Traffic Jams in the Land
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 ai > 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 ax, 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 ax, 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:
- 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.
- replace the period of traffic jams appearing on the segment number x by value y(formally, assign ax = 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, ..., an (2 ≤ ai ≤ 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
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
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。
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
",res);
102 }
103 }
104 return 0;
105 }