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)

//A - Number of Multiples
//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu, zettaichi(absolute)
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
typedef long long ll;
//#define N 998244353
#define pi 3.1415926535897932384626433832

int gcd(int m,int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
int lcm(int m,int n){if((0==m)||(0==n))return 0;return(m*n/gcd(m, n));}
int fact(int x){if(x==1) return(x);else return(x*fact(x-1));}//Kaijou x!
int ruijou(int base,int exponent){int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

long long int llgcd(long long int m,long long int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
long long int lllcm(long long int m,long long int n){if((0==m)||(0==n))return 0;return(m*n/llgcd(m, n));}
long long int llfact(long long int x){if(x==1) return(x);else return(x*llfact(x-1));}//Kaijou
int llruijou(long long int base,long long int exponent){long long int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

int main()
{
  int a,b,d;
  scanf("%d %d %d", &a,&b,&d);
  if(a%d==0){
    printf("%d\n", b/d-a/d+1);
  }
  else{
    printf("%d\n", b/d-a/d);
  }
  return 0;
}