Trailing zeroes in factorial
Trailing zeroes in factorial PROBLEM STATEMENT : Given T testcases and For each testcase an integer N is given as the input. For each input, Find the number of trailing zeroes in N!. Expected Time Complexity: O(logN) (for Finding Number of trailing Zeroes for each testcase) Expected Auxiliary Space: O(1) Example 1: Input: ( ) 5 Output: 1 Explanation: 5! = 120 so the number of trailing zero is 1. Example 2: Input: ( ) 4 Output: 0 Explanation: 4! = 24 so the number of trailing zero is 0. HINT : Trailing 0s in n! = Count of 5s in prime factors of n! = floor(n/5) + floor(n/25) + floor(n/125) + .. SOLUTION : Copy Code #include <iostream> using namespace std; int main() { int t; cin>>t; while(t--) { int n,noOfZeroes=0; cin>>n; Trailing 0s in n! = Count of 5s in prime factors of n! while(n>0) { noOfZeroes+=(n/5); n/=5; } ...