toj_2804_1.cpp

来自「Tianjin University Online Judge 的80多道题目 」· C++ 代码 · 共 107 行

/*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

FILE *fIn = fopen( "toj_2804.in" , "r" );

void scanfNum( char num[] , int & len , char& first , bool& flag )
{
  int i;
  char in;
  num[ 0 ] = first;
  len = 1;
  //  for (  in = getchar(); in != '\n'; in = getchar() , len++ )
  for ( fscanf( fIn , "%c" , &in ); in != '\n' ; fscanf( fIn , "%c" , &in ) , len++ )
    {
      num[ len ] = in - '0';
    }
  if ( fscanf( fIn , "%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';
  fscanf( fIn , "%c" , first );
  first -= '0';
  flag = true;
  while ( flag )
    {
      scanfNum( num , len , first , flag );
      converse( num , len );
      numDi3( num , len );
      printfNum( num , len );
    }
  return 0;
}