Codes Quiz

test

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Bank Customer Feedback Form

calculator

noreply@blogger.com (Unknown) — Mon, 28 Jun 2021 09:48:00 -0700

codextra Calculator

The input element

Click the "Submit" button and the form-data will be sent to a page on the server called "action_page.php".

Akash and GCD 1, Amazon Developer Hiring Challenge. Hackerearth

noreply@blogger.com (Unknown) — Tue, 10 Jan 2017 07:47:00 -0800

This question is taken from Hackerearth.com

Akash is interested in a new function F such that,

F(x) = GCD(1, x) + GCD(2, x) + ... + GCD(x, x)

where GCD is the Greatest Common Divisor.
Now, the problem is quite simple. Given an array A of size N, there are 2 types of queries:
1. C X Y : Compute the value of F( A[X] ) + F( A[X + 1] ) + F( A[X + 2] ) + .... + F( A[Y] ) (mod 10^9 + 7)
2. U X Y: Update the element of array A[X] = Y

Input:
First line of input contain integer N, size of the array.
Next line contain N space separated integers the elements of A.
Next line contain integer Q, number of queries.
Next Q lines contain one of the two queries.

Output:
For each of the first type of query, output the required sum (mod 10^9 + 7).

Constraints:
1 <= N <= 10⁶
1 <= Q <= 10⁵
1 <= A_i <= 5*10⁵

For Update ,
1 <= X <= N
1 <= Y <= 5*10⁵

For Compute ,
1 <= X <= Y <= N

SAMPLE INPUT

SAMPLE OUTPUT

Explanation

A[1] = 3, A[2] = 4, A[3] = 3
F(3) = GCD(1, 3) + GCD(2, 3) + GCD(3, 3) = 1 + 1 + 3 = 5.
F(4) = GCD(1, 4) + GCD(2, 4) + GCD(3, 4) + GCD(4, 4) = 1 + 2 + 1 + 4 = 8.
First query, the sum will be F(3) + F(4) = 5 + 8 = 13 (mod 10^9 + 7).
Second query, the sum will be F(3) + F(4) + F(3) = 5 + 8 + 5 = 18 (mod 10^9 + 7).
Third query, the sum will be F(3) = 5 (mod 10^9 + 7).
Fourth query will update A[1] = 4.
Fifth query, the sum will be F(4) + F(4) + F(3) = 8 + 8 + 5 = 21 (mod 10^9 + 7).
Sixth query, the sum will be F(4) + F(4) = 8 + 8 = 16 (mod 10^9 + 7).

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:1024 KB

Marking Scheme:Marks are awarded if any testcase passes.

Allowed Languages:C, C++, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Scala 2.11.8, Swift, Visual Basic

Solution:

#include <stdio.h>

int gcd (int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}

int compute(int *a, int x, int y)
{
int sum = 0;

for(int i = x; i <= y; i++)
{
sum += a[i-1];
}
printf("%d\n", sum);
return 0;
}

int update(int *a, int x, int y)
{
int sum =0;
for(int j = 1; j < y; j++)
{
sum += gcd(j, y);
}
sum += y;

a[x - 1] = sum;

return 0;
}

int main()
{
int n;
int a[100000];
int nq;
int q[100000];
int i;
int x , y;
char c;
scanf("%d", &n);
for(i = 0; i < n; i++)
{
int t;
scanf("%d", &t);
int j;
for(j = 1; j < t; j++)
{
a[i] += gcd(j, t);
}
a[i] += t;
}
scanf("%d\n", &nq);

for(i = 0; i < nq; i++)
{
scanf("%c %d %d\n", &c, &x, &y);

if(c == 'C')
{
compute(&a, x, y);
}
else if (c == 'U')
{
update(&a, x, y);
}
else
printf("input = %c\n", c);
}
return 0;
}

[Solved] Circular Array Rotation solution HackerRank

noreply@blogger.com (Unknown) — Sat, 7 Jan 2017 14:33:00 -0800

John Watson performs an operation called a right circular rotation on an array of integers,

. After performing one right circular rotation operation, the array is transformed from

Watson performs this operation

times. To test Sherlock's ability to identify the current element at a particular position in the rotated array, Watson asks

queries, where each query consists of a single integer,

, for which you must print the element at index

in the rotated array (i.e., the value of

Input Format

The first line contains

space-separated integers,

, and

, respectively.
The second line contains

space-separated integers, where each integer

describes array element

(where

).
Each of the

subsequent lines contains a single integer denoting

Constraints

Output Format

For each query, print the value of the element at index

of the rotated array on a new line.

