CareerCup

Is it not enough to check weather the array contains elements which are divisible by 3*5*7.

This is a standard question available in cracking the coding interview book..!! Have a look at it..!!

In the book, step 6, when we find the smallest number x among Q3,Q5,Q7, and why if x is in Q5, we only append x*5,x*7 in Q5 and Q7（same to when x is in Q7）, how we can make sure x*3 is ready exist in the list? Thanks, need tips.

#include<iostream>

using namespace std;

int getNthNumber(int number,int Number[]){
    int five,seven,three,nextNumber;
    five=seven=three=0;
    Number[0]=1;

    for(int i=1;i<number;i++){
            nextNumber=min(min(Number[five]*5,Number[seven]*7),min(Number[five]*5,Number[three]*3));
            if(nextNumber==Number[five]*5)
                five++;
            if(nextNumber==Number[seven]*7)
                seven++;
            if(nextNumber==Number[three]*3)
                three++;
            Number[i]=nextNumber;
    }

    return nextNumber;
}

int main(){
    int number=6;int Number[150];
    cout<<getNthNumber(number,Number);
    return 0;
}

//Print out all prime numbers from 1 to n
private static void printOutAllPrimeNumbers(int n)
{
int i = 0, j = 0;

bool prime = true;

for (i = 1; i < n; i++)
{
prime = true;

for (j = 2; j <= i / 2; j++)
{
if (i % j == 0)
prime = false;
}
if (prime == true)
Console.WriteLine(i);
}
}

considering the array as sorted in ascending order...one can check that whether the number has only 3 5 and 7 as factors...if yes then increment the count...when the count reaches k then print that number...

The code to check whether the number i s factor of 3 5 and 7 is--

//code to divide the number by max power of 3 or 5 or 7
int maxdivide(int n, int m)
{
if((n%m) == 0)
n = n/m;

return n;
}


//code to check whether the number is multiple of 3 5 and 7 only
int check(int n)
{
if(n>1)
n = maxdivide(n,3);

if(n>1)
n = maxdivide(n,5);

if(n>1)
n = maxdivide(n,7);

if(n==1)
return 1;
else return 0;
}