UVA 10071 Problem B Back to High School Physics-优快云博客

本文探讨了在给定初始速度和恒定加速度的情况下，如何使用物理公式计算质点在特定时间后的位移，并通过示例代码演示了实现过程。

A particle has initial velocity and constant acceleration. If its velocity after certain time is v then what will its displacement be in twice of that time?

Input

The input will contain two integers in each line. Each line makes one set of input. These two integers denote the value of v (-100 <= v <= 100) and t(0<=t<= 200) ( t means at the time the particle gains that velocity)

Output

For each line of input print a single integer in one line denoting the displacement in double of that time.

Sample Input

0 0
5 12

Sample Output

0
120

题意：t时间后质点的速度为v，那么第二个t时间内质点的位移是多少？要求输出这两个时间t内质点的位移

分析：t时间后质点的速度是v，那么开始时质点的速度为0！第二个t时间后质点的速度为2v，那么由物理公式可知在2t时间内质点的位移是：（0+2v）/2*2t,即质点的位移是2*v*t！即输出的是输入的两个正整数乘积的二倍！

AC源代码（C语言）：

#include<stdio.h>   
  
int main()  
{  
    int v,t;  
    while(scanf("%d %d", &v,&t) != EOF)  
        printf("%d\n", 2*v*t);  
    return 0;  
}

2013-03-14

转载于:https://www.cnblogs.com/fjutacm/archive/2013/02/25/2932706.html