1  
   2  /* 
   3   * A solution for "The Trip" problem.
   4   * UVa ID: 10137
   5   */
   6  #include <stdio.h>
   7  
   8  int main (int argc, const char * argv[]) {
   9  	/* number of students in the trip */
  10  	long numOfStudents;
  11  	
  12  	/* the total sum of money spent */
  13  	double total;
  14  	
  15  	/* the total amount of money to exchange in order to equalize */
  16  	double exchange;
  17  	
  18  	/* the equalized trip amount to be payed by each student */
  19  	double equalizedAmount;
  20  	
  21  	/* difference between the equalized amount and the amount spent */
  22  	double diff;
  23  	
  24  	/* sum of all negative differences */
  25  	double negativeSum;
  26  	
  27  	/* sum of all positive differences */
  28  	double positiveSum;
  29  	
  30  	/* iterator */
  31  	int i;
  32  	
  33  	while(scanf("%ld", &numOfStudents) != EOF) {
  34  				
  35  		/* 0, ends the program */
  36  		if (!numOfStudents) {
  37  			return 0;
  38  		}
  39  		
  40  		/* keeps the amount of money spent by each student */
  41  		double amountSpent[numOfStudents];		
  42  		
  43  		/* clean */
  44  		total = 0;
  45  		negativeSum = 0;
  46  		positiveSum = 0;
  47  				
  48  		for (i = 0; i < numOfStudents; i++) {
  49  			scanf("%lf\n", &amountSpent[i]);
  50  			total += amountSpent[i];
  51  		}
  52  				
  53  		equalizedAmount = total / numOfStudents;
  54  		
  55  		for (i = 0; i < numOfStudents; i++) {
  56  			/* to ensure 0.01 precision */
  57  			diff = (double) (long) ((amountSpent[i] - equalizedAmount) * 100.0) / 100.0;
  58  			
  59  			if (diff < 0) {
  60  				negativeSum += diff;
  61  			} else { 
  62  				positiveSum += diff;
  63  			}
  64  		}
  65  
  66  		/* when the total amount is even, these sums do not differ. otherwise, they differ in one cent */
  67  		exchange = (-negativeSum > positiveSum) ? -negativeSum : positiveSum;
  68  						
  69  		/* output result */
  70  		printf("$%.2lf\n", exchange);
  71  	}
  72  
  73      return 0;
  74  }

   1  
   2  /* 
   3   * A solution for the Minesweeper problem.
   4   * UVa ID: 10189
   5   */
   6  #include <stdio.h>
   7  
   8  #define MINE '*'
   9  #define MATRIX_OFFSET 1
  10  
  11  int main (int argc, const char * argv[]) {
  12  	/* number of lines of the mine field */
  13  	long n;
  14  	
  15  	/* number of colums of the mine field */
  16  	long m;
  17  	
  18  	/* iterators for retrieving the content of each field line*/
  19  	long i, j;
  20  	
  21  	/* field number identifier */
  22  	long fieldNumber = 0;
  23  
  24  	while(scanf("%ld %ld", &n, &m) != EOF) {
  25  			
  26  		/* 0 0, ends the program */
  27  		if (!n && !m) {
  28  			return 0;
  29  		} else {
  30  			if (fieldNumber > 0) {
  31  				/* 
  32  				 * workaround for the extra line on the output, 
  33  				 * which was giving me a Wrong Answer verdict..
  34  				 * so only insert an extra line if there's more
  35  				 * to come :-) 
  36  				 */
  37  				printf("\n");
  38  			}
  39  		}
  40  		
  41  		/* increment field id */
  42  		fieldNumber++;
  43  				
  44  		/* a bidimensional array representing the input mine field */
  45  		char inputField[n + MATRIX_OFFSET + 1][m + MATRIX_OFFSET + 1];
  46  		
  47  		/* a bidimensional array to output the mine field */
  48  		int outputField[n + MATRIX_OFFSET + 1][m + MATRIX_OFFSET + 1];
  49  		
  50  		/* temporary variable to retrieve mine field lines */
  51  		char line[m];
  52  		
  53  		/* init, O(n2) */
  54  		for (i = 0; i < n + MATRIX_OFFSET + 1; i++) {
  55  			for (j = 0; j < m + MATRIX_OFFSET + 1; j++) {
  56  				outputField[i][j] = 0;
  57  			}
  58  		}
  59  		
  60  		/* read mine field, O(n2) */
  61  		for (i = 0; i < n; i++) {
  62  			scanf("%s\n",line);
  63  			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
  64  				inputField[i + MATRIX_OFFSET][j] = line[j - MATRIX_OFFSET];
  65  			}
  66  		}
  67  		
  68  		/* update output, O(n2) */ 
  69  		for (i = MATRIX_OFFSET; i < n + MATRIX_OFFSET; i++) {
  70  			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
  71  				if (inputField[i][j] == MINE) {
  72  				
  73  					/* upper neighbours */
  74  					outputField[i-1][j-1]++;
  75  					outputField[i-1][j]++;
  76  					outputField[i-1][j+1]++;
  77  					
  78  					/* same level neighbours */
  79  					outputField[i][j-1]++;
  80  					outputField[i][j+1]++;
  81  					
  82  					/* lower neighbours */
  83  					outputField[i+1][j-1]++;
  84  					outputField[i+1][j]++;
  85  					outputField[i+1][j+1]++;
  86  				}
  87  			}
  88  		}
  89  		
  90  		/* present output, O(n2) */
  91  		printf("Field #%ld:\n",fieldNumber);
  92  		for (i = MATRIX_OFFSET; i < n + MATRIX_OFFSET; i++) {
  93  			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
  94  				if (inputField[i][j] == MINE)
  95  					printf("%c", MINE);
  96  				else	
  97  					printf("%d",outputField[i][j]);
  98  			}
  99  			printf("\n");
 100  		}
 101  
 102  	}
 103  
 104      return 0;
 105  }