Sample Input

Sample Output

2
3
1

Explanation

After the first rotation, the array becomes

.
After the second (and final) rotation, the array becomes

Let's refer to the array's final state as array

. For each query, we just have to print the value of

on a new line:

, so we print on a new line.
, so we print on a new line.
, so we print on a new line.

The best solution by me:

#include<stdio.h>
#include<stdlib.h>

int main(){
int n;
int k;
int q;
scanf("%d %d %d",&n,&k,&q);
int *a = malloc(sizeof(int) * n);
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
for(int a0 = 0; a0 < q; a0++){
int m;
scanf("%d",&m);
int index = (((m - k) %n )+ n) % n;
printf("%d\n", a[index]);
}

return 0;
}

Given an array of integers, print each element in reverse order as a single line of space-separated integers

noreply@blogger.com (Anonymous) — Tue, 3 Jan 2017 01:09:00 -0800

An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array,

, of size

, each memory location has some unique index,

(where

), that can be referenced as

(you may also see it written as

Given an array,

, of

integers, print each element in reverse order as a single line of space-separated integers.

Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this.

Input Format

The first line contains an integer,

(the number of integers in

).
The second line contains

space-separated integers describing

Output Format

Print all

integers in

in reverse order as a single line of space-separated integers.

Sample Input

4
1 4 3 2

Sample Output

2 3 4 1

Solution:
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int n; 
    scanf("%d",&n);
    int *arr = malloc(sizeof(int) * n);
    for(int arr_i = 0; arr_i < n; arr_i++){
       scanf("%d",&arr[arr_i]);
    }
    for (int arr_i = n - 1; arr_i >= 0; arr_i--)
        {
        printf("%d ", arr[arr_i]);
    }
    return 0;
}

What is modular programming ?

noreply@blogger.com (Anonymous) — Sat, 8 Jun 2013 23:31:00 -0700

Modular programming :

Modular programming is the concept of dividing your whole big programs in to many small programs. Using Modular Programming your program can be divided in to different Functions, files, modules etc..

lets see one example for a function:

Program without modular division :

#include <stdio.h>

#include <conio.h>
void main()

{

int a, b, c, d, avg ;

/*Taking input from user. Four numbers*/

printf("Enter 4 numbers :");

scanf("%d%d%d%d", &a, &b, &c, &d);

/*Here we are doing calculation for our program*/

avg1 = (a+b)/2;

avg2 = (c+d)/2;

avg3 = (b+c)/2;

/*Printing output*/

printf("avg1 = %d", avg1);

printf("avg2 = %d", avg2);

printf("avg3 = %d", avg3);

}

Now this program we can write like this also :
In Header.c file

#include <stdio.h>

In Func.c file

#include "FILE1.c"

int avg(int x, int y)

{

return((x + y )/2);

}

In main.c file

#include "FILE1.c"

#include "FILE2.c"

int a, b, c, d, avg1, avg2, avg3 ;

/*Taking input from user. Four numbers*/

printf("Enter 4 numbers :");

scanf("%d%d%d%d", &a, &b, &c, &d);

/*Here we are doing calculation for our program*/

avg1 = avg(a,b);

avg2 = avg(c,d);

avg3 = avg(b,c);

/*Printing output*/

printf("avg1 = %d", avg1);

printf("avg2 = %d", avg2);

printf("avg3 = %d", avg3);

}

NOTE: If you are working with windows then no need to go for different file. Make it in one file.

For compilation in gcc :
gcc -o exe main.c func.c -I .

Get more details on compilation of Gcc Click here.

Here we have divided our one long program in to modules like functions, subroutines, files.
The benefit of this type of program is that :
1. You can use same function multiple time.

2. Debugging will be easier.

3. When going for writing new program you can use these module also, no need to write once again.

4. Good memory utilities.

Some more benefits are there..If you know some more comment below.

If you have any doubt or need any type of C program or want to contribute...click on Submit program

Implement isalpha

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:56:00 -0800

/*This function will chechk whether the input is an alphabate character or not.*/
int myisalpha(int x)
{
if((x >= 65 && x <=90) || (x >= 97 && x <= 122))
return 1;
else
return 0;
}

Implement strstr

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:46:00 -0800

/*This program will find the substring s2 from the string s1
* and if found the it reutrns the base address of the substring
* if not found it send zero as a character pointer.
* */
char *mystrstr(const char *string, const char *needle)
{
const char *s1;//For retaing address of second sring for checking
const char *tar; //For retaing address of first string
/*Loop untill end*/
for(; *string != '\0'; string++){
for(s1 = string, tar = needle; *tar != '\0'; s1++, tar++){
if(*s1 != *tar)
break;
}
if(*tar == '\0')
return (char *)s1;
}
return NULL;
}

Implement strcasecmp

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:45:00 -0800

/*This function will comapre two string without cases and returns 0 if they are same or the difference of the ascii value if they are difference*/
int mystrcasecmp(const char *s1, const char *s2)
{
while(*s1!='\0' && *s2!='\0'){
if(toupper(*s1) && toupper(*s2)){
s1++;
s2++;
}
else
return(*s1-*s2);
}
return (*s1 - *s2);
}

Inplement strtok

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:44:00 -0800

/*It will tokenize the source string "src" from the place where any of the charcter of the
* target string "tar" will be found. It returns the stating address the token string and
* it should catch by a char pointer
* */

char *mystrtok(char *str, const char *tar)
{
static int i; /*For looping*/
static int len1; /*Holds the length of source string*/
static int len2; /*Holds the length of delimiter string*/
static char *s; /*This variable will holds the address of source string*/
int j ; /*For looping*/
int ptr = 0 ; /*Using for recognizing consecutive delimeter found*/
char *token; /*Variable for storing the token*/
token=(char *)malloc(25*sizeof(char)); /*Creating memory for token*/
if(token == NULL) {
printf("Out of memory space!");
exit(0);
}
/* Defining constraints for first time only*/
if(str != '\0') {
i = 0;
s = str;
len1 = strlen(str);
len2 = strlen(tar);
}
i++;
token=(s + i - 1);
/*tokenizing untill last*/
while(i <= len1+1){
ptr++;
for(j = 0 ; j <= len2 ; j++){
if(*(s + i - 1) == *(tar + j)){
*(token + ptr - 1) = '\0';
if(*token == '\0'){
token++;
ptr = 0;
break;
}
else
return token;
}
if(ptr == 0)
continue;
}
i++;
}
return '\0';
}

Implement Strcspn

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:43:00 -0800

/*
* This program will check the first unmatched characters of
* target string 'tar' to the source string 'src'
* and returns the total number of first unmatched characters.
* */
int mystrcspn(char *src, char *tar)
{
char *t; /*For retaing the address of tar string*/
int count = 0; /*For counting */
for(;*src != '\0'; src++) {

for(t = tar; *t != '\0'; t++) {
if( *src == *t )
return (count);
}
if(*t == '\0')
count++;
}
return (count);
}

Implement strspn

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:43:00 -0800

*
* This program will Count the first matched character
* from target string "tar" to the source string "src"
* and returns the value of count.
* */
int mystrspn(const char *src, const char *tar)
{
const char *t = NULL;/*For retaing the address of target string*/
int count = 0; /*For counting*/
for(;*src != '\0'; src++){
for(t = tar; *t != '\0'; t++)
if(*src == *t){
count++;
break;
}
if(*t == '\0')
return(count);
}
return(count);
}

Implement strrchr

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:41:00 -0800

/*
* This program will search for the occurance of a character
* from the last of the string.
* If found it will return the address of last occurance of that character
* */
char *mystrrchr(const char *s1, int x)
{
const char *p = NULL; /*For copying the address of the string for check*/
while(*s1 != '\0'){
if(*s1 == x)
p = s1;
s1++;
}
return(char *)p;
}

Implement strncmp

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:39:00 -0800

/*This program will compare the n-bit of a string s1 with s2. If both string are same up to n-bits it will return Zero else it will returns the difference of the ascii value of 1st unmatched character.*/

int mystrncmp(const char *s1, const char *s2, int n)

{

while(s1 != '\0' && (*s1++ == *s2++) && --n > 1)

;

return (*s1 - *s2);

}

Implement strncat

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:38:00 -0800

/*This function will append "n" character of source string "src" into target string "tar" & return Zero. */

char *mystrncat(char *tar, const char *src, int n)

{

char *t;/*retaing the address of target string*/

t = tar ;

while (*tar != '\0')

tar++;

/*Copying from src to tar string*/

while (*src != '\0' && (*tar++ = *src++) && (--n > 1));

*tar = '\0';

return t;

}

Implement strncpy

noreply@blogger.com (Anonymous) — Tue, 11 Dec 2012 09:37:00 -0800

/*This function will copy "n" character of the source string "src" to the target string "tar" and return Zero.*/

char *mystrncpy(char *tar, const char *src, int n)

{

char *t = tar; /*for Retaing the address tar string*/

/*Copying string*/

while(*src != '\0' && (*tar++ = *src++) && n--);

*tar = '\0';

return t;

}

Daily quiz no.- 04

noreply@blogger.com (Anonymous) — Sat, 3 Nov 2012 04:55:00 -0700

Daily quiz no.- 04

Answer it!

