Sample programs

Milk man and his bottles?

Program:

#include<stdio.h>
int main()
{
    int n,a[20],c[20]={0},i;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        while(a[i]!=0)
        {
        if(a[i]>=10)
        {
            c[i]+=a[i]/10;
            a[i]=a[i]%10;;
        }
        else if(a[i]>=7)
        {
            c[i]+=a[i]/7;
            a[i]=a[i]%7;;
        }
        else if(a[i]>=5)
        {
            c[i]+=a[i]/5;
            a[i]=a[i]%5;;
        }
        else if(a[i]<5)
        {
            c[i]+=a[i]/1;
            a[i]=a[i]%1;;
        }
        }
    }
    for(i=0;i<n;i++)
    {
        printf("%d\n",c[i]);
    }
    return 0;
}

Mockvita 2 (2018)

Count Bits

Problem Description

Given a sequence of distinct numbers a1, a2, ….. an, an inversion occurs if there are indices i<j such that ai > aj .

For example, in the sequence 2 1 4 3 there are 2 inversions (2 1) and (4 3).

The input will be a main sequence of N positive integers. From this sequence, a Derived derived sequence will be obtained using the following rule. The output is the number of inversions in the derived sequence.

Rule for forming derived sequence

The derived sequence is formed by counting the number of 1s bits in the binary representation of the corresponding number in the input sequence.

Thus, if the input is 3,4,8, the binary representation of the numbers are 11,100 and 1000. The derived sequence is the number of 1sin the binary representation of the numbers in the input sequence, and is 2,1,1

Constraints

N <= 50

Integers in sequence <= 107

Input Format

The first line of the input will have a single integer, which will give N.

The next line will consist of a comma separated string of N integers, which is the main sequence

Output

The number of inversions in the derived sequence formed from the main sequence.

Explanation

Example 1

Input

5

55, 53, 88, 27, 33

Output

8

Explanation

The number of integers is 5, as specified in the first line. The given sequence is 55, 53, 88, 27, 33.

The binary representation is 110111, 110101, 1011000, 11011, 100001and 100001 . The derived sequence is 5,4,3,4,2, 4,3,4,2 (corresponding to the number of 1s bits). The number of inversions in this is 8, namely (5,4),(5,3),(5,4),(5,2),(4,3),(4,2),(3,2),(4,2). Hence the output is 8.

Example 2

Input

8

120,21,47,64,72,35,18,98

Output

15

Explanation

The number of integers is 8. The given sequence is 120,21,47,64,72,35,18,98. The corresponding binary sequence is 1111000,10101,101111,1000000,1001000,100011, 10010,1100010. The derived sequence (number of 1s) is 4,3,5,1,2,3,2,3. The number of inversions is 15, namely (4,3),(4,1),(4,2),(4,3),(4,2),(4,3),(3,1),(3,2),(3,2),(5,1),(5,2),(5,3), (5,2), (5,3),(3,2). Hence the output is 15.

Program: 

#include <stdio.h>

#include <stdio.h>

    int  main()

    {
        int decimal_num,remainder,base=1,

binary=0,no_of_1s=0,a[100],b[100],n,i,c=0,j;
        char ch;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
        scanf("%d", &a[i]);
        if(i<n-1)
        scanf("%c",&ch);
        }
        for(i=0;i<n;i++)
        {
        decimal_num = a[i];

        while (a[i] > 0)

        {

            remainder = a[i] % 2;
            if (remainder == 1)

            {

                no_of_1s++;

            }

            binary = binary + remainder * base;

            a[i] = a[i] / 2;

            base = base * 10;

        }
        b[i]=no_of_1s;
        base=1;
        binary=0;
        no_of_1s=0;
        }
        for(i=0;i<n-1;i++)
            for(j=i+1;j<n;j++)
            {
                if(b[i]>b[j])
                    c++;
            }
        printf("%d",c);  
      return 0;
    }

Finding Sum

Problem Description

You are given a set of N positive integers and another integer P, where P is a small prime. You need to write a program to count the number of subsets of size (cardinality) 3, the sum of whose elements is divisible by P. Since the number K of such sets can be huge, output K modulo 10007 1009 (the remainder when K is divided by 1009)

Constraints

N <= 500

P<50

All integers <=1000

Input Format

First line two comma separated integers N, P

The next line contains N comma separated integers

Output

One integer giving the number of subsets the sum of whose elements is divisible by P. Give result modulo 1009

Explanation

Example 1

Input

4,5

5,10,15,20

Output

4

Explanation

Every non empty subset of the given numbers has sum of its elements a multiple of 5. Since there are 4 subsets of size 3, the output is 4.

