Question 4

Problem : Water Management There are N tubs of water, numbered from 1 to N. Initially there is a few litres of water in each tub. Each water tub has 2 taps attached to it. Incoming Tap with speed x litres / second and Outgoing Tap with speed y litres / second. Let water(i) denote the final volume of water in ith tub. Amit wants to attain such a situation that water(i) < water(i+1) for 1<=i <=N. i.e. water in each tub must be less than water in the tub next to it. He wants to do this as quickly as possible. You task is to find out and tell Amit, what is the minimum number of seconds required to attain in this situation.

  Input Format: First line will contains the number of tubs, denoted by N Next N lines contain a tuple with 3 integers delimited by white space. The tuple contents are 

• Wi - volume of water present initially in ith tub (in litres)
• x - speed of incoming tap of ith tub ( in litres/second)
• y - speed of outgoing tap of ith tub ( in litres/second)

  Output Format: Minimum time in seconds, needed to arrange water in N tubs, in ascending order of volume.
  Constraints: 2 <= N <= 100000 0 <= W <= 1000000000 (1 billion) for each tub 1 <= x <= 10000 for each tub1 <= y <= 10000 for each tub A tap can be used only for integral number of seconds i.e. you cannot use a tap for 0.3 or 0.5 seconds. It can be used either for 1,2,3.... Seconds Capacity of each tub is infinite.Volume of water in any tub cannot be less than zero at any point of time. Any number of taps can be turned on or off simultaneously.   Sample Input and Output 

  Explanation for sample input and output 1: Here we have 3 tubs with following information about each of them

Tub Number	Initial Volume	Incoming Tap speed	Outgoing Tap speed
Tub 1	2	3	3
Tub 2	3	4	5
Tub 3	3	5	5

Initially tub 2 and 3 have same volume of water. So he will just turn on the Incoming Tap at Tub 3 for one second. After one second the water in 3rd tub will be 8 litres. Since 2 < 3 < 8, the answer is 1 second.

  Explanation for sample input and output 2: Here we have 3 tubs with following information about each of them

Tub Number	Initial Volume	Incoming Tap speed	Outgoing Tap speed
Tub 1	6	2	1
Tub 2	4	1	3
Tub 3	4	1	4

As we can see that it is impossible to do the task in one second. But it can be done in two second as below. Turn on the outgoing tap at tub 1. So after two seconds water level will reduce to 4 litres. Turn on the incoming tap at tub 2 only for one second. So water level will be 5 litres. Turn on the incoming tap at tub 3 for two seconds. So water level will be 6 litres. Since 4 < 5 < 6, the answer is 2 seconds.

program code

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
 int i,j=0,n,t=0;
 long *w,*x,*y;
 scanf("%d",&n);
 w=(long*)calloc(n,sizeof(long));
 x=(long*)calloc(n,sizeof(long));
 y=(long*)calloc(n,sizeof(long));
 for(i=0;i<n;i++)
 {
  scanf("%d%d%d",&w[i],&x[i],&y[i]);
 }
 for(i=0;i<n;i++)
 {
  if(w[i]<w[i+1])
  {
   j=0;
   continue;
  }
  else
  {
    goto loop;
  }
 }
 if(j==0)
 {
  goto ans;
 }
loop: for(i=0;i<n;i++)
 {
  t=1;
 cmp:  if(w[i]-y[i]>0&&i==0)
  {
   w[i]=w[i]-y[i];
  }
  else if(w[i]-y[i]>0 && w[i]-y[i]>w[i-1])
  {
   w[i]=w[i]-y[i];
  }
  if(w[i+1]<=w[i])
  {
    w[i+1]=w[i+1]+x[i+1];
    if(w[i+1]<=w[i])
    {
     t++;
     goto cmp;
    }
  }
  if(j<=t)
   j=t;
  }
ans:  printf("%d",j);
  getch();
}

Problem : Super ASCII String Checker

In the Byteland country a string "S" is said to super ascii string if and only if count of each character in the string is equal to its ascii value.

In the Byteland country ascii code of 'a' is 1, 'b' is 2 ...'z' is 26.

Your task is to find out whether the given string is a super ascii string or not.

