Problem:
Score : 100 points

Problem Statement

How many multiples of d are there among the integers between L and R (inclusive)?

Constraints

All values in input are integers.

1 \leq L \leq R \leq 100

1 \leq d \leq 100

Input

Input is given from Standard Input in the following format:

L R d

Output

Print the number of multiples of d among the integers between L and R (inclusive).

Sample Input 1

5 10 2

Sample Output 1

3

Among the integers between 5 and 10, there are three multiples of 2: 6, 8, and 10.

Sample Input 2

6 20 7

Sample Output 2

2

Among the integers between 6 and 20, there are two multiples of 7: 7 and 14.

Sample Input 3

1 100 1

Sample Output 3

100

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//CODING ME ACCHHA KARNEKA HAI TERKO TO BAS PRACTICE KARTE REH
#include <bits/stdc++.h>
#define pragma
using namespace std;
#define ll unsigned long long
#define ld long double
#define pb push_back
#define pf push_front
#define aisort sort(a,a+n)
#define adsort sort(a,a+n,greater<ll>())
#define PI 2 * acos(0)

typedef pair<ll, ll> pii;
typedef pair<ll, pii> piii;
typedef vector<ll> vi;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<piii> viii;
typedef set<ll> Set;

const ll maxs = 1e5+3;
const ll maxs1 = 1e3+4;
const ll mod = 1e9 + 7;

//MODULO OPEARTIONS
    inline ll add(ll a, ll b) { return (mod + a + b) % mod; }
    inline ll sub(ll a, ll b) { return (a - b + mod) % mod; }
    inline ll mul(ll a, ll b) { return (1ll * a * b) % mod; }

//POWER
    ll fastpow(ll x,ll y)
    {ll res = 1;x = x % mod;while (y > 0){ if (y & 1)res = (res * x) % mod; y = y >> 1 ; x = (x * x) % mod; } return res;}

//MODULAR INVERSE
    ll inv(ll x){ return fastpow(x , mod - 2);}
    
//PRIME
    bool isPrime(ll n) { if (n <= 1)  return false; if (n <= 3)  return true;if (n%2 == 0 || n%3 == 0) return false; for (int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false; return true; } 


int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  ll l,r,d;
  cin>>l>>r>>d;
  ll cnt=0;
  for(ll i=l;i<=r;i++)
  {
      if(i%d==0)
      cnt++;
  }
  cout<<cnt<<endl;
}