Comment your answer without watching others comments first! You may explain your answer
Share it with your friend to see what's there answer!

Daily quiz no.- 03

noreply@blogger.com (Anonymous) — Fri, 2 Nov 2012 04:50:00 -0700

Daily quiz no.- 03

How many triangles are there??

Comment your answer without watching others comments first! You may explain your answer
Share it with your friend to see what's there answer!

Daily quiz no.- 02

noreply@blogger.com (Anonymous) — Thu, 1 Nov 2012 10:00:00 -0700

Daily quiz no.- 02

How many squares are in this pic??
What would be the result??

Comment your answer without watching others comments first! You may explain your answer
Share it with your friend to see what's there answer!

C programming Quiz 31-35

noreply@blogger.com (Anonymous) — Wed, 31 Oct 2012 04:30:00 -0700

C programming Quiz 31-35

Quiz 31:
main()
{
static int x=190;
static int y=x;
if(x==y)
printf("equal");
else
printf("unequal");
return 0;
}

Quiz 32:
int increment(int i)
{
i++;
return i;
}
int main()
{
int i;
for(i=0;i<10;increment(i)) printf("%d",i); getch(); }

Quiz 33:
What are the total number of keywords in C?

30
33
32
35

Quiz 34:
What is the output of following program? Assume that stack grows towards lower addresses. Also assume that variables are pushed on to the stack in the order in which they are declared.

main()
{
int x, y;
int *p1 = &x;
int *p2 = &y;
printf("%d\n", p1-p2);
}

Quiz 35:

main()
{
int x = 10;
int y = 20;
if (x <= y == 1) { ++x; } printf("%d\n", x); }

In case of any explanation kindly comment! DON'T FORGET TO JOIN THIS BLOG.. SHARE IT WITH YOUR FRIEND!

Daily quiz no.- 01

noreply@blogger.com (Anonymous) — Wed, 31 Oct 2012 03:07:00 -0700

Daily quiz no-01

What would be the result??

Comment your answer without watching others comments first! You may explain your answer
Share it with your friend to see what's there answer!

C programming Quiz 26-30

noreply@blogger.com (Anonymous) — Sun, 28 Oct 2012 02:28:00 -0700

C programming Quiz 5-10

Quiz 26:
int main()
{
static char s[]="Hai Do It";
int i=0;
char ch;
ch=s[++i];
printf("%c\n",ch);
ch=s[i++];
printf("%c\n",ch);
ch=i++[s];
printf("%c\n",ch);
ch=++i[s];
printf("%c\n",ch);
}

Quiz 27:
main()
{
int c=5;
printf("%d",main||c);
getch();
}

Quiz 28:
main()
{
printf("%d, %d", sizeof("c"), sizeof(100));
}

Quiz 29:
int main()
{
int i = 6;
if( ((++i < 7) && ( i++/6)) || (++i <= 9))
;
printf("%d\n",i);
return 0;
}

Quiz 30:
int main()
{
int cnt = 5, a;
do
{
a /= cnt;

}
while (cnt --);
printf ("%d\n", a);
return 0;
}

In case of any explanation kindly comment! DON'T FORGET TO JOIN THIS BLOG.. SHARE IT WITH YOUR FRIEND!

C programming Quiz 21-25

noreply@blogger.com (Anonymous) — Wed, 24 Oct 2012 12:55:00 -0700

C programming Quiz 21-25

Quiz 21:
int a=1;
printf("%d %d %d", a, ++a, a++);

Quiz 22:
int main()
{

for(putchar('M');putchar('I');putchar('U'))
{
putchar('L');
}
}

Quiz 23:
int main()
{
printf("%d",'+'-'+'-'/'/'/'*'*'/'*');
}

Quiz 24:

int main(){
int i;
for(i = 0; i < 100; i++); { i++ printf("%d", i); }

Quiz 25:
int main()
{
int num=5;
switch(num)
{
default:printf("FACEBOOK\n");
case 1:printf("GOOGLE\n");break;
case 2:printf("GMAIL\n");break;
case 3:printf("ANDROID\n");break;
}
}

In case of any explanation kindly comment! DON'T FORGET TO JOIN THIS BLOG.. SHARE IT WITH YOUR FRIEND!

