Answer / sandeep tiwari

void main()
{
int i,j;
int a[11]={1,1,3,3,3,5,5,5,9,9,9,9};
clrscr();
i=a[0];
printf("%d",i);

for(j=1;j<11;j++)
{
if(a[j]!=a[i])
{
if(a[j-1]==a[j])
continue;
else
printf("%d",a[j]);
i=j;
}
}
getch();
}
}

Answer / raghuram.a

#include<stdio.h>
#include<conio.h>
int getarray(int p[100], int n)
{
int c, i = 1;
for (c=1; c <n; )
if (p[c] != p[i-1])
{
p[i] = p[c];
c++;
i++;
} else
c++;
return i;
}
main()
{
int a[100],n,i,j;
printf("enter n:");
scanf("%d",&n);
printf("enter elements:");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
j=getarray(a,n);
printf("new array is:");
for(i=0;i<j;i++)
printf("\t%d",a[i]);
getch();
return 0;
}

Answer / msvinod

/*This c program aimed at printing the unique numbers in a
sorted array of repeating numbers.*/
#include<stdio.h>
main(){
int len=0,i=0,rep=0; /*Initializing the identifiers*/
printf("Enter length of array :--");
scanf("%d",&len); /*Input for array length*/
int arr[len]; /*Initializing array*/
for(i=0;i<len;i++){ /*Running loop for input of array
elements*/
printf("%dth element :--",i);
scanf("%d",&arr[i]);
}
rep=arr[0]; /*Assigning array first value to rep*/
printf("Unique array :-- [%d,",rep); /*Printing of array
starts here*/
for(i=0;i<len;i++){ /*Running loop to print all the
unique elements of array*/
if(arr[i]!=rep){
rep=arr[i]; /*Assaigning new value to rep*/
printf("%d,",rep);
}
}
printf("]\n"); /*Closing array output*/
}
/*THE END*/

Answer / suraj bhan gupta

#include<stdio.h>
main(){
int a[10],start,i,c=1;
clrscr();
printf(" Enter ten elements in sorted order : ");
for (i=0 ; i<10 ; i++){
scanf("%d",&a[i]);
}

start = a[0];
for (i=1 ; i<10 ; i++){
if (start == a[i]){
c++; // we can eliminate
variable c using only index of array
}else{
printf(" %d no %d times
present",start,c);
start = a[i];
c=1;
}
}
getch();
}

Answer / ram

main ()
{

int a[] = { 1, 1, 3, 3, 3, 5, 5, 5, 9, 9, 9, 9 };
int i = 0, key;
int size;

size = sizeof (a) / sizeof (int);
key = a[i];

for (; i < size; i++)
{
if (key != a[i])
{
printf ("\n %d", a[i]);
key=a[i];
}
}
}

Answer / john

int a[] = { 1, 3, 1, 5, 1, 9, 4, 1, 3, 2, 3, 9 };
int* b;
int i = 0;
int max = 0;
int size = sizeof(a)/sizeof(int);

//find biggest element in array
for(; i < size; i++){
if(a[i] > max)
max = a[i];
}

//fill b with 0's
b = malloc(max*sizeof(int));
for(i = 0; i < max; i++){
b[i] = 0;
}

//use b to check for duplicates
for(i = 0; i < 12; i++){
b[a[i]]++;
if(b[a[i]] == 1)
printf("%d\n", a[i]);
}

Answer / karl miro

/*this program only stores a maximum of 30 unique possitive elements*/
#include<stdio.h>
#include<conio.h>
#define SIZE 30 /*change the value of size to adjust
maximum size of array*/
void main()
{
int array[SIZE],cnt=0,temp,i,flag;
clrscr();
do
{
flag=1;
printf("Please input an element(input -1 to end):");
scanf("%d",&temp);
if(temp==-1)
{break;}
else
for(i=0;i<cnt;i++)
{
if(temp==array[i])
flag=0;
}
if(flag)
{
array[i]=temp;
cnt++;
}
}while(temp!=-1);
printf("The elements are:\n");
for(i=0;i<cnt;i++)
printf(" %d ",array[i]);
}

