USACO 3.2 Spinning Wheels

USACO 3.2 Spinning Wheels

最多转360秒，各个轮子就复位了。所以在360秒内，每过一秒检测一下是否可以透过光即可。
用一个数组记录能透过0-359的轮子的个数，当在某一度轮子的个数达到了5，则说明光可以透过，输出即可。
否则说明是不可能有光透过。

#include  < iostream >
#include  < fstream >

using   namespace  std;

ifstream in( " spin.in " );
ofstream out( " spin.out " );

struct  wedge{
     int  start;
     int  extent;
};

int  speed[ 5 ];
wedge wedges[ 5 ][ 5 ];
int  wedge_num[ 5 ];

void  onesecond()
{
    for ( int  i = 0 ;i < 5 ; ++ i){
        for ( int  j = 0 ;j < wedge_num[i]; ++ j){
           wedges[i][j].start += speed[i];
           wedges[i][j].start %= 360 ;
       }
   } 
}

bool  isok()
{
     int  tmp[ 360 ];

    memset(tmp, 0 , sizeof (tmp));

     for ( int  i = 0 ;i < 5 ; ++ i){
         for ( int  j = 0 ;j < wedge_num[i]; ++ j){
             for ( int  k = 0 ;k <= wedges[i][j].extent; ++ k)
                tmp[(wedges[i][j].start + k) % 360 ] ++ ;
        }
    }

     for ( int  i = 0 ;i < 360 ; ++ i)
         if (tmp[i] == 5 )
             return   true ;

     return   false ;
}

void  solve()
{
     for ( int  i = 0 ;i < 5 ; ++ i){
         in >> speed[i];
         in >> wedge_num[i];
         for ( int  j = 0 ;j < wedge_num[i]; ++ j){
             in >> wedges[i][j].start >> wedges[i][j].extent;
        }
    }

     if (isok()){
         out << 0 << endl;
         return ;
    }

     for ( int  i = 1 ;i <= 360 ; ++ i){
        onesecond();
         if (isok()){
             out << i << endl;
             return ;
        }
    }
     out << " none " << endl;
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}