1922 Ride to School

最新推荐文章于 2021-07-07 07:19:42 发布

原创最新推荐文章于 2021-07-07 07:19:42 发布 · 194 阅读

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探讨了北京大学万柳校区学生Charley的特殊骑行习惯及其对到达燕园时间的影响，通过分析其他学生出发时间和速度，计算Charley到达目的地的最短时间。

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

我想着应该是一道简单的数学题…然后我就卡在了第二个测试的771上
错误代码1：

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
	int n,pub=16200;
	while(cin>>n){
		if(n==0) break;
		int time=999999,t,v;
		for(int i=0;i<n;i++){
			cin>>v>>t;
			if(t>=0){
				t=(pub/v)+t;
				if(t<=time){
					time=t;
				}
			}
		}
		cout<<time<<endl;
	}
	return 0;
}

运算结果770

错误代码2

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
	int n,pub=16200;
	while(cin>>n){
		if(n==0) break;
		double time=999999,t,v;
		for(int i=0;i<n;i++){
			cin>>v>>t;
			if(t>=0){
				t=(pub/v)+t;
				if(t<=time){
					time=t;
				}
			}
		}
		printf("%.0lf\n",time);
	}
	return 0;
}

运算结果770
我寻思着这就是一个四舍五入的问题啊???
终究是我还太菜…

依旧是学习大佬的一天。

今天新学习了一个方法：

函数名: ceil

用法: double ceil(double x);

功能: 返回大于或者等于指定表达式的最小整数

头文件:math.h

返回数据类型:double

AC代码：

 #include<iostream>
#include<cmath>
using namespace std;
int main(){
	int n;
	double pub=16200;
	while(cin>>n&&n){
		int v,t;
		double time=999999,temp;
		for(int i=0;i<n;i++){
			cin>>v>>t;
			if(t>=0){
				temp=(pub/v)+t;
				if(temp<=time){
					time=temp;
				}
			}
		}
		cout<<ceil(time)<<endl;
	}
	return 0;
}