Test your Programming Skill : P-2

Our today's Programming Problem is quite  easier . It uses your Basic mathematics skill of School level But Really This Problem Test your skills .

You Have to Find Number of Zeros at end of factorial of a number without using real factorisation
  example : let No be 15 then 15 ! (15 factorial ) =1*2*3* ..........*13*14*15
                 and you have to find out number of Zero's at end of 15 ! = 3
                as 15!=1307674368000 .

I Gave you some results  to check your Program output .  So hurry Up .


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NUMber (N)	No of Zero at end of N!
100	24
1024	253
23456	8735373
45	10
120	28
1234	305

Solve This problem And Share your Code (try to use codepad.org link ) or Logic To improve your and our Logic for this Code .

Now  solve This Problem :

Now Think when you get a 0 in factorial of any number . that is obvious when we encounter 5 or multiple of 5 multiply by even number (here we neglect number like 10,20,30 .. because they are also multiple of 5 ).

Now even no are more frequent come than 5 or multiple of five so our main concern is 5 not even number .

Think about How you exploit this 5. I am going to tell how I exploit But may be you find some better way to exploit it .If you found please share it with Us .

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Algorithm :

• Now let the Number is N  .
• Divide N by 5 and Note quotient (only integer part ) Ex: 62/5 = 12 (require quotient .) .
• Now divide N by 5^2 =25 and again note quotient .
• again divide N By 5^3=125 and again note quotient .
• divide Until 5^n  greater than N.
• Sum of all quotient is Our required RESULT .   

 Now check Out This simple code for better understanding .

a = input                  b=5                   while(b<a)                              {                                    count=count + a/b;             // a/b  is integer                                    b=b*5;                                 }                 Print(count)

Now Check Out these C and Python pRogram :

Python 3.3.0  Code:

 tst=int(input()) ## No of Test Case  
 while(tst):  ## to solve each test case  
   count=0  
   b=5    ##intialise  
   a=int(input())  
   while(b<=a):  ## condition to break loop  
     count=count+(a//b)  
     b=b*5    
   print(count) ## print result   
   tst=tst-1   

OutPut is Like This :

C CODE :

 #include<stdio.h>  
 main()  
 {  
      int tst,count;  
      long long int b,a;  
      printf("enter No of Test Case : ");  
      scanf("%d",&tst); // No of Test Case  
      while(tst--)  
      {  
           printf("insert N : ");  
           scanf("%lld",&a);  
           count=0;  
           b=5;  
           while(b<=a)  
           {  
                count=count+a/b;  
                b*=5;  
           }  
           printf("%d\n",count);  
      }  
      return 0;  
 }  

Output Is same as for Python : that is obvious ..

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Print n time Without Loop

Test your Programming Skill : P-1

Hope you Enjoy . If You found some better algorithm please share .