POJ2017 Speed Limit【模拟】

Speed Limit

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21880		Accepted: 14948

Description

Bill and Ted are taking a road trip. But the odometer in their car is broken, so they don't know how many miles they have driven. Fortunately, Bill has a working stopwatch, so they can record their speed and the total time they have driven. Unfortunately, their record keeping strategy is a little odd, so they need help computing the total distance driven. You are to write a program to do this computation. 

For example, if their log shows 

Speed in miles perhour	Total elapsed time in hours
20	2
30	6
10	7

this means they drove 2 hours at 20 miles per hour, then 6-2=4 hours at 30 miles per hour, then 7-6=1 hour at 10 miles per hour. The distance driven is then (2)(20) + (4)(30) + (1)(10) = 40 + 120 + 10 = 170 miles. Note that the total elapsed time is always since the beginning of the trip, not since the previous entry in their log.

Input

The input consists of one or more data sets. Each set starts with a line containing an integer n, 1 <= n <= 10, followed by n pairs of values, one pair per line. The first value in a pair, s, is the speed in miles per hour and the second value, t, is the total elapsed time. Both s and t are integers, 1 <= s <= 90 and 1 <= t <= 12. The values for t are always in strictly increasing order. A value of -1 for n signals the end of the input. 

Output

For each input set, print the distance driven, followed by a space, followed by the word "miles"

Sample Input

3
20 2
30 6
10 7
2
60 1
30 5
4
15 1
25 2
30 3
10 5
-1

Sample Output

170 miles
180 miles
90 miles

Source

Mid-Central USA 2004

问题链接：POJ2017 Speed Limit

问题描述：给出一辆车在几个时刻的速度，求车经过的路程，假设车在每两个时刻间均为匀速，不考虑加速对路程的影响。多组用例，每组用例第一行为时刻数n，之后n行每行两个整数分别表示车的速度和此时的时刻，以n=-1结束输入，对于每组用例，输出车总路程 。

解题思路：简单模拟，具体看程序。

AC的C++程序：

#include<iostream>

using namespace std;

int main()
{
	int n;
	while(scanf("%d",&n)&&n!=-1)
	{
		int sum=0,v,t1=0,t2;
		while(n--)
		{
			scanf("%d%d",&v,&t2);
			sum+=v*(t2-t1);
			t1=t2;
		}
		printf("%d miles\n",sum);
	}
	return 0;
}