Example 2

Input

5,5

3,7,12,13,15

Output

4

Explanation

There are 4 subsets of size 3 with sum a multiple of 5: {3, 7, 15}, {12, 13, 15}, {7, 13, 15}, {3, 12, 15}, Hence the output is 4.

Program:

#include <stdio.h>
int c=0;
void combo(int arr[], int n,int m, int r)
{
    int data[r];
    pairs(arr, data, 0, n-1, 0, r,m);
}
void pairs(int arr[], int data[], int start, int end,
                     int index, int r,int m)
{
    int s=0;
    if (index == r)
    {
        for (int j=0; j<r; j++)
        {
            s+=data[j];
        }
        if(s%m==0)
        c++;
        return;
    }
    for (int i=start; i<=end && end-i+1 >= r-index; i++)
    {
        data[index] = arr[i];
        pairs(arr, data, i+1, end, index+1, r,m);
    }
}
int main()
{
    int arr[100],n,i,m;
    char ch;
    scanf("%d",&n);
    scanf("%c",&ch);
    scanf("%d",&m);
    for(i=0;i<n;i++)
    {
        scanf("%d",&arr[i]);
        if(i<n-1)
        scanf("%c",&ch);
    }
    combo(arr, n,m, 3);
    printf("%d",c);
    return 0;
}

Array Product

Problem Description

You are given a list of N integers and another positive integer K. Write a program to compute the number of ways in which the product P of the given N integers can be expressed as a product of K positive integers (not necessarily distinct). The order of the factors in the expression is not important. For example, 1 x 2 x 3 and 2 x 3 x 1 are not counted as different ways of expressing 6 as a product of three integers.

Constraints

The product of the N integers <= 10^9

Each of the N integers <=5000

Input Format

First line contains two space separated integers, N and K

The next line contains N space separated integers

Output

One line containing the number of ways in which the product of the N integers can be expressed as a product of K positive integers

Explanation

Example 1

Input

2 4

2 3

Output

2

Explanation

The product of the given integers is 6. This can be expressed as a product of 4 integers in 2 ways: 1x1x1x6, 1x1x2x3

Example 2

Input

2 3

4 16

Output

7

Explanation

The product is 64. This can be expressed as a product of three integers in the following ways:

1 x 1 x 64

1 x 2 x 32

1 x 4 x 16

1 x 8 x 8

2 x 2 x 16

2 x 4 x 8

4 x 4 x 4

Program: 

#include <stdio.h>
int c=0;
void result(int k,int i,int m)
{
    int j;
    if(k==0)
    {
        if(i<=m)
        {
            printf("%d ",m);
            c++;
        return;
        }
    }
    else
    {
        if(i>m)
            return ;
        for(j=i;j<=m;j++)
        {
            if(m%j == 0)
                result(k-1,j,m/j);
        }
    }
}
int main()
{
   int n,k,a[10],m=1,i,j;
   scanf("%d%d",&n,&k);
   for(i=0;i<n;i++)
   {
       scanf("%d",&a[i]);
       m=m*a[i];
   }
   for(i=1;i<=m;i++)
   {
      /* for(j=i;j<m;j++)
       {
           if(i*j==m)
                printf("\n%d * %d",i,j);
       }*/
        if(m%i==0)
            result(k-2,i,m/i);
   }
   printf("\n%d",c);
   return 0;
}

Bug Crawl

Problem Description

A cube has its 6 faces numbered from 1-6 (three of its faces are shown in the diagram). There are six possible orientations on the cube (U,D,N,E,W,S) as shown in the diagram, of which only 4 are possible on any one face (for example, on face 1, N and S are impossible orientations)

An electronic remote control bug is on a face of a cube at some orientation. On your command it can crawl around the cube. The possible commands are F,B,L,R.

