toj 2804 The Art Gallery Problem

    The Art Gallery Problem

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Time Limit: 1.0 Seconds    Memory Limit: 65536K

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In 1973, Klee posed a fascinating problem - the art gallery problem. Imagine an art gallery room whose floor plan can be modeled by a polygon of n vertices. Klee asked: How many stationary guards are needed to guard the room? Each guard is considered a fixed point that can see in every direction. Of course a guard cannot see through a wall of the room. An equivalent formulation is to ask how many point lights are needed to fully illuminate the room. To let you fully understand this problem, there are two points needed to explain in detail.

Visibility

To make the notion of visibility precise, we say that point x can see point y (or y is visible to x) if and only if the closed segment xy is nowhere exterior to the polygon. Note that this definition permits the line-of-sight to have grazing contact with a vertex, as shown in the following figure.

A set of guards is said to cover a polygon if every point in the polygon is visible to some guard. Guards themselves do not block each other's visibility.

Max over Min Formulation

The phrase "how many" means that we need to find the maximum over all polygons of n vertices, of the minimum number of guards needed to cover the polygon.

For example, the following are two polygons of 12 vertices.

The first polygon needs at least three guards, but the second needs at least four guards. So the number of guards needed when n equals to 12 is at least 4.

You may think how hard this problem is, just like nuanran. But what's surprising is that the conclusion is very simple, it is just n/3, the integer part of n divided by 3.

Although it is such an easy job, nuanran still doesn’t want to do it himself. What a lazy guy! As an excellent student of Tianjin University, can you help him?

Input

The input will contain multiple test cases. Each test case contains a single line with a single integer n, the number of vertices of the polygon. (3 ≤ n ≤ 10¹⁰⁰⁰)

The input will be terminated by the end of file.

Output

For each corresponding n, output a single line with the number of guards needed.

Sample Input

5
100
99999999999999999999

Sample Output

1
33
33333333333333333333

Problem Setter: nuanran

Source:  TJU Programming Contest 2007 Preliminary

Problem ID in problemset: 2804

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#include  < iostream >
#define  MAX 1002
using   namespace  std;
typedef  struct  node
{
     char  num[MAX];
     int  len;
}Num;
void  Change(Num  & ob, int  n)
{
     int  i;
     for (i = 0 ;i < ob.len;i ++ )
        ob.num[i] -= n;
}
void  Rev(Num  & ob)
{
     int  s,e;
     char  temp;
    s = 0 ;
    e = ob.len - 1 ;
     while (s < e)
    {
        temp = ob.num[s];
        ob.num[s] = ob.num[e];
        ob.num[e] = temp;
        s ++ ;
         -- e;
    }
}
void  Mult_ten(Num  & ob)
{
     int  i;
     if (ob.len == 1 && ob.num[ 0 ] == 0 )
         return  ;
     for (i = ob.len;i > 0 ; -- i)
    {
        ob.num[i] = ob.num[i - 1 ];
    }
    ob.num[ 0 ] = 0 ;
    ob.len ++ ;
}
int  Cmp(Num  & a,Num  & b)
{
     if (a.len > b.len)
         return   1 ;
     else   if (a.len < b.len)
         return   - 1 ;
     else
    {
         int  i;
         for (i = a.len - 1 ;i >= 0 ; -- i)
        {
             if (a.num[i] > b.num[i])
                 return   1 ;
             else   if (a.num[i] < b.num[i])
                 return   - 1 ;
        }
    }
     return   0 ;
}
Num Sub(Num  & a,Num  & b)
{
    Num c;
    memset( & c, 0 , sizeof (c));
     int  tw = 0 ,i,l = a.len;
    c.len = a.len;
     for (i = 0 ;i <= l;i ++ )
    {
        c.num[i] = a.num[i] - b.num[i] - tw;
         if (c.num[i] < 0 )
        {
            tw = 1 ;
            c.num[i] += 10 ;
        }
         else
            tw = 0 ;
    }
     while (c.len > 1 &&! c.num[c.len - 1 ])
         -- c.len;
     return  c;
}
Num Div(Num  & a,Num  & b)
{
    Num temp,c;
     int  cnt;
    memset( & c, 0 , sizeof (c));
    memset( & temp, 0 , sizeof (temp));
     for ( int  i = a.len - 1 ;i >= 0 ; -- i)
    {
        cnt = 0 ;
        Mult_ten(c);
        Mult_ten(temp);
        temp.num[ 0 ] = a.num[i];
         while (Cmp(temp,b) >= 0 )
        {
            temp = Sub(temp,b);
            cnt ++ ;
        }
        c.num[ 0 ] = cnt;
    }
     while (c.len > 1 &&! c.num[c.len - 1 ])
         -- c.len;
     return  c;
}
int  main()
{
    Num a,b,c;
    memset( & a, 0 , sizeof (a));
    memset( & b, 0 , sizeof (b));
     while (scanf( " %s " ,a.num) != EOF)
    {
        a.len = strlen(a.num);
        strcpy(b.num, " 3 " );
        b.len = 1 ;
        Change(a, ' 0 ' );
        Change(b, ' 0 ' );
        Rev(a);
        Rev(b);

        c = Div(a,b);
        Rev(c);
        Change(c,( int )( - ' 0 ' ));

        printf( " %s\n " ,c.num);
        memset( & a, 0 , sizeof (a));
        memset( & b, 0 , sizeof (b));
    }
     return   0 ;
}