重设为0，ans[]++，然后在回溯的时候，更新路径上的所有low[]和ans[]。
代码
线段树写的有点烂。。。勉强过了
#include<stdio.h>
#include <cstring>
#define N_max 100005
typedef long long LL;
#define min(a,b) ((a)<(b)?(a):(b))
int lg[N_max << ], rg[N_max << ]
, low[N_max << ]//还要加几次
, sum[N_max << ]//已经加几次
, b[N_max << ]//bi
,ans[N_max<<]//区间的和
; #define lid(x) (x<<1)
#define rid(x) (x<<1|1)
void build(int id,int cl,int cr)
{
lg[id] = cl;
rg[id] = cr;
if (cl == cr)//build的时候完成输入，并初始化low为bi
{
scanf("%d", b + id);
low[id] = b[id]; return;
}
int left = lid(id),right=rid(id),cm= (cl + cr) >> ;
build(left, cl, cm);
build(right, cm + , cr);
low[id] = min(low[left], low[right]);//结点记录最小的low
} //low为0时，向下更新，直接处理整个区间
void addup(int id,int v)
{
sum[id] += v;
low[id] -= v;
if(lg[id]==rg[id])//遇到叶子结点
{
ans[id] += (sum[id]/ b[id]);
low[id] =b[id]- (sum[id] % b[id]);
sum[id] %=b[id] ;
}
if(low[id]<=)//low<=0说明区间内的某个数发生变化，需要继续向下更新
{
addup(lid(id), sum[id]);
addup(rid(id), sum[id]);
sum[id] = ;
ans[id] = ans[lid(id)] + ans[rid(id)];
low[id]=min(low[lid(id)], low[rid(id)]);
}
} inline void pushdown(int id)
{
if (sum[id])
{
sum[lid(id)] += sum[id];
low[lid(id)] -= sum[id];
sum[rid(id)] += sum[id];
low[rid(id)] -= sum[id];
sum[id] = ;
}
}
void add(int id,int cl,int cr,int v)
{
if (lg[id] == rg[id])//叶子结点
{
low[id]--; sum[id]++;//当进入叶子结点时，sum的值其实就是ai
ans[id] += (sum[id] / b[id]);//ai中有几个bi，ans就应该加几
low[id] = b[id] - (sum[id] % b[id]);//low变为bi-ai%bi
sum[id] %= b[id];//ai变为ai%bi
return ;
}
int left = lid(id), right = rid(id);
if(lg[id]==cl&&rg[id]==cr)//遇到匹配的区间
{
low[id]--; sum[id]++;
if(low[id]>)return ;
if ( low[id]<=)//如果low<=0，说明应该往下更新
{
//下面的更新一定是整个结点的，调用addup而不是add
addup(left, sum[id]);
addup(right, sum[id]);
//回溯时更新low和ans
low[id] = min(low[left], low[right]);
ans[id] = ans[left] + ans[right];
//因为sum已经加到子区间上了，把sum改为0
sum[id] =;
return ;
}
} //匹配区间结点
int mid = (lg[id] + rg[id]) >> ;
pushdown(id);
if (cr <= mid)
{
add(left, cl, cr, );
ans[id] = ans[left] + ans[right];
low[id] = min(low[left], low[right]);
return;
}
else if (cl > mid)
{
add(right, cl, cr, );
ans[id] = ans[left] + ans[right];
low[id] = min(low[left], low[right]);
return;
}
else
{
add(left, cl, mid,);
add(right, mid + , cr,);
ans[id] = ans[left] + ans[right];
low[id] = min(low[left], low[right]);
return;
}
} int qry(int id,int cl,int cr)
{
if (lg[id] == cl&&rg[id] == cr)
{
return ans[id];
} int mid = (lg[id] + rg[id]) >> ;
pushdown(id);
if (cr <= mid)
{
return qry(lid(id), cl, cr);
}
else if (cl > mid)
{
return qry(rid(id), cl, cr) ;
}
else
{
return qry(lid(id), cl, mid)+qry(rid(id), mid + , cr);
}
} int n,q;
char str[];
int main() { while (~scanf("%d %d" , &n,&q))
{
memset(lg, , sizeof lg);
memset(rg, , sizeof rg);
memset(low, , sizeof low);
memset(sum, , sizeof sum);
memset(ans, , sizeof ans);
build(, ,n);
int a1, a2;
for(int i=;i<q;++i)
{
scanf("%s", str);
scanf("%d %d", &a1, &a2);
if (str[] == 'a')
add(, a1, a2,);
else if (str[] == 'q')
printf("%d\n",qry(, a1, a2));
}
}
return ;
}