Given an integer n, write a function that returns count of trailing zeroes in n!. Examples: Input: n = 5 Output: 1 Factorial of 5 is 20 which has one trailing 0. Input: n = 20 Output: 4 Factorial of 20 is 2432902008176640000 which has 4 trailing zeroes. Input: n = 100 Output: 24
int first,second;
ReplyDeletecin>>first>>second;
//since int has 2^32-1 values
int x = 2^32-1;
int diff = x-first;
if(diff <second){
cout<<"sum will give overflow";
}else{
cout<<first+second;
}