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最新推荐文章于 2022-12-02 20:39:05 发布

转载最新推荐文章于 2022-12-02 20:39:05 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/sdauyqy/p/3288081.html

本文提供了一种基于初始速度和恒定加速度条件下粒子位移的计算方法，并通过示例展示了如何利用输入的速度和时间来计算粒子在两倍时间内发生的位移。

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Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu
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Status
Description
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

　代码：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int v,t;
    while(cin>>v>>t)
    {
        cout<<2*v*t<<endl;
    }
    return 0;
}