USACO Section 3.2 Spinning Wheels - 纯模拟的大水题

   主要还是题意~~有点长~感觉确实也讲得不是太清晰... 反正这五个轮子有那么一个点是相通过去的就ok了...并且要在整数秒时相通..

   纯模拟..因为不管速度是多少..反正在360秒后都会转回来...那么就模拟这360秒..每动一秒..判断下有无相通的点...

Program:

/*  
ID: zzyzzy12   
LANG: C++   
TASK: spin
*/      
#include<iostream>      
#include<istream>  
#include<stdio.h>      
#include<string.h>      
#include<math.h>      
#include<stack>  
#include<algorithm>      
#include<queue>   
#define oo 1000000000  
#define ll long long   
using namespace std;
struct node
{
      int speed,num,s[6][3];      
}a[6];
int i,j;
void getanswer()
{
      int p,k,t,have[380];
      for (p=0;p<360;p++)
      {
            memset(have,0,sizeof(have));
            for (i=1;i<=5;i++)
              for (j=1;j<=a[i].num;j++)
              {
                    k=(a[i].s[j][0]+p*a[i].speed)%360;
                    t=(k+a[i].s[j][1])%360;
                    while (k!=t)
                    {
                         have[k]++;
                         k++;
                         if (k==360) k=0;      
                    }
                    have[t]++;
              }
            for (i=0;i<360;i++)
            if (have[i]==5) 
            {
                   printf("%d\n",p);
                   return;              
            }
      }
      printf("none\n");
      return;
}
int main()  
{  
      freopen("spin.in","r",stdin);    
      freopen("spin.out","w",stdout);   
      for (i=1;i<=5;i++)
      {
            scanf("%d%d",&a[i].speed,&a[i].num); 
            for (j=1;j<=a[i].num;j++) scanf("%d%d",&a[i].s[j][0],&a[i].s[j][1]); 
      } 
      getanswer();
      return 0;     
}