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)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <stdbool.h>
typedef long long LL;
typedef long double Lf;
#define Rep(i,a,n)for(LL i=(LL)(a);i<(LL)(n);i++)
#define rep(i,n)Rep(i,0,n)
#define Repp(i,l,r,k)for(LL i=(l);i<(r);i+=(k))
#define Ma(x,y) ((x)>(y)?(x):(y))
#define mi(x,y) ((x)<(y)?(x):(y))
#define MOD 1000000007//998244353,1000000007,1000000009 
int upi(const void* a, const void* b) { return*(int*)a<*(int*)b ? -1 : *(int*)a> * (int*)b ? 1 : 0; }
int downi(const void* a, const void* b) { return*(int*)a<*(int*)b ? 1 : *(int*)a> * (int*)b ? -1 : 0; }
void sortupi(int* a, int n) { qsort(a, n, sizeof(int), upi); } //1,2,3,4,...
void sortdowni(int* a, int n) { qsort(a, n, sizeof(int), downi); }
int upLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? -1 : *(LL*)a> * (LL*)b ? 1 : 0; }
int downLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? 1 : *(LL*)a> * (LL*)b ? -1 : 0; }
void sortupLL(LL* a, int n) { qsort(a, n, sizeof(LL), upLL); }
void sortdownLL(LL* a, int n) { qsort(a, n, sizeof(LL), downLL); }
int cmp_char(const void* a, const void* b) { return *(char*)a - *(char*)b; } // a,b,c,d
int cmp_str(const void* a, const void* b) { return strcmp(*(const char**)a, *(const char**)b); } // aaa,aab
LL gcd(LL p, LL q) { while (q) { LL t = p % q; p = q; q = t; }return p; }
LL modpow(LL a, LL n, int m) { LL x = 1; for (a %= m; n; n /= 2)n & 1 ? x = x * a % m : 0, a = a * a % m; return x; }
#define invmodpow(b,p)modpow(b,p-2,p)// a/b≡a*b^{-1}(mod p)...return b^{-1}(mod p)
typedef struct fac { LL n, exp; }F;
LL fact(LL n,F*a){LL x,i,j,c=0;x=n;for(i=2;i*i<=x;i++){for(j=1;x%i==0;j++)x/=i;if(j>1){a[c].n=i;a[c++].exp=j-1;}}if(x-1){a[c].n=x;a[c++].exp=1;}return c;}
//snippet

#define N 1009

int main() {
	int a, b, c, ans=0;
	scanf("%d%d%d", &a, &b, &c);
	Rep(i, a, b + 1)if (i % c == 0)ans++;
	printf("%d", ans);
	return 0;
}