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;

        }

    }

}

}

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");

    }

}

Two string values S1 and S2 are passed as input. The program must print the output as S1 Technologies S2.

Input Format:
The first line denotes the value of S1.
The second line denotes the value of S2.

Output Format:
The first line contains the output as per the description. There must be exactly one space between the words.

Boundary Conditions:
Length of S1 is from 1 to 100.
Length of S2 is from 1 to 100.

Example Input/Output 1:
Input:
Wipro
Bangalore

Output:
Wipro Technologies Bangalore

Example Input/Output 2:
Input:
ABC
Mumbai

Output:
ABC Technologies Mumbai

Program:

print(input()+" Technologies "+input())

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");

}

}

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");

}

}

                                                                          Three Strings

Given N string values, the program must print 3 string values as the output as described in the Example Input/Output section.

Input format:
The first line will contain N denoting the number of string values.
Next N lines will contain the N string values.

Output format:
Three lines containing string values as described in the Example Input/Output section.

Example Input/Output 1:
Input:
3
JOHN
JOHNY
JANARDHAN

Output:
JJOJAN
OHHARD
NNYHAN

Example Input/Output 2:
Input:
4
JOHN
JOHNY
JANARDHAN
MIKESPENCER

Output:
JJOJANMIKE
OHHARDSPE
NNYHANNCER

C++ Program:

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

string s,s1="",s2="",s3="";

int n;

cin>>n;

for(int i=0;i<n;i++)

{

    cin>>s;

    s1+=s.substr(0,i+1);

    s2+=s.substr(i+1,s.length()-2*(i+1));

    s3+=s.substr(s.length()-(i+1));

}

cout<<s1<<"\n"<<s2<<"\n"<<s3;

}

                                                Minimum Distance Between Alphabets (Id-3109)

Given a string S and two alphabets C1 and C2 present in S, find the minimum distance D  between C1 and C2 in S.

Input Format:
The first line will contain S.
The second line will contain C1 and C2 separated by a space.

Output Format:
The first line will contain D.

Boundary Conditions:
2 <= Length of S <= 100

Example Input/Output 1:
Input:
spaceship
c s

Output:
1

c++ Program:

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

string s;

cin>>s;

char a,b;

cin>>a;

cin>>b;

int m=s.length(),flag=0,d=0;

for(int i=0;i<s.length();i++)

{

    if(s[i]==a||s[i]==b)

    {

        if(flag==1&&d!=0)

        {

        if(d<m)

        m=d;

        d=0;

        }

        else

        flag=1;

    }

    if(flag==1)

    d++;

}

cout<<m-1;

}

Given a R*C matrix (R - rows and C- Columns), print the sum of the values in each row as the output.

Input Format:
First line will contain R and C separated by a space.
Next R lines will contain C values, each separated by a space.

Output Format:
R lines representing the sum of elements of rows 1 to R.

Boundary Conditions:
2 <= R, C <= 50

Example Input/Output 1:
Input:
4 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

Output:
10
26
42
58

 C program:

#include<stdio.h>

#include <stdlib.h>

int main()

{

int m,n,t,s;

scanf("%d%d",&m,&n);

for(int i=0;i<m;i++)

{

    s=0;

    for(int j=0;j<n;j++)

{

   scanf("%d",&t);

   s+=t;

}

printf("%d\n",s);

}

}

                                                Head Count - Birds and Animals

In a zoo there are some birds and animals. All birds have two legs and all animals have four legs.
GIven the head count and leg count of both birds and animals taken together, the program must print the head count of birds and animals separated by a space as output.

Input Format:
First line will contain the integer value H representing the head count of both birds and animals taken together.
Second line will contain the integer value L representing the leg count of both birds and animals taken together.

Output Format:
First line will contain the integer values of the head count of birds and animals separated by a space.

Constraints:
0 < H < 1000
1 < L < 2000

Sample Input/Output:

Example 1:
Input:
27
84

Output:
12 15

Explanation:
There are 12 birds and 15 animals.

Example 2:
Input:
114
256

Output:
100 14

Explanation:
There are 100 birds and 14 animals.

C program:

#include<stdio.h>

#include <stdlib.h>

int main()

{

int h,l;

scanf("%d%d",&h,&l);

int a=l/2-h;

printf("%d %d",h-a,a);

}