Kattis - Speed Limit

Speed Limit

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 hour	Total elapsed time in hours
20	2
30	6
10	7

this means they drove 22 hours at 2020 miles per hour, then 6−2=46−2=4 hours at 3030 miles per hour, then 7−6=17−6=1hour at 1010 miles per hour. The distance driven is then 2⋅20+4⋅30+1⋅10=40+120+10=1702⋅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 nn, 1≤n≤101≤n≤10, followed by nn pairs of values, one pair per line. The first value in a pair, ss, is the speed in miles per hour and the second value, tt, is the total elapsed time. Both ss and tt are integers, 1≤s≤901≤s≤90 and 1≤t≤121≤t≤12. The values for ttare always in strictly increasing order. A value of −1−1 for nn 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 1	Sample Output 1
3 20 2 30 6 10 7 2 60 1 30 5 4 15 1 25 2 30 3 10 5 -1

题意

给出某一时刻的时速，求一共走了多少公里

思路

注意时间要减掉前面的时间才能计算当前时速行走的距离

代码

#include<bits/stdc++.h>
using namespace std;
int main() {
    int t;
    while(cin >> t && t != -) {
        int a[], b[];
        for(int i = ; i < t; i++) {
            cin >> a[i] >> b[i];
        }
        int sum = ;
        for(int i = ; i < t; i++) {
            sum += a[i] * (b[i] - b[i - ]);
        }
        printf("%d miles\n", sum);
    }
}