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

102 5 3 2 3 5 3 4 2 410C 10 6A 2 6A 1 3C 3 4A 3 11A 4 9A 5 6C 7 3A 8 10A 2 5

53146244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 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 const int maxn=2e5+10;
  7 struct node{
  8     int l;
  9     int r;
 10     int val;
 11     int t[61];
 12 }e[maxn<<2];
 13 int a[maxn];
 14 void pushup(int cur)//注意！！
 15 {
 16     for(int i=0;i<60;i++)
 17     {
 18         int temp=e[cur<<1].t[i];//左儿子当前时间进入需要花费的时间
 19         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
 20         e[cur].t[i]=temp+e[cur<<1|1].t[nex];
 21     }
 22 }
 23 void build(int l,int r,int cur)
 24 {
 25     e[cur].l=l;
 26     e[cur].r=r;
 27     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
 28     {
 29         for(int i=0;i<60;i++)
 30         {
 31             if(i%a[l]==0)
 32             {
 33                 e[cur].t[i]=2;
 34             }
 35             else
 36                 e[cur].t[i]=1;
 37         }
 38         return;
 39     }
 40     int mid=(l+r)/2;
 41     build(l,mid,cur<<1);
 42     build(mid+1,r,cur<<1|1);
 43     pushup(cur);
 44 }
 45 void update(int tar,int val,int cur)
 46 {
 47     if(e[cur].l==e[cur].r)
 48     {
 49         a[tar]=val;
 50         for(int i=0;i<60;i++)
 51         {
 52             if(i%a[tar]==0)
 53             {
 54                 e[cur].t[i]=2;
 55             }
 56             else
 57                 e[cur].t[i]=1;
 58         }
 59         return;
 60     }
 61     int mid=(e[cur].l+e[cur].r)/2;
 62     if(tar<=mid)
 63         update(tar,val,cur<<1);
 64     else
 65         update(tar,val,cur<<1|1);
 66     pushup(cur);
 67 }
 68 int query(int pl,int pr,int cur,int t)
 69 {
 70     if(pl<=e[cur].l&&e[cur].r<=pr)
 71     {
 72         return t+e[cur].t[t%60];
 73     }
 74     int mid=(e[cur].l+e[cur].r)/2;
 75     if(pl<=mid)
 76         t=query(pl,pr,cur<<1,t);
 77     if(pr>mid)
 78         t=query(pl,pr,cur<<1|1,t);
 79 return t;
 80 }
 81 int main()
 82 {
 83     int n;
 84     cin>>n;
 85     for(int i=1;i<=n;i++)
 86         scanf("%d",&a[i]);
 87     build(1,n,1);
 88     int q,x,y;
 89     char op[10];
 90     cin>>q;
 91     while(q--)
 92     {
 93         scanf("%s %d%d",op,&x,&y);
 94         if(op[0]=='C')
 95         {
 96             update(x,y,1);
 97         }
 98         if(op[0]=='A')
 99         {
100             int res=query(x,y-1,1,0);
101             printf("%d\n",res);
102         }
103     }
104 return 0;
105 }