我做了一个递归函数来查找数字的质因数，但它有一个错误，导致 Turbo c 退出。请帮忙
#include<stdio.h>
#include<conio.h>
int prime(int num);
int primefactor(int num,int i);
void main(void)
{
int num;
printf("Enter a number whose prime factors are to be calculated:");
scanf("%d",&num);
primefactor(num,i);
i=num
getch();
}
int primefactor(int num,int i)
{
if(i==2)
return 1;
if(num%i==0)
{
if(prime(num))
{
printf(",%d",num);
num=num/i;
i++;
}
}
i--;
primefactor(num,i);
return 0;
}
int prime(int num)
{
int i,flag;
for(i=2;i<num;i++)
{
if(num%i==0)
flag=0;
}
return flag;
}
void main(void)
{
int num,i=num; // (*)
printf("Enter a number whose prime factors are to be calculated:");
scanf("%d",&num);
primefactor(num,i);
getch();
}
您认为
i(*)不确定你想要什么
inumnumvoid main(void)
{
int num,i;
printf("Enter a number whose prime factors are to be calculated:");
scanf("%d",&num);
i = num; // assignment goes here.
primefactor(num,i);
getch();
}
（太困了，无法编写好的代码..所以对于任何错误提前表示歉意：p）
更简单的非递归版本
printPrimeFactors(int num) {
for (i = 2; i < sqrt(num); i=getNextPrime()) {
if (num %i)
printf("%d", i);
}
}
如果你必须使用递归
void factorization(int x, int i=2)
{
if(x==1)
return;
if(x%i==0&&isPrime(i))
{
printf("%d ",i);
factorization(x/i,i);
}
else
factorization(x,i+1);
}
C++ 中的完整递归解决方案（对于 C，用 printf 替换 cout 行）：
void printPrimeFactors(int num)
{
static int divisor = 2; // 2 is the first prime number
if ( num == 1 ) //if num = 1 we finished
{
divisor = 2; //restore divisor, so it'll be ready for the next run
return;
}
else if ( num % divisor == 0 ) //if num divided by divisor
{
cout << divisor << " "; //print divisor
printPrimeFactors( num / divisor ); //call the function with num/divisor
}
else //if num not divided by divisor
{
divisor++; //increase divisor
printPrimeFactors( num );
}
}
以低开销函数调用实现素因数分解的最佳方法是。 。 .
void factors(int number)
{
int divisor = 2;
if (number == 1) { cout << "1"; return; }
while ((number % divisor) && (number > divisor)) divisor++;
cout << divisor << ", ";
factors(number / divisor);
}
函数调用（递归）的次数等于质因子的数量，包括1。
我用 C 语言完成了此操作。根据编译器的不同，可能需要在程序中进行微小的更改。
#include<stdio.h>
int primefact(int);
int main()
{
int n;
printf("Enter a number whose prime factors are to be calculated : \n");
scanf_s("%d", &n);
printf("Prime factors of %d are : ");
primefact(n);
printf("\n");
return 0;
}
int primefact(int n)
{
int i=2;
while(n%i!=0)
i++;
printf("%d ", i);
if(n==i)
return 0;
else
primefact(n/i);
}
同意 IVlad - 另外，当 num 为素数时会发生什么？递归函数会被调用多少次，例如数字 = 7？
#include<stdio.h>
#include<stdlib.h>
int ar[10]={0};
int i=0,j=2;
void P(int n)
{
if(n<=1){
return ;
}
else{
if(n%j == 0){
printf("%d\t",j);
n=n/j;
}
else{
j++;
}
P(n);
}
}
int main(void)
{
int n;
printf("Enter n = ");
scanf("%d",&n);
P(n);
printf("\n");
return 0;
}
// recursivePrime.cpp
// Purpose: factor finding for an integer
// Author: Ping-Sung Liao, Kaohsiung,TAIWAN
// Date: 2017/02/02
// Version : 1.0
// Reference:
// http://stackoverflow.com/questions/3221156/recursive-func-to-find-prime-factors
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int primefactor(int num,int i);
int main(void)
{
int num, i;
printf("Enter a number whose prime factors are to be calculated:");
scanf("%d",&num);
i=int ( sqrt (num) );
primefactor(num,i);
system("pause"); // instead of getch()
}
int primefactor(int num,int i)
{ printf("num %d i=%d\n", num, i);
if (i==1)
printf("prime found= %d\n", num); // prime appearing in he variuable num
else if(num%i==0)
{ primefactor( int (num/i) , int ( sqrt(num/i) ) );
primefactor( i , int (sqrt ( i ) ) );
}
else
{ i--;
primefactor(num,i);
}
return 0;
}
#include <`stdio.h`>
void pf(int,int);
int main()
{
int a,i=2;
printf("Enter the Number:\n");
scanf("%d",&a);
pf(a,i);
}
void pf(int x,int y)
{
if(x==1)
return 1;
else
{
if(x%y==0)
{printf("%d\t",y);
pf(x/y,y);
}
else
{
y++;
pf(x,y);
}
}
}
我们不需要编写函数来计算下一个素数。例如，如果 num 是 24，我们不断地将它除以 2，直到它不再能被 2 整除，那么其他 2 的倍数也不能整除该数字。所以最终只有（可能）素数可以完美地整除任何正整数。
这是我的代码：（我已经为迭代逻辑和递归逻辑编写了源代码）
#include<stdio.h>
void pfactors_rec(int, int);
void pfactors(int);
int main()
{
int num;
printf("Enter a positive integer number\n");
scanf("%d", &num);
printf("\nPrime Factors of %d without using recursion\n", num);
pfactors(num);
printf("\nPrime Factors of %d using recursion\n", num);
pfactors_rec(num, 2);
return 0;
}
void pfactors_rec(int num, int count)
{
if(num < 1)
return;
else if(num % count == 0)
{
printf("%d\n", count);
pfactors_rec(num / count, count);
}
else
{
pfactors_rec(num, count+1);
}
}
void pfactors(int num)
{
int count;
for(count = 2; (num > 1); count++)
{
while(num % count == 0)
{
printf("%d\n", count);
num = num / count;
}
}
printf("\n");
}
//JavaScript onliner to find prime factors using recursion
pf = (n,f,i=f?~-n:2,A=[]) => f ? ~-n && !~-i || !!(n%i) && pf(n,1,~-i) : !~-n ? A : !(n%i) && pf(i,1) ? (A=[...A,i], pf(n/i,0,i,A)) : pf(n,0,-~i,A)
//eg. pf(60) = [2,2,3,5]
java实现..
public class PrimeFactor {
public int divisor=2;
void printPrimeFactors(int num)
{
if(num == 1)
return;
if(num%divisor!=0)
{
while(num%divisor!=0)
++divisor;
}
if(num%divisor==0){
System.out.println(divisor);
printPrimeFactors(num/divisor);
}
}
public static void main(String[] args)
{
PrimeFactor obj = new PrimeFactor();
obj.printPrimeFactors(90);
}
}