Showing posts with label code vita 2017 solutions. Show all posts
Showing posts with label code vita 2017 solutions. Show all posts

One Egg - Code Vita 2017 | Round-2

One Egg

Problem

"One Egg" is an egg supply company which supplies eggs to retailers. They have M classes of eggs. Each class can have N number of eggs (N can be same or can vary class to class).  They accept an order via mail for X eggs. In response, they confirm if they can supply the eggs with a "Thank you" note and the number of eggs or with a "Sorry" note and the numbers of eggs they can supply. They also mention the breakdown of eggs by class they will supply. The ordered eggs are adjusted against the different classes with the most number of eggs adjusted first then the balance is adjusted against the second highest and so on.   The company is a bit superstitious as well. If the number of eggs ordered is greater than or equal to the total number of eggs in stock then they retain one egg and responds back with the "Sorry" note with total number of eggs in stock minus one and breakdown of eggs by class.  

Note: If the classes have same number of eggs then class entered first should be selected to adjust. 

Input Format

 First line contains two space-separated integers denoting the respective values of M (the number of classes of eggs) and X, the number of eggs ordered  The following M lines contain an integer each indicating the number of eggs available in each class 

Output Format  

First line should be, if X is less than total number of Eggs then Print  " Thank you, your order for X eggs is accepted"  Else if X is greater than or equal to total number of Eggs then print "  " Sorry, we can only supply (total number of Eggs in stock -1) eggs"  T hen M lines with 3 columns:  First column - Number of eggs available in each class  Second column - Eggs allocated against each class for that order  Third column - Balance Eggs against each class

Constraints

 1 ≤ M ≤ 20  N ≥ 1  X ≥ 1
 
 
Example 1  
Input  
5 150  
50  
15  
80  
10  
5  

Output  
Thank you, your order for 150 eggs are accepted  
5 0         50        0  
1 5         15        0                      
8 0        80        0  
1 0        5          5          
5           0          5  

Explanation 
 Total order of 150 eggs is less than the total number of Eggs 50+15+80+10+5 = 160. Hence the Thank you message.  150 was first adjusted against Class with first highest number of eggs 80. Balance of 150-80 = 70 was adjusted against second highest class of 50. Balance of 70-50 = 20 then adjusted against 15. Balance of 20-15 = 5 then adjusted against 10 leaving behind 5 eggs in that class.
  
Example 2
  
Input 
 
4 250  
80  
50  
70  
20  

Output 

Sorry, we can only supply 219 eggs  
8 0        80        0  
5 0        50        0                      
7 0        70        0  
2 0        19        1           

Explanation 
Total order of 250 eggs was greater than the total number of eggs 80+50+70+20 = 220. Hence the sorry message.  250 was first adjusted against Class with first highest number of eggs 80. Balance of 250-80 = 170 was adjusted against second highest class of 70.   Balance of 170-70 = 100 was then adjusted against 50. Balance of 100-50 = 50 then adjusted against 20. Since Balance is greater than 
last class of egg all but one egg is left in that last class.

Program

#include <stdio.h>
int main() {
int m,x,i,a[1000],sum=0,s;
scanf("%d %d",&m,&x);
for(i=0;i<m;i++)
{
scanf("%d",&a[i]);
sum=sum+a[i];
}
if(sum>x)
    printf("Thank you, your order for %d eggs are accepted\n",x);
else
{
    printf("Sorry, we can only supply %d eggs\n",sum-1);
    x=sum-1;
}
for(i=0;i<m;i++)
{
    if(x>=a[i])
    {
        printf("%d\t%d\t%d\n",a[i],a[i],0);
        x=x-a[i];
    }
    else if(x<a[i])
    {
        s=a[i]-x;
     printf("%d\t%d\t%d\n",a[i],x,s);
     x=0;
     }
     
     else if(x==0)
     printf("%d\t%d\t%d\n",a[i],0,a[i]);
    
}
return 0;
}


Output

Thank you, your order for 150 eggs are accepted
50 50 0
15 15 0
80 80 0
10 5 5
5 0 5


You can also run it on an online IDE: https://ide.geeksforgeeks.org/1NJLg1Fo5N

Your feedback are most welcome! If you have any doubt contact me or leave a comment!! Cheers!!!
 

Counting Rock Samples - Code Vita 2017 | round 1

Counting Rock Samples

Problem


 Juan Marquinho is a geologist and he needs to count rock samples in order to send it to a chemical laboratory. He has a problem: The laboratory only accepts rock samples by a range of its size in ppm (parts per million).
Juan Marquinho receives the rock samples one by one and he classifies the rock samples according to the range of the laboratory. This process is very hard because the number of rock samples may be in millions.
Juan Marquinho needs your help, your task is develop a program to get the number of rocks in each of the ranges accepted by the laboratory. 

Input 

 An positive integer S (the number of rock samples) separated by a blank space, and a positive integer R (the number of ranges of the laboratory); A list of the sizes of S samples (in ppm), as positive integers separated by space R lines where the ith line containing two positive integers, space separated, indicating the minimum size and maximum size respectively of the ith range.

Output 

 R lines where the ith line containing a single non-negative integer indicating the number of the samples which lie in tin the ith range. 

Constraints 

10 ≤ S ≤ 10000 1 ≤ R ≤ 1000000 1≤size of each sample (in ppm) ≤ 1000
Example 1
Input: 10 2 
           345 604 321 433 704 470 808 718 517 811 
           300 350 
           400 700
Output: 2 4
Explanation: 
There are 10 samples (S) and 2 ranges ( R ). The samples are 345, 604,…811. The ranges are 300-350 and 400-700. There are 2 samples in the first range (345 and 321) and 4 samples in the second range (604, 433, 470, 517). Hence the two lines of the output are 2 and 4
Example 2
Input: 20 3 
           921 107 270 631 926 543 589 520 595 93 873 424 759 537 458 614 725 842 575 195 
           1 100 
           50 600 
           1 1000
Output: 1 12 20
Explanation: 
There are 20 samples, and 3 ranges. The samples are 921, 107 … 195. The ranges are 1-100, 50-600 and 1-1000. Note that the ranges are overlapping. The number of samples in each of the three ranges are 1, 12 and 20 respectively. Hence the three lines of the output are 1, 12 and 20.

Program

#include <stdio.h>
int main() {
int a[1000],s,i,j,t,l1,l2,c=0;
scanf("%d",&s);
scanf("%d",&t);
for(i=0;i<s;i++)
scanf("%d",&a[i]);
for(i=0;i<t;i++)
{
scanf("%d %d",&l1,&l2);
for(j=0;j<s;j++)
{
    if((a[j]>=l1)&&(a[j]<=l2))
    c++;
}
printf("%d\n  ",c);
c=0;
}
return 0;
}

Output

20 3 
921 107 270 631 926 543 589 520 595 93 873 424 759 537 458 614 725 842 575 195 
1 100 
50 600 
1 1000

1
12
20

You can also run it on an online IDE: https://ide.geeksforgeeks.org/fkaX23kbLX

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!! Cheers!!!

Related Links: Kth factor of N
                       

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