SHERLOCK HOMES

• Problem Description

  
  There is a mysterious temple in Mysteryland. The door of the temple is always closed.It can only be opened by a unique procedure.There are two boxes and N items outside the temple.Sherlock homes visits the temple many times.Each time Sherlock holmes visits the temple,the number of items N outside the door of the temple is changed but each time he anyhow manages to know the cost of those N items.The door of the temple can only be opened if those "N items" are distributed in those two boxes such that the sum of cost of items in one box is equal to the sum of cost of items in other box.Sherlock homes is trying to do such a distribution so as to open the door of the temple.you have to tell whether the door the temple can be opened or not.
  
  INPUT
  
  the first line contain the number of test cases i.e the number of time sherlock homes visits the temple. Next lines contains the description of those test cases.For the first line contain number of items "N".The second line contains cost of those N items.
  
  OUTPUT
  
  output "YES" if the door of the temple can be opened otherwise output "NO".
  
  Constraint:
  
  1 <= testcases <= 10
  
  1 <= N <= 100
  
  1 <= cost <= 100 
  Explanation
  there are three testcases testcase 1:the items can be distributed in the boxes as 4,3,3 ,5,5 so output is YES testcase 2: the items can be distributed in the boxes as 4,1 3,2 so output is YES testcase 3: such a distribution is not possible.

• CODING ARENA

• #include <stdio.h>

  int main()

  {

    int t,n,i,a[100],b,f=1,sum=0,s=0;

    scanf("%d",&t);

    while(t>0)

    {

      scanf("%d",&n);

      i=0;

      while(i<n)

      {

        scanf("%d",&a[i]);

        i++;

      }

      i=0;

      while(i<n)

      {

        sum=sum+a[i];

        i++;

      }

      if(sum%2!=0)

      {

        f=0;

      }

      else

      {

        i=0;

        b=sum/2;

        while(i<n)

        {

          if(a[i]==b)

          {

            f=1;

            break;

          }

          else

            if(a[i]>b)

            {

              f=0;

              break;

            }

          else

          {

            if((s+a[i])<=b)

            {

              s=s+a[i];

              if(s==b)

              {

                f=1;

                break;

              }

            }

          }

          i++;

        }

      }

      if(f==0)

      {

        printf("NO\n");

      }

      else

      {

        printf("YES\n");

      }

      f=1;

      sum=0;

      s=0;

      t--;

    }

    return 0;

  }

• Test Case 1

  
  Input (stdin)

  3
  
  5
  
  4 3 5 5 3
  
  4
  
  1 2 3 4
  
  2
  
  1 2
  
  

  Expected Output

  YES
  
  YES
  
  NO

• Test Case 2

  
  Input (stdin)

  2
  
  10
  
  1 1 1 1 1 1 1 1 1 1
  
  3
  
  1 2 1
  
  

  Expected Output

  YES
  
  YES