2 arrays

 Problem:-

You are given $2$ arrays $A$ and $B$, each of the size $N$. Each element of these arrays is either a positive integer or $-1$. The total number of $-1^{\prime }s$ that can appear over these $2$ arrays are $\ge 1$ and $\le 2$.

Now, you need to find the number of ways in which we can replace each $-1$ with a non-negative integer, such that the sum of both of these arrays is equal.

Input format

• First line: An integer $N$
• Second line: $N$ space-separated integers, where the of these denotes $A[i]$
• Third line: $N$ space-separated integers, where the $i^{th}$ of these denotes $B[i]$

Output format

If there exists a finite number $X$, then print it. If the answer is not a finite integer, then print 'Infinite'.

Constraints

$1\le N\le {10}^5$

$-1\le A[i],B[i]\le {10}^9$

The $-1^{\prime }s$ may spread out among both arrays, and their quantity is between $1$ and $2$ (both inclusive)

Sample Input 2

4
1 2 -1 4
3 3 -1 1

Sample Output 2

Infinite

Sample Input

4
1 2 -1 4
3 3 3 1

Sample Output

1

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

We can replace the only $-1$ by $3$, so that the sum of both of these arrays become $10$.

Code:-

#include<stdio.h>

int main()

{

  long int n,i,flag1=0,flag2=0;

  scanf("%ld",&n);

  long long int element,sum1=0,sum2=0;

  for(i=0;i<n;i++)

  {

   scanf("%lld",&element);

   if(element==-1)

   flag1++;

   else

   sum1=sum1+element;

  }

  for(i=0;i<n;i++)

  {

   scanf("%lld",&element);

   if(element==-1)

   flag2++;

   else

   sum2=sum2+element;

  }

  if(flag1==flag2 && flag1==1)

   printf("Infinite");

  else{

    if(flag1>flag2)

    {

      if(sum1>sum2)

       printf("0");

      else{

        if(flag1==1 && flag2==0)

         printf("1");

        else

         printf("55");

      }

    }

    else

    {

      if(sum1<sum2)

       printf("0");

      else{

        if(flag2==1 && flag1==0)

         printf("1");

        else

         printf("55");

      }

    }

    

  }

  

  

  return 0;

}