Question

我做了以下练习：

设计一个程序，查找从1到1000的所有数字，其素数因子加在一起时，总和为素数（例如，12的素数因子为2,2和3，总和为7，其中是素数）。实现该算法的代码。

我的代码：

#include <iostream> 
#include <string>

//function prototype
int PrimeFactor(int number);
bool isPrime(int PrimeOrNot);

int main() {
    int i;
    for (i = 0; i < 1000; i++)
        PrimeFactor(i);
}

int PrimeFactor(int number) {
    int i;
    int firstsplit = 0;
    int secondsplit = 0;
    int nextSplit = 0;
    int afternextsplit = 0;
    int sum = 0;

    for (i = 2; i < number; i++) //loop that find the number that split it to two number , the purpose is for buiding the factor tree. 
    {
        if (number % i == 0) {
            firstsplit = number / i;
            secondsplit = number / firstsplit;

            if (isPrime(firstsplit)) 
                sum = sum + firstsplit;

            if (isPrime(secondsplit)) 
                sum = sum + secondsplit;

            break;
        }
    }

    if (isPrime(firstsplit) == false) //check if the splitted number is not already a factor from previous loop.
    {
        while (isPrime(firstsplit) == false) {
            for (i = 2; i < firstsplit; i++) //loop that continue to split the splitted number for the purpose of prime factors.
            {
                if (firstsplit % i == 0) {
                    nextSplit = firstsplit / i;
                    afternextsplit = firstsplit / nextSplit;

                    if (isPrime(nextSplit)) 
                        sum = sum + nextSplit;

                    if (isPrime(afternextsplit)) 
                        sum = sum + afternextsplit;
                }
            }
        }
    }
}

bool isPrime(int PrimeOrNot) {
    int i;
    for (i = 0; i < PrimeOrNot; i++) {
        if (PrimeOrNot % i == 0) 
            return false;
    }
    return true;
}

问题：

我认为我的程序似乎不那么有效，因为有很多变量，我想它可以用更少的时间来完成。任何人都可以通过一个素因子树找到另一种方法来帮助我吗？

Prime树示例：

Answer 1

这是解决任务的一种有效方法。我正在使用筛子，您可以参考https://www.quora.com/What-is-the-sieve-in-c++/answer/Ankit-Vallecha

#include<iostream>
#include<cmath>

using namespace std;

#define N 1000
int flag[N+1];  // flag[i] =0 means i is prime, flag[i] =1 means i is not prime
int primes[N+1]; // primes[i] contains the (i+1)th prime, like prime[0]=2, prime[1]=3, prime[2]=5...
int len=0; // len determines the length of primes array filled

void sieve(){ // this function calculates all the primes less than N in primes array and number of primes is in len.

    for( int i=2; i<=sqrt(N); i++ ){
        if( flag[i] == 0 )
            for( int j=2*i; j<=N; j+=i )
                flag[j] = 1;
    }
    // now after this first loop flag[i] is ready to determine prime

    for( int i=2; i<=N; i++ )
        if( flag[i] == 0 )
            primes[len++] = i;

    // now primes contain the primes
}

bool isFactorSumPrime( int x ){ // this functions check whether sum of prime factors is prime or not
    int sum = 0; // sum of factors
    for( int i=0; i<len; i++ ){
        if( x==1 ) break; // x has no prime factors
        while( x%primes[i] == 0 ){  // if primes[i] is factor of x
            sum += primes[i];
            x /= primes[i];
        }
    }
    return !flag[sum];
}

int main(){
    sieve();
    for( int i=2; i<=N; i++ )
        if( isFactorSumPrime(i) == true )
            cout << i << endl;
    return 0;
}

跳入c ++（在素因子树上练习）

1 个答案: