Code For Alls..!

get the code!!

Showing posts with label c. Show all posts
Showing posts with label c. Show all posts

Monday 9 October 2017

Interlace odd / even from A to B.

October 09, 2017 c , skillrack 1 Comments
                                          Interlace odd / even from A to B.
Interlace odd / even from A to B.
Two numbers A and B are passed as input. The program must print the odd numbers from A to B (inclusive of A and B) interlaced with the even numbers from B to A.
Input Format:
The first line denotes the value of A.
The second line denotes the value of B.
Output Format:
The odd and even numbers interlaced, each separated by a space.
Boundary Conditions:
1 <= A <= 9999999
A <  B <= 9999999
Example Input/Output 1:
Input:
5
11
Output:
5 10 7 8 9 6 11
Explanation:
The odd numbers from 5 to 11 are 5 7 9 11
The even numbers from 11 to 5 (that is in reverse direction) are 10 8 6
So these numbers are interlaced to produce 5 10 7 8 9 6 11
Example Input/Output 2:
Input:
4
14
Output:
14 5 12 7 10 9 8 11 6 13 4
Explanation:
The odd numbers from 4 to 14 are 5 7 9 11 13
The even numbers from 14 to 4 (that is in reverse direction) are 14 12 10 8 6 4
So these numbers are interlaced to produce 14 5 12 7 10 9 8 11 6 13 4
(Here as the even numbers count are more than the odd numbers count we start with the even number in the output)
Example Input/Output 3:
Input:
3
12
Output:
3 12 5 10 7 8 9 6 11 4
Explanation:
The odd numbers from 3 to 12 are 3 5 7 9 11
The even numbers from 12 to 3 (that is in reverse direction) are 12 10 8 6 4
So these numbers are interlaced to produce 3 12 5 10 7 8 9 6 11 4
program:
#include<stdio.h>
#include <stdlib.h>

int main()
{
int a,b;
scanf("%d\n%d",&a,&b);
int o,e;
if(b%2==0)
e=b;
else
e=b-1;
if(a%2==1)
o=a;
else
o=a+1;
int p=0,t=b-a+1;
while(p<t)
{
    if(a%2==0&&b%2==0)
    {
        if(e>=a)
        {
            printf("%d ",e);
            p++;
            e=e-2;
        }
        if(o<=b)
        {
            printf("%d ",o);
            p++;
            o=o+2;
        }
    }
    else
    {
        if(o<=b)
        {
            printf("%d ",o);
            p++;
            o=o+2;
        }
        if(e>=a)
        {
            printf("%d ",e);
            p++;
            e=e-2;
        }
    }
}
}
By at 1 comment:
Labels: ,

Saturday 7 October 2017

Print Calendar Month in Words

October 07, 2017 c , skillrack 0 Comments
                                            Print Calendar Month in Words
The program must accept an integer value N and print the corresponding calendar month in words.
1 - January, 2 - February, .... , 11 - November, 12 - December
If any value apart from 1 to 12 is passed as input, the program must print "Invalid Input".
Input Format:
The first line denotes the value of N.
Output Format:
The first line contains the output as per the description. (The output is CASE SENSITIVE).
Boundary Conditions:
1 <= N <= 12
Example Input/Output 1:
Input:
5
Output:
May
Example Input/Output 2:
Input:
12
Output:
December
Example Input/Output 3:
Input:
105
Output:
Invalid Input
program:
#include<stdio.h>
#include <stdlib.h>

int main()
{
int n;
scanf("%d",&n);
    switch(n)
    {
        case 1:
            printf("January");
            break;
        case 2:
            printf("February");
            break;
        case 3:
            printf("March");
            break;
        case 4:
            printf("April");
            break;
        case 5:
            printf("May");
            break;
        case 6:
            printf("June");
            break;
        case 7:
            printf("July");
            break;
        case 8:
            printf("August");
            break;
        case 9:
            printf("September");
            break;
        case 10:
            printf("October");
            break;
        case 11:
            printf("November");
            break;
        case 12:
            printf("December");
            break;
        default:
            printf("Invalid Input");
    }

}
By at No comments:
Labels: ,

Friday 6 October 2017

Perfect Number

October 06, 2017 c , skillrack 0 Comments
Perfect Number
Given a positive integer N as the input, the program must print yes if N is a perfect number. Else no must be printed.
Input Format:
The first line contains N.
Output Format:
The first line contains yes or no
Boundary Conditions:
1 <= N <= 999999
Example Input/Output 1:
Input:
6
Output:
yes
Example Input/Output 2:
Input:
8
Output:
no