Command F : it moves to the adjacent face in the same direction it is facing

.Command B : it turns around by 180° and moves forward by one face.

Command L: it turns left and moves forward by one face

Command R: it turns right and moves forward by one face.

The faces 4,5,6 are opposite to faces 1,2,3 respectively.

For example, if it has orientation U on face 1, on command L, it goes to face 5 and has orientation N, and from there on command F , it will move to face 4 and will have orientation E.

Given a sequence of commands, and the final position (face and orientation) of the bug (after executing the commands in the sequence), we need to determine the initial position (face and orientation) of the bug.

Constraints

The length of the string of command letters <=50

Input Format

One string of 2 characters giving the face and orientation of the bug after it executes the instructions. The first character is a number between 1 and 6, denoting the face, and the second character is the orientation (from the set {U, D, N, S, E, W}) of the bug after executing the commands

One string of command letters. This is a sequence of letters from the set of valid commands {F, B, L, R}

Output

One string of two characters denoting the position of the bug before it executes the instructions. The first character gives the face number (1,2,3,...,6) and the second giving the orientation (E,W,U,D,N,S ) the bug was facing before it executed the instructions

Explanation

Example 1

Input

1U

FFF

Output

3N

Explanation

If the bug starts at 3N, it will move to 4D, 6S and 1U if a command F is given at each position. Hence, if it starts at 3N, after 3 consecutive F commands, it will be at 1U, which is the given final position. Hence the output is 3N.

Example 2

Input

4W

LRB

Output

3E

Explanation

If the bug starts at 3E, after the L command, it goes to 4D, and after the R command, it goes to 2S. From there, on the B command, it goes to 4W, which is the end position. Hence the output is 3E.

Program:

#include<stdio.h>
#include<string.h>

int main()
{
    char c1,c2,a[6],com[]={"WDSEUNWDSEUN"},s[51],b[10];
    int i=0,j,k,t,n;
    scanf("%c%c",&c1,&c2);
    scanf("%s",s);
    n=strlen(s);
    t=c1-48;
    for(i=n-1;i>=0;i--)
    {
        for(j=t-1,k=0;j<t+4;j++,k++)
            a[k]=com[j];
        for(j=0;j<5;j++)
        {
            if(c2==a[j])
            {
                c2=com[t+1];
                if(s[i]=='B')
                    c2=com[t+4];
                t=t+j+1;
                if(t>6)
                    t-=6;
                if(s[i]=='R' || s[i]=='L')
                {
                    if(s[i]=='R')
                    {
                        switch(c2)
                        {
                            case 'U':   strcpy(b,"WSUEND"); break;
                            case 'D':   strcpy(b,"ENDWSU"); break;
                            case 'E':   strcpy(b,"UENDWS"); break;
                            case 'W':   strcpy(b,"DWSUEN"); break;
                            case 'S':   strcpy(b,"SDENUW"); break;
                            case 'N':   strcpy(b,"NUWSDE"); break;
                        }
                    }
                    else if(s[i]=='L')
                    {
                        switch(c2)
                        {
                            case 'D':   strcpy(b,"WSUEND"); break;
                            case 'U':   strcpy(b,"ENDWSU"); break;
                            case 'W':   strcpy(b,"UENDWS"); break;
                            case 'E':   strcpy(b,"DWSUEN"); break;
                            case 'N':   strcpy(b,"SDENUW"); break;
                            case 'S':   strcpy(b,"NUWSDE"); break;
                        }
                    }
                c2=b[t-1];
                }
                break;  
            }
        }
    }
    printf("%d%c\n",t,c2);
    return 0;
}

Building Blocks

Problem Description

The Mathematics teacher wanted to introduce a new competition to the students to sharpen their skills in optimization. He drew a rectangular M ×N grid on the ground and filled it with some non-negative integers on each cell of the grid. The cell named (i,j) is at the intersection of i th row and j th column. He gave the following challenge to the students:

1. On each cell of the grid, you can pile any number of cube blocks.

2. Each layer must be rectangular (with no gaps) and rest p supported completely by the immediate below layer.

3. The number of blocks on each cell should not exceed the number written on the cell on the ground.

4. In each layer, the cell above (1,1) must be covered.

The challenge is to pile up the maximum number of blocks subject to the above conditions.

