Ride to School - POJ 1922 水题

Ride to School

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 18666 Accepted: 7529

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

420025-1552719030240221022340

Sample Output

780771

题意：每次给你n个人的速度和出发时间，小明（就叫小明吧）从0时间开始出发，每次都跟着最快的人走，问他到目的地需要多长时间。

思路：计算时间>=0出发的人到的最短时间，对于时间<0的来说，能追上的就能超过，追不上的就追不上了。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;int main(){ int n,i,j,v,k,p,time;  double t;  while(~scanf("%d",&n) && n>0)  { time=1000000000;    while(n--)    { scanf("%d%d",&v,&k);      t=(4.5/v*3600+k);      p=0;      while(t>0.0)      { p++;t--;      }      //printf("%d\n",p);      if(k>=0 && p<time)       time=p;    }    printf("%d\n",time);  }}