POJ 2017 Speed Limit （直叙式的简单模拟编程题目动态属性很少，难度小）

Speed Limit

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17578		Accepted: 12361

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

Sample Output

170 miles

180 miles

90 miles

Source

Mid-Central USA 2004

题目分析：输入n行，每行是速度和运行总时间。也就是说：以样例为例：运行到2小时末的时候都是20的速度，运行到6小时末的时候期间的都是30的速度，==中间4小时是30的速度

运行到7小时末的时候是10的速度，也就是说（7-6）的1小时是10的速度，输出总的行驶距离。

代码：

//直叙式的简单模拟题

#include <stdio.h>

#include <string.h>

int main()

{

    int n;

    int i, j;

    int a[20], b[20];

    while(scanf("%d", &n)&&n!=-1)

    {

        for(i=0; i<n; i++)

        {

            scanf("%d %d", &a[i], &b[i] );

        }

        int ans=0, t=0;

        for(j=0; j<n; j++)

        {

           ans=ans+a[j]*(b[j]-t);

           t=b[j];

        }

        printf("%d miles\n", ans );

    }

    return 0;

}

巴特西

POJ 2017 Speed Limit （直叙式的简单模拟编程题目动态属性很少，难度小）

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巴特西

POJ 2017 Speed Limit （直叙式的简单模拟 编程题目 动态属性很少，难度小）

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POJ 2017 Speed Limit （直叙式的简单模拟编程题目动态属性很少，难度小）