[原创]UVAlive 7308 Tom and Jerry 【几何数学】

2016-06-26 19:17:17 Tabris_ 阅读数：497

博客爬取于2020-06-14 22:44:22
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/51763961

题目链接：这里是传送阵。。

Tom and Jerry are very fond of cat and mice games, which might be rather obvious to you. Today they
are playing a very complicated game. The goals are simple as usual though, Jerry would be running
and Tom would have to catch Jerry.
However, today Jerry is running on a perfect circular
path with radius R meters, at a constant speed of
V m/s. Initially Tom is sitting at the very center of
the circle. He wants to catch Jerry as soon as possible,
but we all know, Tom is not very intelligent. Instead
of calculating an optimal direction to catch Jerry, he is
just running towards Jerry.这里写图片描述
As Jerry is also moving, the path Tom has taken
start to look like a curve (see picture above). At any
given moment, Tom’s position is between Jerry’s current
position and the center of the circle. Tom is also
moving at a constant speed of V m/s, same speed as
Jerry. Find the time (in seconds) Tom would need to
catch Jerry.
Input
Input file has T (T ≤ 10000) test cases, each case consists of two integer R and V . Here, 0 < R, V ≤
10000.
Output
For each test case, print the case number and the time Tom will need to catch Jerry. Floating point
rounding error lower than 10−5 will be ignored by the judge.
Sample Input
4
45 100
5 1547
1000 10000
5668 5467
Sample Output
Case 1: 0.70685835
Case 2: 0.00507691
Case 3: 0.15707963
Case 4: 1.62854830

题目大意 ： 就是 Jerry在半径为R的圆上 Tom在圆心 Tom面向Jerry追 Jerry和Tom的速度都是V 问多长时间Tom能追上Jerry

既然Tom 总是面向Jerry 追的 所以Tom的路径就是一个半径为R/2 的半圆 Jerry的路径就是一个半径为R的四分之一圆 时间就可求了

t=2 * PI * r/4 / v;

附本题代码

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# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <limits.h>
# include <malloc.h>
# include <ctype.h>
# include <math.h>
# include <string>
# include <iostream>
# include <algorithm>
# include <map>
# include <vector>
# include <set>

using namespace std;

const double eps = 1e-7;
const double Pi = acos(-1);


int main()
{
    int t,p=0;
    scanf("%d", &t);
    while(t--)
    {
        int r,v;
        scanf("%d%d",&r,&v);
        long double t;
        t=Pi*r/v/2;

        printf("Case %d: %llf\n",++p,t);
    }
    return 0;
}