暑期实践日志（二）

这道题的思路是，将题目中所描述的齿轮看成是圆，其中两个齿轮相互咬合则是两个圆相切，那么根据判断两个圆相切的方法来判断两个齿轮间是否咬合。其中还会用到齿轮传动的规律，即传动轮的速度由主动轮来决定，根据该题的描述可知所有齿轮中只有一个是主动轮，那么其他齿轮的速度都都它决定，随度的比值就是R1/Ri。

代码：

#include <iostream>

#include <cstdio>

#include <cmath>

#include <cstring>

#include <cstdlib>

#include <algorithm>

using namespace std;

int x[1003];

int y[1003];

int r[1003];

int flag[1003];

int queue[1003];

int gcd(int a,int b)

{

    return b==0?a:gcd(b,a%b);

}

int main()

{

    //freopen("all.in","r",stdin);

    //freopen("out.txt","w",stdout);

    int T;

    scanf("%d",&T);

    while(T--)

    {

        int n;

        scanf("%d",&n);

        for(int i=1;i<=n;i++)

        {

            scanf("%d%d%d",&x[i],&y[i],&r[i]);

            flag[i]=0;

        }

        flag[1]=1;

        queue[0]=1;

        int head=0,tail=1;

        while(head!=tail)

        {

            for(int i=2;i<=n;i++)

            {

                if(queue[head]==i)

                    continue;

                if(flag[i]!=0)

                    continue;

                int t1=(x[queue[head]]-x[i])*(x[queue[head]]-x[i])+(y[queue[head]]-y[i])*(y[queue[head]]-y[i]);

                int t2=(r[queue[head]]+r[i])*(r[queue[head]]+r[i]);

                if(t1==t2)

                {

                    flag[i]=-flag[queue[head]];

                    queue[tail]=i;

                    tail++;

                }

            }

            head++;

        }

        for(int i=1;i<=n;i++)

        {

          if(flag[i]==0)

            printf("not moving
");

          else if(flag[i]==1)

          {

            int t=gcd(r[1],r[i]);

            if(r[i]/t==1)

                printf("%d ",r[1]/t);

            else

            {

                printf("%d/",r[1]/t);

                printf("%d ",r[i]/t);

            }

            printf("clockwise
");

          }

          else

          {

            int t=gcd(r[1],r[i]);

            if(r[i]/t==1)

                printf("%d ",r[1]/t);

            else

            {

                printf("%d/",r[1]/t);

                printf("%d ",r[i]/t);

            }

            printf("counterclockwise
");

          }

        }

    }

    return 0;

}