Answer / senthil

// correcting Cfuzz answer

#include <stdio.h>
#include <stdlib.h> // for rand function

#define ROWS 50
#define COLS 50

int main(void)
{
int A[ROWS][COLS];
int i=0, j=0;
int zero_cnt = 0;
int one_cnt = 0;

/* Initializing*/
for(i=0; i < ROWS; i++) {
for(j=0; j < COLS; j++) {
A[i][j] = rand() % 2;
}
}

// here one loop is sufficient
for(i=0; i < ROWS; i++) {
printf("A[%d][%d] = %2d\n", i, i, A[i][i]);
if(A[i][i] == 0)
{
zero_cnt++;
}
else //if(A[i][i] == 1)
{
one_cnt++;
}
}
printf("\nNumber of zeros in the diagonal = %d", zero_cnt);
printf("\nNumber of ones in the diagonal = %d", one_cnt);
}

Answer / pushpender singh

#include<stdio.h>

main
{
int a,b,c;
a[50][50];
b[][];
for(c=0;c>50;c+a)
{
a=a+c;
b=a;
}
printf("%d",b[][]);
}
}

Answer / cfuzz

/* THIS CODE IS WRONG...CUZ I CAN'T COUNT ZEROS*/


#include <stdio.h>

#define ROWS 50
#define COLS 50

int count_occur(int A[], int num_elements, int value);


int main(void)
{
int A[ROWS][COLS];
int i=0, j=0;
int num_occ, value=0;

/* Initializing*/

for(i=0; i < ROWS; i++) {
for(j=0; j < COLS; j++) {
A[i][j] = 0;
A[i][j] = 1;
A[i][j] = rand() % 2;

}
}

for(i=0; i < ROWS; i++) {
for(j=0; j < COLS; j++) {

if (i == j){
printf("%2d", A[i][j]);
}
}



}

for(value=0; value<1; value++)
{

num_occ = count_occur(A, 50, value);

if (value = 1){

printf("\n\nThe value %d was found %d times.\n", value,
num_occ);
}

else if (value = 0){

printf("\n\nThe value %d was found %d times.\n", value,
num_occ);
}


}

}

int count_occur(int A[], int num_elements, int value)
/* checks array a for number of occurrances of value */
{
int i, count=0;
for (i=0; i<num_elements; i++)
{
if (A[i] == value)
{
++count; /* it was found */
}
}
return(count);
}

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