UESTC 1171 Ants Run!

Ants Run!

Time Limit: 1000 ms Memory Limit: 65535 kB 

 

 

Description

Professor Yang likes to play with ants when he is free. What? Are you asking why he plays with ants instead of others? Ah, because ant is the only non-plant living thing which can be found in Qingshuihe Campus of UESTC apart from human beings.
This time, Professor Yang caught several ants after finishing his lecture for freshmen. At the beginning of the game, he puts N ants around a plate and numbers them in clockwise order. The ants are so obedient that they run clockwise under the guide of Professor Yang on the boundary of the plate which is a circle. When one ant catches up with its previous ant, the game is over. Knowing the speed of ants, Professor Yang wants you to help him to adjust the distance between adjacent ants to make the game last longer. 

Input

The first line of the input is T (no more than 10000), which stands for the number of test cases you need to solve. 
Each test case begins with “N R”(without quotes) representing the number of ants participating the game is N and the radius of the circle is R cm. The next line lists N integers and the i-th number is the speed (cm/s) of the i-th ant in clockwise direction. All numbers are positive integer not larger than 20.

Output

If the game can last forever, print “Inf” in a single line, otherwise please output the longest time in seconds each game can last, which should be printed accurately rounded to three decimals.

Sample Input

2
3 1
1 1 1
3 1
3 2 1

Sample Output

Inf
3.142

Source

The 8th UESTC Programming Contest Preliminary

 

 

 

解题思路：
一群蚂蚁围成一圈以各自的速度沿顺时针方向跑，求大家互不相遇的最长时间。
我们最优的策略是把每一只比它前面蚂蚁跑的快的蚂蚁尽量往后放。也就是，对于总路程，除以任意相邻两只蚂蚁后面比前面跑的快的蚂蚁的速度差总和。

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define N 105
#define PI 3.1415926
double a[N];
int main()
{
    int i,n,T;
    double r,tmp;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%lf",&n,&r);
        for(i=0;ia[i])
                tmp+=a[i-1]-a[i];
        }
        if(a[n-1]>a[0])
            tmp+=a[n-1]-a[0];
        if(tmp==0)
            printf("Inf\n");
        else
            printf("%.3f\n",PI*2*r/tmp);
    }
    return 0;
}