I thought of adding this small code snippets to thelibrary which just reverses the content of the array. Not sort in ascending/descending order, but put last array entry to first, etc. EG.. if array consists of {2,3,4,7,12,98},, need to output {98, 12,7,4,3,2} Code: #include <iostream.h> #define ARR_SIZE 11 int main() { int arr[ARR_SIZE] = {1,2,3,4,5,6,7,8,9,0,11}; int i = 0; // Dispay the content of the array initially cout<<"Array content as input"<<endl; for(i=0;i<ARR_SIZE;i++) cout<<arr[i]<<"\t"; cout<<endl; // Swap the array elements each from the first and the last. // It handles automatically the odd and the even no of elements for(i=0;i<ARR_SIZE/2;i++) { int temp = arr[i]; arr[i] = arr[ARR_SIZE - i-1]; arr[ARR_SIZE - i-1] = temp; } // Dispay the content of the array after the swap. cout<<"Array content as output"<<endl; for(i=0;i<ARR_SIZE;i++) cout<<arr[i]<<"\t"; cout<<endl; return 0; } Enjoy

Your code is very complex. Just update it. by this code. Code: #include<stdio.h> int main() { int a[10]={1,2,3,4,5,6,7,8,9,10},temp; int *ptr1=&a[0]; int *ptr2=&a[9]; while(ptr1<ptr2) { temp=*ptr1; *ptr1=*ptr2; *ptr2=temp; ++ptr1; --ptr2; } for(int i=0;i<10;++i) printf("%d\n",a[i]); return 0; }

I see your one more complex but yes more efficient for ADT's as you are not moving the data but just the pointers.

What's the target of program. Just Reverse the content of Array. That code is doing that. In my program , a[9] & a[0] , a[8] & a[1], a[7] & a[2] , a[6] & a[3], a[5] & a[4] are swapping by pointer. Not too much efficient but this algorithm can be made generic for string as well as any data type may be user data or may enbuilt.

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Hello! Need some help in working out with this problem using PHP.. Is there anyone who has code for this one????Urgent haha thanks thanks Euclid’s Algorithm: Create a program that will implement Euclid’s GCD (Greatest Common Divisor) algorithm: Assume you wish to find the GCD of 2047 and 391. 1. Divide the larger by the smaller and note the remainder: 2047/391 = (391 X 5) + 92 2. Divide the remainder (92) into the previous divisor (391): 391/92 = (92 X 4) + 23 3. Repeat steps 1 and 2 until the remainder is 1 or zero. 3a Divide the remainder (23) into the previous divisor (92): 92/23 = (23 X 4) + 0 4. When the remainder is zero the last divisor is the GCD! 23 X 89 =2047 and 23 X 17 = 391. Therefore 89/17 = 2047/391 5. When the remainder is 1 the two numbers have NO common divisor and are relatively prime. Example: Assume you wish to find the GCD of 8191 and 1023. 8191/1023 = (1023 X 8) + 7 1023/7 = (7 X 146) + 1 The remainder is 1 therefore these two numbers have NO common divisor!