For example if the bottom grid was as follows:

1	1	0
1	1	1
1	1	0

the maximum number of blocks you can pile is 6 with one layer covering all the cells from (1,1) to (3,2).

Constraints

1 <= M,N <= 50

Maximum value in each cell is 50

Input Format

The first line will contain two comma separated integers M,N giving the size of the grid, where M is the number of rows and N is the number of columns.

The next M lines will each contain comma separated N non-negative integers giving the numbers in the grid cells.

Output

One line containing the number of blocks that can be piled according to the rules.

Explanation

Example 1:

Input:

3,4

5,4,9,3

4,3,5,6

2,2,1,1

Output:

32

Explanation:

One example of the maximum number of blocks that could be piled on the grid is shown below:

5	4	4	3
3	3	3	3
1	1	1	1

The total number of blocks is 32. Hence the output is 32.

Example 2:

Input

4,7

27,26,28,14,15,38,0

38,40,35,2,20,43,39

18,48,43,2,47,18,26

38,2,29,23,14,31,32

Output

242

Explanation

The number of rows is 4. The number of columns is 7. The values in the cells of the grid are

27	26	28	14	15	38	0
38	40	35	2	20	43	39
18	48	43	2	47	18	26
38	2	29	23	14	31	32

One possible maximal placing of the blocks is

27	26	26	2	2	2	0
27	26	26	2	2	2	0
18	18	18	2	2	2	0
2	2	2	2	2	2	0

As there are 242 blocks in this maximal placing, the output is 242.

Program: 

#include<stdio.h>
int a[10][10];
void del(int i,int j,int s,int f,int r,int c)
{
    int k,l;
    if(s==0)
    {
        for(k=0;k<r;k++)
            for(l=j;l<c;l++)
                a[k][l]=f-1;
    }  
    else
    {
        for(k=i;k<r;k++)
            for(l=0;l<c;l++)
                a[k][l]=f-1;
    }
}
int rsum(int i,int j,int r,int c)
{
    int k,l,s=0;
    for(k=0;k<i;k++)
        for(l=0;l<c;l++)
            s+=a[k][l];
    return s;      
}
int csum(int i,int j,int r,int c)
{
    int k,l,s=0;
    for(k=0;k<r;k++)
        for(l=0;l<j;l++)
            s+=a[k][l];
    return s;
}
int main()
{
    int i,j,k,r,c,max=0,r1,c1,s,t1,t2,sum=0;
    char ch;
    scanf("%d%c%d",&r,&ch,&c);
    r1=r;
    c1=c;
    for(i=0;i<r;i++)
        for(j=0;j<c;j++)
        {
            scanf("%d",&a[i][j]);
            if(j<c-1)
            scanf("%c",&ch);
            if(a[i][j] > max)
                max=a[i][j];
        }
    for(k=1;k<=max;k++)
    {
        for(i=0;i<r1;i++)
            for(j=0;j<c1;j++)
            {
                if(a[i][j] < k)
                {
                    t1=rsum(i,j,r1,c1);
                    t2=csum(i,j,r1,c1);
                    s=(t1 > t2) ? 1 : 0;
                    del(i,j,s,k,r1,c1);
                    if(s==1)
                        r1=i;
                    else
                        c1=j;
                }
            }
    }
    for(i=0;i<r;i++)
    {
        for(j=0;j<c;j++)
        sum+=a[i][j];   
    }
    printf("%d",sum);
    return 0;
}

Mock vita 1 (2018)

Triangles

Problem Description

Constraints

N<=50

-89 <= angle for any line <=90

Input Format

The first line of the input consists of a single integer, N.

The next line consists of a series of integers (positive, zero or negative), each corresponding to a separate line, and giving the angle that it makes with the x axis (measured in degrees and in anticlockwise direction).

