toj 2970 Hackle Number

2970.   Hackle Number

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Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 183    Accepted Runs: 81

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When I was in college one of my chemistry professor invented a machine, which can make a benzin chain. A benzin chain is made by carbon atom and look like following:

Benzin chain	Benzin chain	Benzin chain
with one cycle	with two cycle	with three cycle

One of the disadvantages of that machine is it is very costly and if we do not supply it proper number of carbon then it cannot complete its operation and destroy the entire atom. Its need 1000$ for make a benzin chain of any length or unsuccessful operation. Acutely it was very inefficient machine. But he has no way rather than using it. So he asked me to make a program for him that can complete a proper number of carbon and its volume in standard temperature and pressure. I solve it that time. But now I am in reverse situation. He again ask me to extend that program so that if he give me a number of atom I have to find is it possible to build a carbon chain with that given number of atom. Would you please help me?

Input

The input contains a several test cases. Each test case will contain a single integer not more than 100 digit in one line. The EOF indicates the end of file.

Output

There will be one line output for one input. Each of the output line contain one of the two words "Possible." Or "Not possible.". If it is possible to make a carbon chain with the given number M then print "Possible." Otherwise print "Not possible.".

Sample Input

6
10
14
15
100

Sample Output

"Possible."
"Possible."
"Possible."
"Not possible."
"Not possible."

Problem Setter: M H Rasel.

Source: CUET Easy Contest

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// 规律:f(n)=4*n+2,如果满足就是可能的,否则不可能
#include  < iostream >
using   namespace  std;
typedef  struct  node
{
     char  num[ 105 ];
     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;
    }
     return ;
}
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;
}
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)
{
    Num c;
    memset( & c, 0 , sizeof (c));
     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 Mod(Num  & a,Num  & b)
{
    Num temp;
    memset( & temp, 0 , sizeof (temp));
     for ( int  i = a.len - 1 ;i >= 0 ; -- i)
    {
        Mult_ten(temp);
        temp.num[ 0 ] = a.num[i];
         while (Cmp(temp,b) >= 0 )
        {
            temp = Sub(temp,b);
        }
    }
     return  temp;
}
int  main()
{
    Num a,b,ans,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, " 2 " );
        b.len = 1 ;
        Change(a, ' 0 ' );
        Change(b, ' 0 ' );
        Rev(a);
        Rev(b);
        c = Sub(a,b);
        Change(c, - ' 0 ' );
        Rev(c);

        strcpy(b.num, " 4 " );
        b.len = 1 ;
        Change(c, ' 0 ' );
        Change(b, ' 0 ' );
        Rev(c);
        Rev(b);
        ans = Mod(c,b);
        Change(ans, - ' 0 ' );
        Rev(ans);
         if (ans.len == 1 && ans.num[ 0 ] == ' 0 ' )
            printf( " \ " Possible.\ " \n " );
         else
            printf( " \ " Not possible.\ " \n " );
        memset( & a, 0 , sizeof (a));
        memset( & b, 0 , sizeof (b));
    }
     return   0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/19/1460369.html