HOJ 2258 Rotating

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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 <stdio.h>#include <math.h>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;}