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
Input
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
Output
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 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\n",res); 102 } 103 } 104 return 0; 105 }
原文地址:https://www.cnblogs.com/1013star/p/9597530.html