仅使用质数2、3和5生成序列，然后显示第n个项(C ++) [英] Generating a sequence using prime numbers 2, 3, and 5 only, and then displaying an nth term (C++)

问题描述

我正在研究一个问题，该问题要求使用质数2、3和5生成一个序列，然后在序列中显示第n个数字.因此，如果我要求程序显示第1000个数字，则应该显示它.

I'm working on a problem that asks to generate a sequence using prime numbers 2, 3, and 5, and then displaying then nth number in the sequence. So, if I ask the program to display the 1000th number, it should display it.

我不能使用数组或类似的东西，而只能使用基本的决策和循环.

I can't be using arrays or anything like that, just basic decisions and loops.

我开始研究它并撞到墙...这就是我得到的:

I started working on it and hit a wall... here's what I got:

#include <iostream>

using namespace std;
int main() {
    unsigned int n=23;
    for(int i=2; i<n; i++){
        if(i%2==0){
            cout<<i<<", ";
        }else if(i%3==0){
            cout<<i<<", ";
        }else if(i%5==0){
            cout<<i<<", ";
        }
    }

    return 0;
}

不幸的是，该代码无法满足要求.它显示诸如14之类的数字，其中包括质数7....这些数字只能除以3个指定的质数(2,3,5).

Unfortunately, that code doesn't do what's required. It displays numbers such as 14, which includes a prime number 7.... The numbers can only be divided by the 3 specified primes (2,3,5).

我找到了一些我想理解的信息，但到目前为止还不确定如何实现它……也许使用了许多for()循环?因此，看来我必须使用2 ^ n * 3 ^ m * 5 ^ k的概念，其中n + m + k> 0.

I found some information that I'm trying to understand, and so far not sure how to implement it... maybe using lots of for() loops? So, it appears I have to use the concept of 2^n * 3^m * 5^k where n+m+k>0.

我想我必须通过一个数字来进行测试，首先检查它是否可以被2 ^ 1 * 3 ^ 0 * 5 ^ 0完全整除，然后再进行2 ^ 0 * 3 ^ 1 * 5 ^ 0的整除，然后2 ^ 0 * 3 ^ 0 * 5 ^ 1，依此类推...只是不确定从哪里开始.

I guess I have to run a number through a test where it checks to see first if it's fully divisible by 2^1 * 3^0 * 5^0, then 2^0 * 3^1 * 5^0, then 2^0 * 3^0 * 5^1, and so on... Just not sure where to begin.

推荐答案

选中此项.

#include <iostream>
using namespace std;

int IsPrime(int var);
int CheckifPrimeGreaterThaFive(int Num);
int GetFactors(int Num)
{
    int i =0,j=0;
    for (i =2,j=0; i <= Num; i++)
    {
        if (Num%i == 0)
        {
           if (1 == CheckifPrimeGreaterThaFive(i))
           {
                 return 1;
              }
        }
    }
    return 0;
}

int CheckifPrimeGreaterThaFive(int Num)
{
   if ((Num != 2 && Num != 3 && Num != 5) && IsPrime(Num))
   {

           return 1;
   }

    return 0;
}

int IsPrime(int var)
{
    for (int i = 2; i <= var/2; i++)
    {
        if (var % i == 0)
           return 0;
    }
    return 1;
}


int main() {
    int n=98;
    int i, FactorsCount=0;

    for(i=2; i<n; i++)
    {
        if (0 == GetFactors(i))
        {  
           cout<<" "<<i;
        }   
    }
    return 0;
}

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