Output

The output is a single integer giving the number of triangles formed by the lines

Test Case

TestCase 1
8,3
D,C,E,F,G,H
C,A,E
D,C,B,E

A,B

TestCase 2
8,3
D,C,E,F,G,H
C,A,B,E
D,B

N/A

Explanation

Example 1

Input

5

20,-20,0,50,50

Output

7

Explanation

There are 5 lines, with angles at 20,-20,0, 50 and 50 degrees with the x axis. The figure looks like this

There are 7 triangles, those formed by (L1,L2,L3),(L1,L2,L5), (L1,L2,L4), (L1,L3,L4), (L1,L3,L5), (L2,L3,L5), (L2,L3,L4). Hence the output is 7.

Example 2

Input

5

50,-50,50,-50,50

Output

0

Explanation

There are 5 lines with angles 50,-50,50,-50 and 50 degrees. The figure looks like this

As L1,L3 and L5 are parallel, and L2 and L4 are parallel, no triangles are formed by any set of three lines. Hence, the output is 0.

Program:-

#include<stdio.h>

int main()

{

    int n,a[100],i,k,j,count=0;

   char b[50];

     scanf("%d",&n);

      for(i=0;i<n;i++)

      {

          scanf("%d",&a[i]);

        if(i!=n-1)

           scanf("%c",&b);

      }

     for(i=0;i<n;i++)

     {

         for(j=i+1;j<n;j++)

         {

           for(k=j+1;k<n;k++)

           {

               if(a[i]!=a[j]&&a[i]!=a[k]&&a[k]!=a[j])

                  count++;

             

           }

         }

     }

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

  return 0;

}

Finding Product

Problem Description

You are given a set of N positive integers and two small prime numbers P and Q (not necessarily distinct). You need to write a program to count the number of subsets of the set of N numbers whose product is divisible by PQ (the product of P and Q). Since the number K of such sets can be huge, output K modulo 1009 (the remainder when K is divided by 1009).

Constraints

N <= 300

P,Q <=50

The integers are <= 10000

Input Format

First line three comma separated integers N, P,Q

The next line contains N comma separated integers

Output

One integer giving the number of subsets the product of whose elements is divisible by PQ. Give the result modulo 1009.

Test Case

TestCase 1
8,3
D,C,E,F,G,H
C,A,E
D,C,B,E

A,B

TestCase 2
8,3
D,C,E,F,G,H
C,A,B,E
D,B

N/A

Explanation

Example 1

Input

4,5,7

5,49,10,27

Output

6

Explanation

N is 4, P is 5, Q is 7. We need to find subsets of the numbers given so that the product of the elements is divisible by 35 (the product of 5 and 7). These subsets are (5,49),(5,49,10),(5,49,27),(5,49,10,27), (49,10),(49,10,27). There are 6 subsets, and the output is 6.

Example 2

Input

4,11,13

3,7,12,13

Output

0

Explanation

N is 4, P is 11, Q is 13. We need to find subsets of the numbers given so that the product of the elements is divisible by 143 (the product of 11 and 13).As none of the N numbers is divisible by 11 (a prime number), there are no subsets for which the product of the elements is divisible by 143. Hence the output is 0.

 

Program:

#include <stdio.h>
int t;
static int count=0;
void combinationUtil(int arr[], int n, int r,
                     int index, int data[], int i);
void printCombination(int arr[], int n, int r)
{
    int data[r];
    combinationUtil(arr, n, r, 0, data, 0);
}
void combinationUtil(int arr[], int n, int r, int index,
                     int data[], int i)
{
    long int s;
    if (index == r) {
        s=1;
        for (int j = 0; j < r; j++)
        {
            s=s*data[j];
        }
        if(s%t==0)
        count++;
        return;
    }
    if (i >= n)
        return;
    data[index] = arr[i];
    combinationUtil(arr, n, r, index + 1, data, i + 1);
    combinationUtil(arr, n, r, index, data, i + 1);
}
int main()
{
    int arr[100],n,p,q,i;
    char ch;
    scanf("%d",&n);
    scanf("%c",&ch);
    scanf("%d",&p);
    scanf("%c",&ch);
    scanf("%d",&q);
    for(i=0;i<n;i++)
    {
    scanf("%d",&arr[i]);
    if(i<n-1)
    scanf("%c",&ch);
    }
    t=p*q;
    for(i=2;i<=n;i++)
    printCombination(arr, n, i);
    printf("%d ",count);
    return 0;
}

