HOJ 2258 Rotating

http://acm.hit.edu.cn/hoj/problem/view?id=2258

正n边形外接圆半径R=1

沿贴地的那一边滚动m次

求最开始在最左下那个顶点的移动轨迹长度

 

正n边形内角A=2*pi/n

易知该正n边形滚动n次之后回到初始状态

以正五边形为例

五边形如上图滚动4次

 

每次滚动半径如左图

可以计算出一个滚动周期内的顶点走过的距离

其余见代码

#include 
#include 

const double pi = acos(-1);

int main()
{
    int t, n, m, i, c;
    double angle, r, arclength[1024], circle, totalength;

    scanf("%d",&t);
    while (t--)
    {
        scanf("%d %d", &n, &m);

        angle = 2*pi/n;
        c = m / n;
        m = m % n;
        totalength = 0, circle = 0;

        for (i = 1; i <= n / 2; i++)
        {
            r = sqrt(2 - 2 * cos(i*angle) );
            arclength[i] = arclength[n-i] = angle * r;
        }
        arclength[n] = 0;
        
        for (i = 1; i <= n; i++)
            circle += arclength[i];
        for (i = 1; i <= m; i++)
            totalength += arclength[i];

        totalength += c*circle;
        printf("%.2lf\n", totalength);
    }

    return 0;
}