C programming Quiz 16-20

noreply@blogger.com (Anonymous) — Wed, 24 Oct 2012 12:13:00 -0700

C programming Quiz 16-20

Quiz 16:
main()
{
float a = 0.7;
if (a < 0.7) printf ("C"); else printf("C++"); }

Quiz 17:
main()
{
char near * near *ptr1;
char near * far *ptr2;
char near * huge *ptr3;
printf("%d %d %d",sizeof(ptr1),sizeof(ptr2),sizeof(ptr3));
}

Quiz 18:
main( ) {
int z = 4;
printf( “%d”, printf(“ %d %d “, z, z)); }

Quiz 9:
int main()
{
int x,a=2;
x=++a,++a,a++;
printf("%d %d",x,a);
return 0;
}

Quiz 20:
program1::
main()
{int i=4,x;
x=++i + ++i + ++i;
printf("%d",x);
}

program2::
main()
{
int i=4;
printf("%d",++i + ++i + ++i);
}

In case of any explanation kindly comment!

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C programming Quiz 10-15

noreply@blogger.com (Anonymous) — Fri, 19 Oct 2012 00:31:00 -0700

C programming Quiz 10-15

<--Previous Page number: 1,2,3 Next-->

Quiz 10:

main()
{
int j,ans;
j = 4;

ans = count(4);
printf("ans=%d\n",ans);
}
int count(int i)
{
if ( i < 0) return(i);
else
return( count(i-2) + count(i-1));
}

Quiz 11:

‎‎#include
int main()
{
int a=3, b = 5;
printf(&a["hi!Hello! how is this? %s\n"], &b["junk/super"]);
printf(&a["WHAT%c%c%c %c%c %c !\n"], 1["this"],2["beauty"],0["tool"],0["is"],3["sensitive"],4["CCCCCC"]);
return 0;
}

Quiz 13:
int main(){
int num,a=15;
num=- - - -a--;
printf("%d , %d",num,a);
return 0;}

Quiz 14:
int main(){
int x,i=2;
x=~-!++i;
printf("%d",x);
return 0;}

Quiz 15:

main()
{
char a =0xAA ;
int b ;
b = (int) a ;
b = b >> 4 ;
printf("%x",b);
}

<--Previous Page number: 1,2,3 Next-->

In case of any explanation kindly comment!

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Codes Quiz

test

Compound Interest Calculator

Principal*

Rate*

Time*

calculator

The input element

Akash and GCD 1, Amazon Developer Hiring Challenge. Hackerearth

[Solved] Circular Array Rotation solution HackerRank

Given an array of integers, print each element in reverse order as a single line of space-separated integers

What is modular programming ?

Modular programming :

Implement isalpha

Implement strstr

Implement strcasecmp

Inplement strtok

Implement Strcspn

Implement strspn

Implement strrchr

Implement strncmp

Implement strncat

Implement strncpy

Daily quiz no.- 04

Daily quiz no.- 03

Daily quiz no.- 03

Daily quiz no.- 02

C programming Quiz 31-35

C programming Quiz 31-35

Daily quiz no.- 01

C programming Quiz 26-30

C programming Quiz 5-10

C programming Quiz 21-25

C programming Quiz 21-25

C programming Quiz 16-20

C programming Quiz 16-20

C programming Quiz 10-15

C programming Quiz 10-15

main()
{
int j,ans;
j = 4;

ans = count(4);
printf("ans=%d\n",ans);
}
int count(int i)
{
if ( i < 0) return(i);
else
return( count(i-2) + count(i-1));
}

Codes Quiz

test

Compound Interest Calculator

Principal*

Rate*

Time*

calculator

The input element

Akash and GCD 1, Amazon Developer Hiring Challenge. Hackerearth

[Solved] Circular Array Rotation solution HackerRank

Given an array of integers, print each element in reverse order as a single line of space-separated integers

What is modular programming ?

Modular programming :

Implement isalpha

Implement strstr

Implement strcasecmp

Inplement strtok

Implement Strcspn

Implement strspn

Implement strrchr

Implement strncmp

Implement strncat

Implement strncpy

Daily quiz no.- 04

Daily quiz no.- 03

Daily quiz no.- 03

Daily quiz no.- 02

C programming Quiz 31-35

C programming Quiz 31-35

Daily quiz no.- 01

C programming Quiz 26-30

C programming Quiz 5-10

C programming Quiz 21-25

C programming Quiz 21-25

C programming Quiz 16-20

C programming Quiz 16-20

C programming Quiz 10-15

C programming Quiz 10-15

main() { int j,ans; j = 4; ans = count(4); printf("ans=%d\n",ans); } int count(int i) { if ( i < 0) return(i); else return( count(i-2) + count(i-1)); }

main()
{
int j,ans;
j = 4;

ans = count(4);
printf("ans=%d\n",ans);
}
int count(int i)
{
if ( i < 0) return(i);
else
return( count(i-2) + count(i-1));
}