Question

I’m trying to write a program with do..while loop for: list all the numbers from 100 downwards that are dividable by 7 and 5 but not 11.
can you help me?

Answer 1

This code should work, I haven’t tested it or timed it and there are probably better ways of doing it speed wise, but:

int counter = 0;// Counts the index of the array
int num = 100;// the iterator like 'i' in a for loop
int nums [100];// the array to store all the numbers
do {
    if (num % 7 == 0 && num % 5 == 0 && num % 11 != 0) {
        nums[counter] = num;
        counter++;
    }
    num--;
} while(num >= 0)

Also here is a page with some stuff about do…while if you just need to study about it:

http://www.keil.com/support/docs/1950.htm

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Edit (sehe) I couldn’t let my hard work go to waste, could I:

For absolute fun and kicks, here is my pythonist response in C++0x

#include <boost/range/adaptors.hpp>
#include <boost/range/irange.hpp>
#include <boost/phoenix/phoenix.hpp>
#include <iostream>

using boost::adaptors::filtered;
using boost::phoenix::arg_names::arg1;
using boost::irange;

int main(int argc, const char *argv[])
{
    for (auto i : irange(200,1,-1) | 
        filtered(!((arg1 % 5) | (arg1 % 7)) && (arg1 % 11)))
            std::cout << i << std::endl;
}