31.(C) Codeforces 617 A--->>>An elephant

 #include<stdio.h>

/*A. Elephant

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

An elephant decided to visit his friend. It turned

out that the elephant's house is located at the point

0 and his friend's house is located at point x(x > 0)

of the coordinate line. In one step the elephant can

move 1, 2, 3, 4 or 5 positions forward. Determine,

what is the minimum number of steps he needs to make

in order to get to his friend's house.

Input

The first line of the input contains an integer

x (1 ≤ x ≤ 1 000 000) — The coordinate of the friend's house.

Output

Print the minimum number of steps that the elephant needs

to make to get from point 0 to point x.

Examples

inputCopy

5

output copy

1

inputCopy

12

output copy

3

Note

In the first sample, the elephant needs to make

one step of length 5 to reach the point x.

In the second sample, the elephant can get to point

x if he moves by 3, 5, and 4. There are other ways to

get the optimal answer but the elephant cannot reach

x in less than three moves.

*/

int main(){

    int a,n;

    scanf("%d",&a);

    n=a/5;

    if(a%5==0)// if the distance divisble by 5 then

    // it will print n as number of step

    {

        printf("%d",n);

    }

    else{

        printf("%d",n+1);// if there remains any

remainder then the step .

        // will be n+1 .bcz, the remainder will

be always 1,2,3 or 4.

        //1,2,3,4 indicate just one step

    }

// one thing to clarify here :

//if a<5 then the value of n will be fractional

// value but it will just take 0 as an integer value

// as we have declared n as int

Example:

4/5=0.8 but it will just take 0

and 0+1=1 .So, if we enter anything less than

5 then it will print just 1 step.

   

   

    return 0;

}