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  • 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。
     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
    ",ceil(min));
    21     }
    22     return 0;
    23 }
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  • 原文地址:https://www.cnblogs.com/jhldreams/p/3761371.html
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