toj_2804.cpp

Tianjin University Online Judge 的80多道题目 .

/*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 ≤ 10^1000)

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( n / 3 )
Sample Input

5
100
99999999999999999999

Sample Output

1
33
33333333333333333333

Problem Setter: nuanran


Source: TJU Programming Contest 2007 Preliminary*/
#include<cstdio>
#include<cstring>

#define MAXLEN 1010

void scanfNum( char num[] , int & len , char& first , bool& flag )
{
  int i;
  char in;
  bool flagIn;
  num[ 0 ] = first;
  len = 1;
  flagIn = true;
  //  for (  in = getchar(); in != '\n'; in = getchar() , len++ )
  while ( flag && flagIn )
    {
      if ( scanf( "%c" , &in ) != EOF  )
	{
	  if ( in != '\n' )
	    num[ len ] = in - '0';
	  else
	    flagIn = false;
	}
      else
	flag = false;
    }
}
	  
      
/*
      if ( scanf( "%c" , &first ) != EOF )
	{
	  first = first - '0';
	  flag = true;
	}
      else
	flag = false;
	}*/

  void converse( char num[] , int const len )
  {
    int i , temp;
    for ( i = 0; i <= len / 2; i++ )
      {
	temp = num[ i ];
	num[ i ] = num[ len - 1 - i ];
	num[ len - 1 - i ] = temp;
      }
  }

  void numDi3( char num[] , int const len )
  {
    int i , base , residual , digit;
    base = 3;
    for ( residual  = 0 , i = len - 1 ; i >= 0 ; i-- )
      {
	digit = residual * 10 + num[ i ];
	residual = digit % base;
	num[ i ] = digit / base;
      }
  }

  void printfNum( char num[] , int const len )
  {
    int i;
    if ( num[ len - 1 ] != 0 )
      //    printf( "%c" , num[ len - 1 ] );
      putchar( num[ len - 1 ] + '0' );
    for ( i = len - 2; i >= 0; i-- )
      {
	putchar( num[ i ] + '0' );
      }
    putchar( '\n' );
  }      
  
int main()
{
  char num[ MAXLEN ]  , first;
  int len;
  bool flag;
  first = getchar() - '0';
  flag = true;
  while ( flag )
    {
      scanfNum( num , len , first , flag );
      converse( num , len );
      numDi3( num , len );
      printfNum( num , len );
    }
  return 0;
}