[Challenge:1] Multiples of 3 and 5

Pikami · 11-25-2016, 10:08 PM

(11-25-2016, 09:01 PM)Vi-Sion Wrote: @Pikami you share the solution here in this thread .
I think you missed the thread with the rules and the explanation so here you go https://sinister.li/Thread-Competition-R...scoreboard.

Thanks, that explained a lot.

Here is my solution in C++ (Tested on windows):

Code:
#include <iostream>
#include <vector>

int getSum();           //A function that gets the sum of all the multiples of 3 or 5 below 1000
int reverseNum(int a);  //A function that reverses a number
int sortNum(int a);     //A function for sorting a number


int main(){
    int sum = getSum();
    std::cout<<"The sum of all the multiples of 3 or 5 below 1000: "<<sum<<std::endl;
    std::cout<<"This number reversed is: "<<reverseNum(sum)<<std::endl;
    std::cout<<"This number sorted number is: "<<sortNum(sum)<<std::endl;
}

int getSum(){
    int sum = 0;
    for(int i=1;i<1000;i++)
        if(i%3==0 || i%5==0)
            sum+=i;
    return sum;
}

int reverseNum(int a){
    int newNum = 0; //The new number
    while(a>0){
        newNum*=10;
        newNum += a%10;
        a/=10;
    }
    return newNum;
}

int sortNum(int a){
    int n = 0;          //The number of digits
    std::vector<int> D; //The digits
    //Get the digits
    while(a>0){
        D.push_back(a%10);
        a/=10;
        n++;
    }
    //Sort the vector
    for(int i=0;i<n;i++)
        for(int o=i;o<n;o++)
            if(D[o]<D[i])
                std::swap(D[o],D[i]);
    //Build the new number
    int newNum = 0;
    for(int i=0;i<n;i++)
        newNum=newNum*10+D[i];
    return newNum;
}

I don't know if I was allowed to use built-in sorting algorithms so I've written my own.

Here's the output:

Code:
The sum of all the multiples of 3 or 5 below 1000: 233168
This number reversed is: 861332
This number sorted number is: 123368