Ride to School
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 18704 | Accepted: 7552 |
Description
Many graduate students of Peking University are
living in Wanliu Campus, which is 4.5 kilometers from the main campus – Yanyuan.
Students in Wanliu have to either take a bus or ride a bike to go to school. Due
to the bad traffic in Beijing, many students choose to ride a bike.
We
may assume that all the students except "Charley" ride from Wanliu to Yanyuan
at a fixed speed. Charley is a student with a different riding habit – he
always tries to follow another rider to avoid riding alone. When Charley gets
to the gate of Wanliu, he will look for someone who is setting off to Yanyuan.
If he finds someone, he will follow that rider, or if not, he will wait for
someone to follow. On the way from Wanliu to Yanyuan, at any time if a faster
student surpassed Charley, he will leave the rider he is following and speed up
to follow the faster one.
We assume the time that Charley gets to the
gate of Wanliu is zero. Given the set off time and speed of the other students,
your task is to give the time when Charley arrives at Yanyuan.
Input
There are several test cases. The first line of
each case is N (1 <= N < = 10000) representing the number of riders
(excluding Charley). N = 0 ends the input. The following N lines are
information of N different riders, in such format:
Vi [TAB] Ti
Vi
is a positive integer <= 40, indicating the speed of the i-th rider (kph,
kilometers per hour). Ti is the set off time of the i-th rider, which is an
integer and counted in seconds. In any case it is assured that there always
exists a nonnegative Ti.
Output
Output one line for each case: the arrival time of
Charley. Round up (ceiling) the value when dealing with a fraction.
Sample Input
4
20 0
25 -155
27 190
30 240
2
21 0
22 34
0
Sample Output
780
771
Source
Beijing 2004 [email protected]
简单模拟，这题目开始看的时候觉得有些复杂，
看到了讨论区别人的思路，但是自己想想想总觉得有问题，可能出题的人也没注意到，或者说题目没叙说清楚吧，不过这题精度要是卡的很准的。
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 using namespace std;
5
6 int main()
7 {
8 int n;
9 const double distance = 4.5;
10 while(scanf("%d",&n)!=EOF&&n!=0)
11 {
12 double x,t,v,min = 1e100;
13 for(int i =0 ;i<n;i++)
14 {
15 scanf("%lf%lf",&v,&t);
16 x =distance*3600/v+t;
17 if(t>=0&&x<min)
18 min = x;
19 }
20 printf("%.0lf\n",ceil(min));
21 }
22 return 0;
23 }