TestVita | Code Vita 2015 | round 2

TestVita

Problem:

TCS is working on a new project called "TestVita". There are N modules in the project. Each module (i) has completion time denoted in number of hours (Hi) and may depend on other modules. If Module x depends on Module y then one needs to complete y before x.

As Project manager, you are asked to deliver the project as early as possible.
Provide an estimation of amount of time required to complete the project.

Input Format:

First line contains T, number of test cases.

For each test case: 

1. First line contains N, number of modules.
2. Next N lines, each contain:
   • (i) Module ID
   • (Hi) Number of hours it takes to complete the module
   • (D) Set of module ids that i depends on - integers delimited by space.

Output Format:

Output the minimum number of hours required to deliver the project.

Constraints:

1. 1 <= T <= 10

2. 0 < N < 1000; number of modules

3. 0 < i <= N; module ID 

4. 0 < Hi < 60; number of hours it takes to complete the module i

5. 0 <= |D| < N; number of dependencies

6. 0 < Dk <= N; module ID of dependencies

Sample Input and Output

SNo.	Input	Output
1	1 5 1 5 0 2 6 1 3 3 2 4 2 3 5 1 3	16

Program:

#include <stdio.h>

int main() {

int n,m[10],t[10],d[10],q,a,i,sum=0;

scanf("%d",&a);

for(q=1;q<=a;q++)

{

scanf("%d",&n);

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

    scanf("%d %d %d",&m[i],&t[i],&d[i]);

    for(i=0;i<n-1;i++)

    {

            if(d[i]==d[i+1])

            {

                if(t[i]<t[i+1])

                t[i]=0;

                else

                t[i+1]=0;

            }

    }

   for(i=0;i<n;i++)

   {

       sum=sum+t[i];

   }

    printf("%d",sum);

}

return 0;

}

Output:

1
5
1 5 0
2 6 1
3 3 2
4 2 3
5 1 3

16

You can also run it on an online IDE: https://ide.geeksforgeeks.org/eTqOGzjCr5

Your feedback are welcomed! If you have any doubts you can contact me or comment below! Cheers!

 Related Link: Bank Compare

Jumble With Numbers

Jumble With Numbers

Problem

In NASA, two researchers, Mathew and John, started their work on a new planet, but while practicing research they faced a mathematical difficulty. In order to save the time they divided their work.

So scientist Mathew worked on a piece and invented a number computed with the following formula:
A Mathew number is computed as follows using the formula:
H(n) = n(2n-1)
And scientist John invented another number which is built by the following formula which is called John number.
T(n) = n(n+1)/2
Now Mathew and John are jumbled while combining their work. Now help them combine their research work by finding out number in a given range that satisfies both properties. Using the above formula, the first few Mathew-John numbers are:
1 6 15 28 …

Input Format:

Input consists of 3 integers T1,T2,M separated by space . T1 and T2 are upper and lower limits of the range. The range is inclusive of both T1 and T2. Find Mth number in range [T1,T2] which is actually a Mathew-John number.

Line 1

T1 T2 M,where T1 is upper limit of the range, T2 is lower limit of the range and M ,where Mth element of the series is required

Constraints:

0 < T1 < T2 < 1000000

Output Format:

Print Mth number from formed sequence between T1 and T2(inclusive).

Line 1

For Valid Input,print Print Mth number from formed sequence between T1 and T2 Or No number is present at this index For Invalid Input,print Invalid Input

Sample Input and Output:

SNo.	Input	Output
1	90 150 2	120
2	20 80 6	No number is present at this index
3	-5 3 a	Invalid Input

Program:

#include <stdio.h>

int main() {

int t1,t2,m,i,j,c=0,a,b,th[100],k=0;

scanf("%d %d %d",&t1,&t2,&m);

if(t1>0&&t2>0&&m>0)

{

for(i=1;i<=t2/2;i++)

{

    a=i*((2*i)-1);

    for(j=1;j<=t2/2;j++)

    {

        b=j*(j+1)/2;

        if((a==b)&&(a>=t1&&a<=t2))

        th[k++]=a;

    }

}

   if(k<m)

    printf("No number is present at this index");

    else

    printf("%d",th[m-1]);

}
         else
         printf("Invalid Input");

return 0;

}

Output:

90 150 2

120

You can also run this in online IDE: https://ide.geeksforgeeks.org/ZndEODl1gU

Your comments and feedbacks are welcomed! If you have any doubts you can leave it on the comment! Cheers!!

Related Links: Test vita

Consecutive Prime Sum | Code Vita 2016 | round 1

Problem Description:

Some prime numbers can be expressed as Sum of other consecutive prime numbers.
For example

5 = 2 + 3
17 = 2 + 3 + 5 + 7
41 = 2 + 3 + 5 + 7 + 11 + 13

Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.
Write code to find out number of prime numbers that satisfy the above mentioned property in a given range.
Input Format:

First line contains a number N

Output Format:
Print the total number of all such prime numbers which are less than or equal to N.

Sample Input and Output

SNo.	Input	Output	Comment
1	20	2	(Below 20, there are 2 such numbers: 5 and 17). 5=2+3 17=2+3+5+7
2	15	1

Program:

#include <stdio.h>
int prime(int b)
{
    int j,cnt;
   cnt=1;
     for(j=2;j<=b/2;j++)
     {
         if(b%j==0)
         cnt=0;
     }
     if(cnt==0)
     return 1;
     else
     return 0;
}
int main() {
 int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
 scanf("%d",&n);
 for(i=2;i<=n;i++)
 {
     cnt=1;
     for(j=2;j<=n/2;j++)
     {
         if(i%j==0)
         cnt=0;
     }
     if(cnt==1)
     {
        a[k]=i;
        k++;
        }
 }
 for(i=0;i<k;i++)
 {
     sum=sum+a[i];
    c= prime(sum);
    if(c==1)
    count++;
 }
 printf("%d",count);
 return 0;
}

Output:

20

2

You can also run it on a online IDE: https://ide.geeksforgeeks.org/XcOKzTd4Ik

If you have any doubt you can contact me or comment it below! Your comments and feedbacks are also welcomed!! Cheers!!

Related Links: Jumble with Numbers