2176Speed Limit（水题）

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 per hourTotal elapsed time in hours202306107

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 integern, 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 fort are always in strictly increasing order. A value of -1 forn signals the end of the input.

Output

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

Example input:Example output:3
20 2
30 6
10 7
2
60 1
30 5
4
15 1
25 2
30 3
10 5
-1170 miles
180 miles
90 miles

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Source: Mid-Central USA 2004

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//s=v*t;注意第一次的时间无需减前面的时间
#include<iostream>
using namespace std;
int main()
{
int speed[10],time[10];
int s,i;
int num;
while(cin>>num&&num!=-1)
{  
   s=0;
       for(i=0;i!=num;i++)
  cin>>speed[i]>>time[i];
  for(i=1;i!=num;i++)
  s+=speed[i]*(time[i]-time[i-1]);


  s+=speed[0]*time[0];
  cout<<s<<" miles"<<endl;
}
return 0;
}