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.
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 one line for each case: the arrival time of
Charley. Round up (ceiling) the value when dealing with a fraction.
/***********************************************************************************************************************
题意：有一痴汉(笑) 骑车非要跟在其他人后面 给出他们的速度和出发时间 求出到4.5KM的地方的时间
思路：画个v-t图 很容易看出来总时间就是最早到的那个同学到达的时间减去这痴汉(笑)出发的时间
***********************************************************************************************************************/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
int a[10000+10];
int main()
{
//freopen("data.in" , "r" , stdin);
int n;
while(scanf("%d", &n) && n)
{
int ans = 0x7fffffff;
for(int i = 0 ; i < n ; i ++)
{
int v , t;
scanf("%d %d", &v , &t);
if(t < 0)
continue;
int temp =(int)ceil(4.5 / v * 3600.0) + t;
ans = min(ans , temp);
}
printf("%d\n" , ans);
}
return 0;
}
