I banged out a brute-force iterative solution pretty quickly in C.

It's not general, but could be made so pretty easily.

  #include <stdio.h>
  
  #define NUM_COINS		5
  #define GOAL_IN_CENTS	100
  
  unsigned char ucCoinVal[] = { 1, 5, 10, 25, 50 };
  
  unsigned char isGoodSet( unsigned char aucCoins[] );
  
  int main( void )
  {
  	int i;
  	unsigned char max[NUM_COINS];
  	unsigned char aucCoins[NUM_COINS];
  	int iTotal = 0;
  
  	for ( i = 0; i < NUM_COINS; i++ )
  	{
  		max[i] = GOAL_IN_CENTS / ucCoinVal[i];
  		printf( "\nCoin %3d cents has max %3d for goal", ucCoinVal[i], max[i] );
  	}
  
  	printf( "\n" );
  
  	for ( aucCoins[0] = 0; aucCoins[0] <= max[0]; aucCoins[0]++ )
  	{
  		for ( aucCoins[1] = 0; aucCoins[1] <= max[1]; aucCoins[1]++ )
  		{
  			for ( aucCoins[2] = 0; aucCoins[2] <= max[2]; aucCoins[2]++ )
  			{
  				for ( aucCoins[3] = 0; aucCoins[3] <= max[3]; aucCoins[3]++ )
  				{
  					for ( aucCoins[4] = 0; aucCoins[4] <= max[4]; aucCoins[4]++ )
  					{
  						if ( isGoodSet( aucCoins ) )
  						{
  							iTotal++;
  							printf( "\nGood set : %3d %3d %3d %3d %3d",
  								    aucCoins[0], aucCoins[1], aucCoins[2], aucCoins[3], aucCoins[4] );
  						}
  					}
  				}
  			}
  		}
  	}
  
  	printf( "\nFound %d solutions.\n", iTotal );
  
  }
  
  // Function to determine if we have a good set of coins (adds up to the goal).
  unsigned char isGoodSet( unsigned char aucCoins[] )
  {
  	int i;
  	int iSum = 0;
  	
  	for ( i = 0; i < NUM_COINS; i++ )
  	{
  		iSum += ( aucCoins[i] * ucCoinVal[i] );
  	}
  
  	return ( iSum == GOAL_IN_CENTS ) ? 1 : 0;
  }