Answer / karthikeyan s

/* Using LinkedHashSet concept */
/* No duplicate checking */
import java.util.*;

class Main
{
public static void main(String args[])
{
Scanner obj=new Scanner(System.in);
LinkedHashSet <Integer> l1=new LinkedHashSet<Integer>();
int n=obj.nextInt();
int a=0;
while(n-->0)
{
a=obj.nextInt();
l1.add(a);
}
System.out.println(l1 + " ");
}
}

Answer / neo

didn't find the posts good enough, was looking for a very
structured code, professional:

<CODE>
{
int i, j, iLen;
char *sTmpStr;

iLen = (int) strlen(sStr);
sTmpStr = (char *)malloc(sizeof(char) * iLen);

for(i = 0;i < iLen; i++)
sTmpStr[i] = sStr[i];

for(i = 0, j = iLen - 1; i < iLen/2; i++, j--)
sStr[i] = sTmpStr[j];
free(sTmpStr);
return;
}


int main()
{
char sStr1[] = {"This is an awful string"};
char sStr2[] = {"This is a K...O...O...L string"};

printf("String 1: %s\n", sStr1);
StrRevNoTempVar(sStr1);
printf("String 1 Rev: %s\n", sStr1);

printf("String 2: %s\n", sStr2);
StrRevNoTempVar(sStr2);
printf("String 2 Rev: %s\n", sStr2);

return 0;
}
</CODE>

Answer / neo

Crap <Posted Wrong Piece>, excuse.

<CODE>
// Write, efficient code for extracting unique elements
from a sorted list of array.
// e.g. (1, 1, 3, 3, 3, 5, 5, 5, 9, 9, 9, 9) -> (1, 3, 5,
9).

#include <stdio.h>


void RemoveDumInSortedList(int *iArr, int iSize)
{
int i, j, k, iCurrNo = 0;

for(i = 0, j = i + 1; j < iSize;)
{

if(iArr[j] == iArr[i])
{
j++;
}
else
{
iArr[++i] = iArr[j];
j++;
}
}

for(j = 0; j <= i; j++)
printf("ARRAY[%d] = %d\n", j, iArr[j]);
}


int main()
{
int iArr1[] = {1, 1, 3, 3, 3, 5, 5, 5, 9};
int iArr2[] = {1, 2, 2, 3, 4, 5, 5};

RemoveDumInSortedList(iArr2, sizeof(iArr2)/sizeof
(int));

return 0;
}
</CODE>

void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof(farthest)); }

2 Answers  

#define a 10 int main() { printf("%d..",a); foo(); printf("%d..",a); return 0; } void foo() { #undef a #define a 50 }

3 Answers  

pls anyone can help me to write a code to print the values in words for any value.Example:1034 to print as "one thousand and thirty four only"

2 Answers  

#include <stdio.h> int main(void) { int a=4, b=2; a=b<<a+b>>2 ; printf("%d",a); return 0; }

0 Answers   Student,

main() { char *p; p="Hello"; printf("%c\n",*&*p); }

1 Answers  

struct aaa{ struct aaa *prev; int i; struct aaa *next; }; main() { struct aaa abc,def,ghi,jkl; int x=100; abc.i=0;abc.prev=&jkl; abc.next=&def; def.i=1;def.prev=&abc;def.next=&ghi; ghi.i=2;ghi.prev=&def; ghi.next=&jkl; jkl.i=3;jkl.prev=&ghi;jkl.next=&abc; x=abc.next->next->prev->next->i; printf("%d",x); }

1 Answers  

create a login program that ask username and password. if you input username or password 3 times wrong, the program will terminate else the program will prompt a message "congratulations"

2 Answers  

main( ) { char *q; int j; for (j=0; j<3; j++) scanf(“%s” ,(q+j)); for (j=0; j<3; j++) printf(“%c” ,*(q+j)); for (j=0; j<3; j++) printf(“%s” ,(q+j)); }

1 Answers  

programming in c lanugaue programm will errror error with two header file one as stdio.h and other one is conio.h

1 Answers  

main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }

1 Answers  

main() { int i = 100; clrscr(); printf("%d", sizeof(sizeof(i))); } a. 2 b. 100 c. 4 d. none of the above

5 Answers   HCL,

main() { char *p = "hello world"; p[0] = 'H'; printf("%s", p); } a. Runtime error. b. “Hello world” c. Compile error d. “hello world”

5 Answers   HCL,