toj_2872.cpp

Tianjin University Online Judge 的80多道题目 .

/*2872.   Barbara Bennett's Wild Numbers 
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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 292   Accepted Runs: 140

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A wild number is a string containing digits and question marks (like 36?1?8). A number X matches a wild number W if they have the same length, and every non-question mark character in X is equal to the character in the same position in W (it means that you can replace a question mark with any digit). For example, 365198 matches the wild number 36?1?8, but 360199, 361028, or 36128 does not. Write a program that reads a wild number W and a number X from input, both of length n, and determines the number of n-digit numbers that match W and are greater than X.
Input
There are multiple test cases in the input. Each test case consists of two lines of the same length. The first line contains a wild number W, and the second line contains an integer number X. The length of input lines is between 1 and 10 characters. The last line of input contains a single character #.
Output
For each test case, write a single line containing the number of n-digit numbers matching W and greater than X (n is the length of W and X).
Sample Input
36?1?8
236428
8?3
910
?
5
#
Sample Output
100
0
4


Source: Asia - Tehran 2006

*/
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>

int getHead( char const str[] , char const len )
{
  char head[ 20 ];
  memcpy( head , str , len );
  head[ len ] = '\0';
//  printf( "head = %s\n" , head );
  return atoi( head );
}
int main()
{
  char w[ 20 ] , x[ 20 ] , position[ 20 ] , m[ 20 ] , xM[ 20 ] ,  flag , temp;
  int  i , j , k , headW , headX , nOfM , lenXM , lenW , xMN , result , power10;
  while ( scanf( "%c" , &flag ) != EOF && flag != '#' )
    {
      ungetc( flag , stdin );
      scanf( "%s" , w );
      scanf( "%s" , x );
      lenW = strlen( w );
      for ( nOfM = 0 , i = 0; i < lenW; i++ )
	{
	  if ( w[ i ] == '?' )
	    {
	      m[ nOfM ] = i;
	      nOfM++;
	    }
	}
      headW = getHead( w , m[ 0 ] );
      headX = getHead( x , m[ 0 ] );
      //      printf( "%d %d\n" , headW , headX );
      if ( headW < headX )
	result = 0;
      else if ( headW == headX )
	{
	  for ( i = 0; i < nOfM; i++ )
	    {
	      xM[ i ] = x[ m[ i ] ];
	    }
	  xM[ lenW ] = '\0';
	  xMN = atoi( xM );
	  result = pow( 10 , nOfM ) - 1 - xMN;
	}
      else if ( headW > headX )
	{
	  result = pow( 10 , nOfM );
	}
      printf( "%d\n" , result );
    }
  return 0;
}