Largest Integer

#include<stdio.h>
int main()
{
    int a[20][20],i,j,m,n,f=0,c=0,r,p,

k;
    char ch;
    scanf("%d%c%d",&m,&ch,&n);
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
        {
            scanf("%d",&a[i][j]);
            if(j<n-1)
            scanf("%c",&ch);
           
        }
    }
   
    while(f!=1)
    {
        j=n-1;c=-1;
     for(k=m-1;k>=0;k--)
     {
        if(a[k][j]>c)
        {
            c=a[k][j];
            r=k;
          
        }
       
     }
     for(p=j;p>0;p--)
     {
         a[r][p]=a[r][p-1];
     }
     a[r][p]=-1;
     if(c!=-1)
     printf("%d",c);
     else
     break;
    }
}

Code vita Round 1(2015)

Reverse Gear?

 #include <stdio.h>
int main()
{
    int F[20],B[20],T[20],D[20],i,n,j=1,b=0,f=0,t;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    scanf("%d%d%d%d",&F[i],&B[i],&T[i],&D[i]);
    while(j<=n)
    {
    for(i=1;D[j]>0;i++)
    {
        if(i%2==1)
        {
        D[j]=D[j]-B[j];
        b++;
        }
        else
        {
        D[j]=D[j]+F[j];
        f++;
        }
    }
    b=b*B[j];
    f=f*F[j];
    t=(b+f)*T[j]+(D[j]*T[j]);
    printf("%d ",t);
    b=0;f=0;
    j++;
    }
    return 0;
}

Catch 22?

#include <stdio.h>
int main()
{
    int F[20],B[20],T[20],FD[20],BD[20],i,n,j=1,b=0,f=0,t;
    char ch;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    scanf("%d %d %d %d %d",&F[i],&B[i],&T[i],&FD[i],&BD[i]);
    while(j<=n)
    {
    if(B[j]==F[j]&&F[j]<FD[j]&&B[j]<BD[j])
    printf("No Ditch\n");
    else
    {
    for(i=1;FD[j]>0&&BD[j]>0;i++)
    {
        if(i%2==0)
        {
        FD[j]=FD[j]+B[j];
        BD[j]=BD[j]-B[j];
        b++;
        }
        else
        {
        FD[j]=FD[j]-F[j];
        BD[j]=BD[j]+F[j];
        f++;
        }
    }
    b=b*B[j];
    f=f*F[j];
    if(BD[j]<0)
    {
    t=(b+f)*T[j]+(BD[j]*T[j]);
    ch='B';
    }
    else
    {
    t=(b+f)*T[j]+(FD[j]*T[j]);
    ch='F';
    }
    printf("%d %c\n",t,ch);
    b=0;f=0;
    }
    j++;
    }
    return 0;
}

Saving for a rainy day?

#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("%d\n",c);

 getch();

}

           (or)
#include<stdio.h>
int main()
{
    float m,r,r1,x;
    int t,i;
    scanf("%f%d%f",&m,&t,&r1);
    r=r1/12;
    x=m;
    for(i=t;i>0;i--)
    {
        x=(x*100)/(100+r);
        if(i!=1)
            x=x+m;
    }
    printf("%.0f",x);
    return 0;
}

Sheldon Cooper and his beverage paradigm?

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
int main()
{
 int i,n,q,l,k,j,*a=NULL;
 scanf("%d",&n);
 a=(int*)calloc(n,sizeof(int));
 for(i=0;i<n;i++)
 {
  scanf("%d",&a[i]);
 }
 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("True\n");
     goto end;
    }
   }
  k=k-a[j];
        }
 k=k-a[i];
  }
  printf("False");
      end:getch();
      exit(0);
return 0;
}

Credit and Risk calculator? 

