for more codevita round 1 solutions comment below indicating the question

Question 8

Problem : Milk Man and His Bottles A Milkman serves milk in packaged bottles of varied sizes. The possible size of the bottles are {1, 5, 7 and 10} litres. He wants to supply desired quantity using as less bottles as possible irrespective of the size. Your objective is to help him find the minimum number of bottles required to supply the given demand of milk.

  Input Format: First line contains number of test cases N Next N lines, each contain a positive integer L_iwhich corresponds to the demand of milk. 

  Output Format: For each input L_i, print the minimum number of bottles required to fulfill the demand 

 Constraints: 1 <= N <= 1000 L_i > 0 1 <= i <= N Sample Input and Output

SNo.	Input	Output
1	2 17 65	2 7

  Explanation: Number of test cases is 2

1. In first test case, demand of milk is 17 litres which can be supplied using minimum of 2 bottles as follows
   • 1 x 10 litres and
   • 1 x 7 litres
2. In second test case, demand of milk is 65 litres which can be supplied using minimum of 7 bottles as follows
   • 6 x 10 litres and
   • 1 x 5 litres

 here is the code in  c

#include<stdio.h>

#include<conio.h>

void main()

{

int i,j,r,nb,q,n;

clrscr();

printf("no of tests : ");

scanf("%d",&n);//no of test cases

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

{

printf("quantity of milk =");

scanf("%d",&q);

nb=0;

r=q%10;

if(q>=10)

{

if(r==2||r==9||r==3||r==4)

{

j=q/10;

nb=nb+j-1;

q=q%10+10;

}

else

{

j=q/10;

nb=nb+j;

q=q-(j*10);

}

}

if(q>=7)

{

j=q/7;

nb=nb+j;

q=q%7;

}

if(q>=5)

{

j=q/5;

nb=nb+j;

q=q%5;

}

nb=nb+q;

printf("minimum no of bottles = %d\n",nb);

}

getch();

}

hope this helps

Question 2

Problem : Saving for a rainy day By nature, an average Indian believes in saving money. Some reports suggest that an average Indian manages to save approximately 30+% of his salary. Dhaniram is one such hard working fellow. With a view of future expenses, Dhaniram resolves to save a certain amount in order to meet his cash flow demands in the future. Consider the following example. Dhaniram wants to buy a TV. He needs to pay Rs 2000/- per month for 12 installments to own the TV. If let's say he gets 4% interest per annum on his savings bank account, then Dhaniram will need to deposit a certain amount in the bank today, such that he is able to withdraw Rs 2000/- per month for the next 12 months without requiring any additional deposits throughout. Your task is to find out how much Dhaniram should deposit today so that he gets assured cash flows for a fixed period in the future, given the rate of interest at which his money will grow during this period.
  Input Format: First line contains desired cash flow M Second line contains period in months denoted by T Third line contains rate per annum R expressed in percentage at which deposited amount will grow
  Output Format: Print total amount of money to be deposited now rounded off to the nearest integerConstraints: M > 0 T > 0 R >= 0 Calculation should be done upto 11-digit precision   Sample Input and Output 

SNo.	Input	Output
1	500 3 12	1470
2	6000 3 5.9	17824
3	500 2 0	1000

here is my code freinds  (this problem gave me some trouble)

by looking at the code you may think this was easy but frankly this took the most amount of my time to get it right(because of  a small trick)

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float m,p,r;

double x,b;

int c;

scanf("%f%f%f",&p,&m,&r);

b=1200+r*(m-1);

x=p*m*1200/b;

c=(int)x;

printf("the amount to be deposited is :%d\n",c);

getch();

}

codevita 2015 solutions in c

Question 1

Problem : Catch-22 A robot is programmed to move forward F meters and backwards again, say B meters, in a straight line. The Robot covers 1 meter in T units of time. On Robot's path there is a ditch at a distance FD from initial position in forward direction as well as a ditch at a distance BD from initial position in backward direction. This forward and backward movement is performed repeatedly by the Robot. Your task is to calculate amount of time taken, before the Robot falls in either ditch, if at all it falls in a ditch. Input Format: First line contains total number of test cases, denoted by N Next N lines, contain a tuple containing 5 values delimited by space F B T FD BD, where 

1. F denotes forward displacement in meters
2. B denotes backward displacement in meters
3. T denotes time taken to cover 1 meter
4. FD denotes distance from Robot's starting position and the ditch in forward direction
5. BD denotes distance from Robot's starting position and the ditch in backward direction

  Output Format: For each test case print time taken by the Robot to fall in the ditch and also state which ditch he falls into. Print F for forward and B for backward. Both the outputs must be delimited by whitespace OR Print No Ditch if the Robot does not fall in either ditch

  Constraints: First move will always be in forward direction 1 <= N <= 100 forward displacement > 0 backward displacement > 0 time > 0 distance of ditch in forward direction (FD) > 0 distance of ditch in backward direction (BD) > 0 All input values must be positive integers only   Sample Input and Output 

SNo.	Input	Output
1	3 9 4 3 13 10 9 7 1 11 13 4 4 3 8 12	63 F 25 F No Ditch
2	5 8 4 7 11 22 4 5 4 25 6 4 9 3 6 29 7 10 6 24 12 10 10 1 9 7	133 F 216 B 231 B 408 B 9 F

here is my code in c

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i;
int f,b,fd,bd,t,c=0,tc=0;
clrscr();
printf("enter no of test cases ");
scanf("%d",&n);
if(n>=1 && n<=100)
{
printf("f b t fd bd\n");
for(i=0;i<n;i++)
{
scanf("%d %d %d %d %d",&f,&b,&t,&fd,&bd);
if(fd>0 && bd>0 &&  t>0 && f>0 && b >0)
{
if(f>=fd)
{
c=fd*t;
printf("output:%d F\n",c);
}
else if(f==b)
{
printf("output:NO Ditch\n");
}
else if(f>b)
{
while(!0)
{
tc=tc+f;
c=c+f;
if(c>fd) break;
c=c-b;
tc=tc+b;
}
tc=tc-(c-fd);
printf("output:%d F\n",tc*t);
}
else
{

while(!0)
{
tc=tc+f;
c=c-f;
if(c>bd) break;
c=c+b;
tc=tc+b;
}
tc=tc-(c-bd);
printf("output:%d B\n",tc*t);
}
}
}
}
}