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

int main()
{
int n,s=0;
scanf("%d",&n);
for(int i=1;i<=n/2;i++)
{
    if(n%i==0)
    {
        s=s+i;
    }
}
if(s==n)
{
    printf("yes");
}
else
{
    printf("no");
}
}
By at No comments:
Labels: ,

Monday 25 September 2017

Pattern Printing - Diamond Numbers

September 25, 2017 c , skillrack 0 Comments
                                                   Pattern Printing - Diamond Numbers
Pattern Printing - Diamond Numbers
Given an integer N as the input, print the pattern as given in the Example Input/Output section.
Input Format:
The first line contains N.
Output Format:
2N-1 lines containing the desired pattern.
Boundary Conditions:
2 <= N <= 50
Example Input/Output 1:
Input:
3
Output:
0 0 1 0 0
0 2 0 8 0
3 0 0 0 7
0 4 0 6 0
0 0 5 0 0
Example Input/Output 2:
Input:
5
Output:
0 0 0 0 1 0 0 0 0
0 0 0 2 0 16 0 0 0
0 0 3 0 0 0 15 0 0
0 4 0 0 0 0 0 14 0
5 0 0 0 0 0 0 0 13
0 6 0 0 0 0 0 12 0
0 0 7 0 0 0 11 0 0
0 0 0 8 0 10 0 0 0
0 0 0 0 9 0 0 0 0
c program:
#include<stdio.h>
#include <stdlib.h>

int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<2*n-1;i++)
{
    for(int j=0;j<2*n-1;j++)
    {
        if(i==n-j-1)
        {
            printf("%d ",i+1);
        }
        else if(i==j-n+1)
        {
            printf("%d ",4*n-i-3);
        }
        else if(i==j+n-1)
        {
            printf("%d ",i+1);
        }
        else if(n-1==i+j+2-2*n)
        {
          printf("%d ",4*n-i-3);  
        }
        else
        {
            printf("0 ");
        }
    }
    printf("\n");
}
}
By at No comments:
Labels: ,

Tuesday 19 September 2017

Day 6: Let's Review-hackerrank-solution

September 19, 2017 c , day 2 Comments
                                             Day 6: Let's Review-hackerrank-solution
Day 6: Let's Review-hackerrank-solution
Task
Given a string, SS, of length NN that is indexed from 00 to N−1N−1, print its even-indexed and odd-indexed characters as 22 space-separated strings on a single line (see the Sample below for more detail).
Note: 00 is considered to be an even index.
Input Format
The first line contains an integer, TT (the number of test cases).
Each line ii of the TT subsequent lines contain a String, SS.
Constraints
  • 1≤T≤101≤T≤10
  • 2≤length of S≤100002≤length of S≤10000
Output Format
For each String SjSj (where 0≤j≤T−10≤j≤T−1), print SjSj‘s even-indexed characters, followed by a space, followed by SjSj‘s odd-indexed characters.
Sample Input
2
Hacker
Rank
Sample Output
Hce akr
Rn ak
Explanation
Test Case 0: S=“Hacker”
S[0]=“H”
S[1]=“a”
S[2]=“c”
S[3]=“k”
S[4]=“e”
S[5]=“r”
The even indices are 0, 2, and 4, and the odd indices are 1, 3, and 5. We then print a single line of 2 space-separated strings; the first string contains the ordered characters from S‘s even indices (Hce), and the second string contains the ordered characters from S‘s odd indices (akr).
Test Case 1: S=“Rank”
S[0]=“R”
S[1]=“a”
S[2]=“n”
S[3]=“k”
The even indices are 0 and 2, and the odd indices are 1 and 3. We then print a single line of 2 space-separated strings; the first string contains the ordered characters from S‘s even indices (Rn), and the second string contains the ordered characters from S‘s odd indices (ak).


C program:

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

int main() {

   int t;
    scanf("%d",&t);
    while(t)
        {
        char a[10000];
        scanf("%s",a);
        int i=0;
        while(a[i]!='\0')
        {if(i%2==0)
            printf("%c",a[i]);
        i++;}
       i=0;
        printf(" ");
        while(a[i]!='\0')
        {if(i%2!=0)
            printf("%c",a[i]);
        i++;}
        t--;
        printf("\n");
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}
By at 2 comments:
Labels: , ,