 #include<stdio.h>
int min(int a,int b)
{
    if(a<b)
    return a;
    return b;
}
int main()
{
    float n[10],sv[10],csv[10],r[10],cr[10],sv1[10],r1[10],cv[10],ce[10],ca[10];
    int i,t,flag=0;
    scanf("%d",&t);
    for(i=0;i<t;i++)
    {
        scanf("%f%f%f%f%f",&n[i],&sv[i],&csv[i],&r[i],&cr[i]);
    }
    for(i=0;i<t;i++)
    {
        sv1[i]=sv[i]-csv[i];
        r1[i]=r[i]-cr[i];
        cv[i]=n[i]*(min(sv[i],sv1[i]));
        ce[i]=cv[i]/2;
        ca[i]=ce[i]*((float)min(r1[i],r[i])/100);
    }
for(i=0;i<t;i++)
{
    if(csv[i]<0)
    {
        if(cr[i]<=0)
        flag=1;
    }
    else
    flag=1;
    if(flag==1&&(sv1[i]>=20&&sv1[i]<=10000)&&(r1[i]>=0.01&&r1[i]<=99.99)&&(n[i]>=20000&&n[i]<=10000000)&&(sv[i]>=20.00&&sv[i]<=10000.00)&&(-2000.00<=csv[i]&&csv[i]<=2000.00)&&(0.01<=r[i]&&r[i]<=99.99)&&(-10.00<=cr[i]&&cr[i]<=10.00))
    printf("%.2f\n%.2f\n%.2f\n%.2f\n%.2f\n",sv1[i],r1[i],cv[i],ce[i],ca[i]);
    else
    printf("Invalid input\n");
    flag=0;
}
return 0;
}

Code Vita Round 1(2016)

Min product array?

 #include <stdio.h>
#include<stdlib.h>
int main()
{
int maxDiff = 0, minimumSum = 0,product,A[10],B[10],k,n,diff,temp,i;
scanf("%d",&n);
scanf("%d",&k);
for(i=0;i<n;i++)
scanf("%d",&A[i]);
for(i=0;i<n;i++)
scanf("%d",&B[i]);
for(i=0;i<n;i++)
{
product = A[i] * B[i];
if ( product < 0 && B[i] < 0 )
temp = (A[i] +  2  * k ) * B[i];
else if( product < 0 && A[i] < 0)
temp = (A[i] - 2 * k) * B[i];
else if( product > 0 && A[i] < 0)
temp = (A[i] + 2 * k) * B[i];
else if (product > 0 && A[i] > 0)
temp = (A[i] - 2 * k)  * B[i];
diff =  abs(product - temp);
if( diff > maxDiff )
maxDiff = diff;
minimumSum = minimumSum + product;
}

minimumSum = minimumSum - maxDiff;
printf("%d",minimumSum);
return 0;
}

Consecutive prime sum?

 #include<stdio.h>
#include<math.h>
int main()
{
        int n,c,i,j,sum=0,f=-1,k,l;
        scanf("%d",&n);
        for(i=1;i<n;i++)
        {
            c=0;
            for(j=1;j<i;j++)
            {
                if(i%j==0)
                  c++;
            }
            if(c==1)
            {
               k=0;
               sum=sum+i;
               printf(" %d ",sum);
               for(l=1;l<sum;l++)
               {
                   if(sum%l==0)
                     k++;
               }
               if(sum<=n)
               {
                   if(k==1)
                     f++;
                    
               
               }
            
                else
                 break;
                
              //  printf("  %d  ",sum);
            }
        }
        printf("\n%d",f);
       
       
        return 0;
} 

Logic pyramid?

#include<stdio.h>
#include<math.h>
int main()
{
       int n,i,m,j,k;
       scanf("%d",&n);k=2;
       for(i=1;i<=n;i++)
       {
           j=i;
           while(j>0)
           {
             m=(2*(k*k))-k;
             printf("%.5d ",m);
             j--;
           }
           printf("\n");
       }
return 0;
}

Code Vita Round 1(2017)

Chakravyuha?

