poj1922

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 Preliminary@POJ

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。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 using namespace std;

 int main()
 {
     int n;
     const double distance = 4.5;
     while(scanf("%d",&n)!=EOF&&n!=)
     {
         double x,t,v,min = 1e100;
         for(int i = ;i<n;i++)
         {
             scanf("%lf%lf",&v,&t);
             x =distance*/v+t;
             if(t>=&&x<min)
                 min = x;
         }
         printf("%.0lf\n",ceil(min));
     }
     return ;
 }