Write a c program for the following

Find the fixed point of an array ...i.e return i if a[i]=i..

OBJECTIVE: Find the "fixed point" of an array

Input: an array A of *distinct* integers in ascending order.

(Remember that integers can be negative!) The number of

integers in A is n.

Output: one position i in the list, such that A[i]=i, if any exists.

Otherwise: "No".

method 1 :

If A [ i ] > i , we can ignore the right half of the array because

in right half ,for all j > i , we must have A [ j ] > j since A [ i ] >

i, as all the numbers are distinct.

Similarly , if A [ i ] < i, we can ignore the left half of the array

because in left half, for all j < i , we must have A [ j ] < j since A

[ i ] < i ,as all the numbers are distinct.

method 2:

n = size of array

take it=n/2

iterate till it>0&&it<n

if(A[it]>it)

it=it/2

else if(A[it]<it)

it+=it/2

else

return it

if loop returns nothing return error msg

I have tried to write the following code...but got infinite loop error

#includCode:e<stdio.h> #include<math.h> main() { int a[20]; int n,i,mid,low,high,j,temp; printf("\nEnter the Limit : "); scanf("%d",&n); printf("\nEnter the numbers : "); for(i=0;i<n;i++) scanf("%d",&a[i]); low=0; high=n-1; do { mid= (low + high) / 2; if(a[mid]==mid ) { printf("Present = %d \n",a[mid]); } else if(a[mid]< mid) low = mid + 1; else high = mid - 1; continue; } while( low <= high); }