zoukankan      html  css  js  c++  java
  • Poj 2017 Speed Limit(水题)

    一、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"
    二、题解
            太难得题做得头痛,太容易的题呢又做得没意思,哎!
    三、java代码
    import java.util.Scanner;
    
    public class Main {
    	 
        public static void main(String[] args) { 
          Scanner sc = new Scanner(System.in);
          int n,i,j,a[],b[],sum;
          while((n=sc.nextInt())!=-1){
        	  a=new int[n];
        	  b=new int[n];
        	  sum=0;
        	  for(i=0;i<n;i++){
        		  a[i]=sc.nextInt();
        		  b[i]=sc.nextInt();
        	  }
        	  sum=a[0]*b[0];
        	  i=0;
    		  j=i+1;
        	  while(j<n){
        		  sum+=(b[j]-b[i]) * a[j];
        		  i++;
        		  j++;
        	  }
        	  System.out.println(sum+" miles");
          }
        } 
    } 
    


  • 相关阅读:
    [WPF]Win10便签软件
    [WPF]使用Fody提高效率
    [WPF]限制程序单例运行
    [WPF]创建系统栏小图标
    Run Performance Testing Which Was Distributed To Multiple Test Agents
    FxZ,C#开发职位面试测试题(30分钟内必须完成)
    BYS推荐MS前端PhoneCall面试问题整理-2
    BYS推荐MS前端PhoneCall面试问题整理-1
    LxNx前端F2F面试问题整理
    复杂DIV交错布局
  • 原文地址:https://www.cnblogs.com/AndyDai/p/5135306.html
Copyright © 2011-2022 走看看