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;
}