Speed Limit

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

問題連結：POJ2017 Speed Limit

問題描述：給出一輛車在幾個時刻的速度，求車經過的路程，假設車在每兩個時刻間均為勻速，不考慮加速對路程的影響。多組用例，每組用例第一行為時刻數n，之後n行每行兩個整數分別表示車的速度和此時的時刻，以n=-1結束輸入，對於每組用例，輸出車總路程 。

解題思路：簡單模擬，具體看程式。

AC的C++程式：

#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;
}