Input Format:

First line contains number of test cases T, followed by T lines, each containing a string "S".

Output Format:

For each test case print "Yes" if the String "S" is super ascii, else print "No"

Constraints:

1<=T<=100

1<=|S|<=400, S will contains only lower case alphabets ('a'-'z').

Sample Input and Output

SNo.	Input	Output
1	2 bba scca	Yes No

Explanation:

In case 1, viz. String "bba" -
The count of character 'b' is 2. Ascii value of 'b' is also 2.
The count of character 'a' is 1. Ascii value of 'a' is also 1.
Hence string "bba" is super ascii.

program code:

#include<stdio.h>
#include<string.h>
#include<conio.h>
void main()
{
 int a[26],l,i,k,j,c;
 char s[50];
 clrscr();
 printf("enter the number of test cases\n");
 scanf("%d",&j);
 for(l=0;l<j;l++)
 {
  for(i=0;i<26;i++)
  {
   a[i]=0;
  }
  printf("enter the string\n");
  scanf("%s",&s);
  for(i=0;i<strlen(s);i++)
  {
   k=(int)s[i]-97;
   a[k]++;
  }
  for(i=0;i<26;i++)
  {
   if(a[i]==i+1||a[i]==0)
    c=1;
   else
   {
    c=0;
    break;
   }
  }
  if(c!=0)
   printf("\tyes\n");
  else
   printf("\tno\n");
 }
 getch();
}

Question 3

Problem : Sheldon Cooper and his beverage paradigm Sheldon Cooper, Leonard Hofstadter and Penny decide to go for drinks at Cheese cake factory. Sheldon proposes to make a game out of this. Sheldon proposes as follows, 

• To decide the amount of beverage they plan to consume, say X.
• Then order for a random number of different drinks, say {A, B, C, D, E, F} of quantities {a, b, c, d, e, f} respectively.
• If quantity of any three drinks add up to X then we'll have it else we'll return the order. E.g. If a + d + f = X then True else False

You are given 

1. Number of bottles N corresponding to different beverages and hence their sizes
2. Next N lines, contain a positive integer corresponding to the size of the beverage
3. Last line consists of an integer value, denoted by X above

Your task is to help find out if there can be any combination of three beverage sizes that can sum up to the quantity they intend to consume. If such a combination is possible print True else False
  Input Format: 

1. First line contains number of bottles ordered denoted by N
2. Next N lines, contains a positive integer Ai, the size of the ith bottle
3. Last line contains the quantity they intend to consume denoted by X in text above

  Output Format: True, if combination is possible False, if combination is not possible

  Constraints: N >= 3 Ai > 0 1 <= i <= N X > 0   Sample Input and Output 

SNo.	Input	Output
1	6 1 4 45 6 10 8 22	True
2	4 1 3 12 4 14	False

  Explanation for sample input and output 1: The sum of 2nd, 5th and 6th beverage size is equal to 22. So the output will be True.
  Explanation for sample input and output 2: Since no combination of given beverage sizes sum up to X i.e. 14, the output will be False

program code:(hint:set of all possible combinations with some condition)

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
 int i,n,q,l,k,j,*a=NULL;
 clrscr();
 printf("enter the no of entries: ");
 scanf("%d",&n);
 a=(int*)calloc(n,sizeof(int));
 for(i=0;i<n;i++)
 {
  scanf("%d",&a[i]);
 }
 printf("enter the quantity to be consumed\n");
 scanf("%d",&q);
 for(i=0;i<n;i++)
 {
  if(a[i]>q)
  {
   a[i]=a[n-1];
   n--;
   i--;
  }
 }
 for(i=0;i<n;i++)
 {
        k=a[i];  //selecting 1st element
        for(j=i+1;k+a[j]<q&&j<n;j++)
        {
   k=k+a[j];      //2nd element decided
   for(l=j+1;l<n;l++)
   {
    if(k+a[l]==q)         //checking for 3rd element
    {
     printf("\tyes\n");
     goto end;
    }
   }
  k=k-a[j];
        }
 k=k-a[i];
  }
  printf("\tno");
      end:getch();
      exit(0);
}