How many in the interval [a; b] are the numbers that are divisible by 7 without remainder

Write a program that reads numbers a, b (100 <a, b <10000) from the input data and determines how many in the interval [a; b] are numbers that are divisible by 7 without remainder. Show them on the screen and count them

#include <iostream>

using namespace std;

int main() {
    int a, b;
    int sum=0;
    cin >> a >> b;
    for (int i = a; i <= b; ++i) {
        if (i % 7 ==0) {
            i++;
            cout << sum << ' ';
        }
    }
    return 0;
}
 3
c++c++14
Author: MSDN.WhiteKnight, 2018-06-19
Source

4 answers

Code:

#include <iostream>

using namespace std;

int main() {
    int a, b;
    int number = 0;
    cin >> a >> b;
    for (int i = a; i <= b; i++) {
        if (i % 7 == 0) {
            number++;
            cout << i << ' ';
        }
    }

    cout << "\n";
    cout << "number: " << number << "\n";

    return 0;
}

Link to the code to play with: https://ideone.com/ctv16G

 2
Author: gil9red, 2018-06-19 15:36:12 
int deltaA = (a % 7 == 0)? 0 : (7 - (a % 7));
int count = (b - (b % 7) - (a + deltaA)) / 7 + 1;
if (count < 0)
  count = 0;


function bySeven(a, b) {
  var deltaA = (a % 7 == 0)? 0 : (7 - (a % 7));
  var count = (b - (b % 7) - (a + deltaA) ) / 7 + 1;
  return Math.max(count, 0);
}

console.log(bySeven(43, 25));
console.log(bySeven(43, 48));
console.log(bySeven(25, 43));
console.log(bySeven(25, 30));
console.log(bySeven(14, 14));
console.log(bySeven(13, 14));
console.log(bySeven(14, 15));
 4
Author: Igor, 2018-06-19 15:58:59

Align a to a border multiple of 7, with rounding up

a = (a + 6) / 7 * 7;

Align b to a border multiple of 7, with rounding down

b = b / 7 * 7;

Calculate the number of numbers that are multiples of 7 in the resulting interval [a, b] (assuming that a <= b)

n = (b - a) / 7 + 1;
 1
Author: AnT, 2018-06-22 19:07:12 
#include <iostream>
short a, b, c;
int main() {
std::cin >> a >> b;
for(a = a; a <= b; a++) {
if(a % 7 == 0) {
std::cout << a << ' ';
c++;
}
    }
std::cout << std::endl << c;
        }
 1
Author: Cormentor, 2018-06-23 07:23:06