c++ - 使用筛法进行素因数分解

Question

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已关闭 10 年前。

我的代码有什么问题吗？
我试图找到素因数，但在获取输入后它崩溃了，它不起作用。

我现在应该做什么？我在生成素数后使用单独的方法生成素数，我只是在 main 中调用素数函数。

#include<iostream>
#include<cstdio>
#include<math.h>
#define SIZE 1000000
using namespace std;
long p[SIZE],input;
long List[SIZE];  // saves the List
long listSize;   // saves the size of List

void prime(void)
{
       long i,j;
       p[0]=1;
       p[1]=1;
       for(i=2;i<=sqrt(SIZE);i++)  // prime generate part
            if(p[i]==0)
                for(j=2;j*i<=SIZE;j++)
                    p[i*j]=1;
}

void primeFactorize( long n )
{
    listSize = 0;   // Initially the List is empty, we denote that size = 0
    long sqrtN = long( sqrt(n) ); // find the sqrt of the number
    for( long i = 0; p[i] <= sqrtN; i++ ) { // we check up to the sqrt
        if( n % p[i] == 0 ) { // n is multiple of prime[i]
            // So, we continue dividing n by prime[i] as long as possible
            while( n % p[i] == 0 ) {
                n /= p[i]; // we have divided n by prime[i]
                List[listSize] = p[i]; // added the ith prime in the list
                listSize++; // added a prime, so, size should be increased
            }
            // we can add some optimization by updating sqrtN here, since n
            // is decreased. think why it's important and how it can be added
        }
    }
    if( n > 1 )
    {
        // n is greater than 1, so we are sure that this n is a prime
        List[listSize] = n; // added n (the prime) in the list
        listSize++; // increased the size of the list
    }
}
int main()
{
    prime();    
    while(scanf_s("%ld",&input),input)
    {
          if(input==1)
                printf("1 = 1\n");
            else if(input==-1)
                printf("-1 = -1 x 1\n");
            else
            {
                primeFactorize( input );

            }
    for( long i = 0; i < listSize; i++ ) // traverse the List array
                printf("%d ", List[i]);

    }
    return 0;
}

Answer 1

void prime(void){
    long i,j;
    p[0]=1;
    p[1]=1;
    for(i=2;i<=(long)sqrt((double)SIZE);i++)
        if(p[i]==0)
            for(j=2;j*i<SIZE;j++)//<=SIZE ---> <SIZE
                p[i*j]=1;
}

void primeFactorize( long n ){
    listSize = 0;
    long sqrtN = (long)sqrt((double)n);//
    for( long i = 2; i <= sqrtN; i++ ) { //i = 0 ---> i = 2 and p[i] ---> i (Same below)
        while( n % i == 0 ) {//possible [if statement] is omitted
            n /= i;//!
            List[listSize] = i;//!
            listSize++;
        }
    }
    if( n > 1 )
    {
        List[listSize] = n;
        listSize++;
    }
}