The time complexity of the  following code snippet isc=0; while (n>=1) {for(i=1;i<=n;i++){ c++;}n=n/2;}

The given code snippet consists of a while loop and a nested for loop. The outer while loop runs as long as n is greater than or equal to 1, and in each iteration, n is halved. The inner for loop runs from 1 to n.

The time complexity of this code can be calculated as follows:

1. In the first iteration of the while loop, the for loop runs n times.
2. In the second iteration, n is halved, so the for loop runs n/2 times.
3. In the third iteration, n is halved again, so the for loop runs n/4 times.
4. This continues until n is less than 1.

So, the total number of iterations is n + n/2 + n/4 + n/8 + ... until n < 1. This is a geometric series, and the sum of this series is less than 2n.

Therefore, the time complexity of the code is O(n), where n is the initial value of n.