Traffic Jams in the Land（线段树好题）

Traffic Jams in the Land

 CodeForces - 498D

　　Some country consists of (n + 1) cities, located along a straight highway. Let's number the cities with consecutive integers from 1 to n + 1 in the order they occur along the highway. Thus, the cities are connected by n segments of the highway, the i-th segment connects cities number i and i + 1. Every segment of the highway is associated with a positive integer a_i > 1 — the period of traffic jams appearance on it.

In order to get from city x to city y (x < y), some drivers use the following tactics.

Initially the driver is in city x and the current time t equals zero. Until the driver arrives in city y, he perfors the following actions:

• if the current time t is a multiple of a_x, then the segment of the highway number x is now having traffic problems and the driver stays in the current city for one unit of time (formally speaking, we assign t = t + 1);
• if the current time t is not a multiple of a_x, then the segment of the highway number x is now clear and that's why the driver uses one unit of time to move to city x + 1 (formally, we assign t = t + 1 and x = x + 1).

You are developing a new traffic control system. You want to consecutively process qqueries of two types:

1. determine the final value of time t after the ride from city x to city y (x < y) assuming that we apply the tactics that is described above. Note that for each query t is being reset to 0.
2. replace the period of traffic jams appearing on the segment number x by value y(formally, assign a_x = y).

Write a code that will effectively process the queries given above.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of highway segments that connect the n + 1 cities.

The second line contains n integers a₁, a₂, ..., a_n (2 ≤ a_i ≤ 6) — the periods of traffic jams appearance on segments of the highway.

The next line contains a single integer q (1 ≤ q ≤ 10⁵) — the number of queries to process.

The next q lines contain the descriptions of the queries in the format c, x, y (c — the query type).

If c is character 'A', then your task is to process a query of the first type. In this case the following constraints are satisfied: 1 ≤ x < y ≤ n + 1.

If c is character 'C', then you need to process a query of the second type. In such case, the following constraints are satisfied: 1 ≤ x ≤ n, 2 ≤ y ≤ 6.

Output

For each query of the first type output a single integer — the final value of time t after driving from city x to city y. Process the queries in the order in which they are given in the input.

Examples

Input

10
2 5 3 2 3 5 3 4 2 4
10
C 10 6
A 2 6
A 1 3
C 3 4
A 3 11
A 4 9
A 5 6
C 7 3
A 8 10
A 2 5

Output

5
3
14
6
2
4
4
题意：某个国家有(n+1)(n+1)个城市(1≤n≤10^5)，城市之间有高速公路，其中第ii段高速公路是从城市i通往城市(i+1)的，而且第i条道路有一个属性值ai(2≤ai≤6)表示这段城市的拥堵状态，当我们要从城市x到城市y时候，定义时刻t从0开始，如果当前时刻t是ax的倍数，那么当前车辆就不能行驶，只能停在原地，等待当前这一秒过去（相当于从x到y需要花费2秒），否则花费1秒。
现在有两类操作，q(1≤q≤10^5)次查询： 

• 1 x y查询从城市x到城市y所需要耗费的时间；
• 2 x y修改第x个城市的拥堵值ax为y。

题解：因为题目给出2 ≤ y ≤ 6，而2-6的最小公倍数是60，相当于第0s进入和第60s进入某一个城市花费的时间相同，所以可以以60为一个周期，对每一个结点创建一个大小为60的时间数组，记录从0-59任何一个时间进入该结点所花费的时间。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=2e5+10;
struct node{
    int l;
    int r;
    int val;
    int t[61];
}e[maxn<<2];
int a[maxn];
void pushup(int cur)//注意！！
{
    for(int i=0;i<60;i++)
    {
        int temp=e[cur<<1].t[i];//左儿子当前时间进入需要花费的时间     
        int nex=(i+temp)%60;//右儿子需要的时间为左儿子花费的时间加上进入左儿子的初始时间
        e[cur].t[i]=temp+e[cur<<1|1].t[nex];
    }
}
void build(int l,int r,int cur)
{
    e[cur].l=l;
    e[cur].r=r;
    if(l==r)//对每一个结点创建一个时间数组记录任何时间进入的花费
    {
        for(int i=0;i<60;i++)
        {
            if(i%a[l]==0)
            {
                e[cur].t[i]=2;
            }
            else
                e[cur].t[i]=1;
        }
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,cur<<1);
    build(mid+1,r,cur<<1|1);
    pushup(cur);
}
void update(int tar,int val,int cur)
{
    if(e[cur].l==e[cur].r)
    {
        a[tar]=val;
        for(int i=0;i<60;i++)
        {
            if(i%a[tar]==0)
            {
                e[cur].t[i]=2;
            }
            else
                e[cur].t[i]=1;
        }
        return;
    }
    int mid=(e[cur].l+e[cur].r)/2;
    if(tar<=mid)
        update(tar,val,cur<<1);
    else
        update(tar,val,cur<<1|1);
    pushup(cur);
}
int query(int pl,int pr,int cur,int t)
{
    if(pl<=e[cur].l&&e[cur].r<=pr)
    {
        return t+e[cur].t[t%60];
    }
    int mid=(e[cur].l+e[cur].r)/2;
    if(pl<=mid)
        t=query(pl,pr,cur<<1,t);
    if(pr>mid)
        t=query(pl,pr,cur<<1|1,t);
return t;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    build(1,n,1);
    int q,x,y;
    char op[10];
    cin>>q;
    while(q--)
    {
        scanf("%s %d%d",op,&x,&y);
        if(op[0]=='C')
        {
            update(x,y,1);
        }
        if(op[0]=='A')
        {
            int res=query(x,y-1,1,0);
            printf("%d
",res);
        }
    }
return 0;
}