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>


#define ll long long


//gcdやり直す

void swap (int *x, int *y) {
  int temp;    

  temp = *x;
  *x = *y;
  *y = temp;
}




/***
* pivotを決め、
* 全データをpivotを境目に振り分け、
* pivotの添え字を返す
***/
int partition (int array[], int left, int right) {
  int i, j, pivot;
  i = left;
  j = right + 1;
  pivot = left;   // 先頭要素をpivotとする

  do {
    do { i++; } while (array[i] < array[pivot]);
    do { j--; } while (array[pivot] < array[j]);
    // pivotより小さいものを左へ、大きいものを右へ
    if (i < j) { swap(&array[i], &array[j]); }
  } while (i < j);

  swap(&array[pivot], &array[j]);   //pivotを更新

  return j;
}

/* クイックソート */
int dn(const void*a,const void*b){return*(int*)b-*(int*)a;}


int modpow(int a,int n,int mod){
  int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }

  return ans;
}




/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/

int main(void){
  int l,r,d;
  scanf("%d%d%d",&l,&r,&d);
  int ans=r-l+1;
  for(int i=l;i<=r;i++){
    ans-=!!(i%d);
  }
  printf("%d\n",ans);
}