Answer / lee

#include<stdio.h>
#include <stdlib.h>
#include<conio.h>
int main()
{
int A[50][50],i,j,z=0,k,o=0;
for( i = 0 ; i < 50 ; i++ )
{
for( j = 0 ; j < 50 ; j++ )
{
A[i][j]=rand() % 2;
}
}

printf("The diagonal elements are : \n");
for(i=0;i<50;i++)
{
for(j=0;j<50;j++)
{
if(i==j)
{
printf("%d\t",A[i][j]);
if(A[i][j]==0)
z++;
else
o++;
}
}
}

printf("The no. of zeroes : \t %d\nThe no. of ones : \t %d",z,o);
getch();
return 0;
}

Answer / cfuzz

/* this is my code, only thing I'm missing is counting
zeros...WHAT THE HELL am I doing WRONG?*/

#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] = 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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3 Answers  

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