#include <stdio.h>

int main()
{
    int a[10][10],n,i,j,d=0,c=1;
    scanf("%d",&n);
    for(i=0;i<10;i++)
        for(j=0;j<10;j++)
        {
            if(i>=n || j>=n)
                a[i][j]=-1;
            else
            a[i][j]=0;
        }   
    for(c=1,i=0,j=0;c<=n*n;c++)
    {
        if(a[i][j]==0 && d==0)
        {
            a[i][j]=c;
            j++;
        }
        else if(a[i][j]==0 && d==1)
        {
            a[i][j]=c;
            i++;
        }
        else if(a[i][j]==0 && d==2)
        {
            a[i][j]=c;
            j--;
        }
        else if(a[i][j]==0 && d==3)
        {
            a[i][j]=c;
            i--;
        }
        if(a[i][j]!=0)
        {
            if(d==0)
            {
                i=i+1;
                j=j-1;
                d=1;
            }
            else if(d==2)
            {
                i=i-1;
                j=j+1;
                d=3;
            }
            else if(d==1)
            {
                i=i-1;
                j=j-1;
                d=2;
            }
            else if(d==3)
            {
                i=i+1;
                j=j+1;
                d=0;
            }
        }
    }
    for(i=0;i<n;i++)
    {
        printf("\n");
        for(j=0;j<n;j++)
            printf("%d ",a[i][j]);
    }
    return 0;
}

Spiral

The prime numbers are written in a spiral form staring at (0,0) and moving as shown in the diagram below. The numbers
shown on the right column and the bottom row are the column numbers and row numbers respectively (y and x coordinate
frames).

The objective is to find the position (x and y coordinates) of a given prime.
Input Format:
The input consists of multiple lines.
The first line gives the number of primes (N) in this test case.
The next N lines contain one prime in each line.
Output Format:
The output consists of N lines.
Each consists of a space separated pair of integers giving the x and y coordinates of the corresponding prime in the input.
Constraints:
N≤10
Each prime < 1000000
Example 1
Input
2
3
7
Output
1 0
0 1
Explanation
There are 2 primes in this test case (N=2). The primes are 3 and 7. The coordinates of these in the spiral is (1,0) and (0,1).
The output hence has these in space separated form.
Example 2
Input
3
5
11
13
Output
1 1
1
1
1
0
Explanation
There are 3 primes in this test case (N=2). The primes are 5, 11 and 13. The coordinates of these in the spiral is (1,1), (1,1)
and (1,0).
The output hence has these in space separated form.

Program:

#include<stdio.h>
int prime[100];
int a[9][10],num,search[100];
void create()
{
    int i=4,j=4,n=3,m=5,n1=5,m1=3,k=0,dire=0;
    //creating an array size of 9*10 which is shown in program
    while(i!=-1)
    {
        a[i][j]=prime[k];
        k++;
        if(dire==0)//Right direction set to 0
        {
            j++;
            if(j==m)
            {
            m++;
            dire=1;
            }
        }
        else if(dire==1)//Up direction set to 1
        {
            i--;
            if(i==n)
            {
                n--;
                dire=2;
            }
        }
        else if(dire==2)//Left direction set to 2
        {
            j--;
            if(j==m1)
            {
                m1--;
                dire=3;
            }
        }
        else if(dire==3)//Down direction set to 3
        {
            i++;
            if(i==n1)
            {
                n1++;
                dire=0;
            }
        }
    }
    //Serahing the element postion in the Array
    for(k=0;k<num;k++)
    {
    for(i=0;i<9;i++)
    {
        for(j=0;j<10;j++)
        {
        if(a[i][j]==search[k])
        {
            printf("%d %d\n",j-4,4-i);//printing the position as required
        }
        }
    }
    }
   
}
int main()
{
    int n, i = 3, count, c,j=0;
    n=100;
    //Reading Input
    scanf("%d",&num);
    for(j=0;j<num;j++)
    scanf("%d",&search[j]);
    //calculating 100 prime numbers and storing into array
    j=0;
        if ( n >= 1 )
        {
          prime[j]=2;
          j++;
        }
        for ( count = 2 ; count <= n ;  )
        {
          for ( c = 2 ; c <= i - 1 ; c++ )
          {
             if ( i%c == 0 )
                break;
          }
          if ( c == i )
          {
             prime[j]=i;
             j++;
             count++;
          }
          i++;
        }
        //Calling function to search and create the elements
        create();
return 0;
}