我想找到一个数的所有因子。
目前我有这样的代码:
{
int n;
int i=2;
scanf("%d",&n);
while(i<=n/2)
{
if(n%i==0)
printf("%d,",i);
i++;
}
getch();
}
有没有什么办法可以改善它?
我想找到一个数的所有因子。
目前我有这样的代码:
{
int n;
int i=2;
scanf("%d",&n);
while(i<=n/2)
{
if(n%i==0)
printf("%d,",i);
i++;
}
getch();
}
有没有什么办法可以改善它?
首先,你的代码应该有条件i ≤ n/2,否则可能会漏掉其中一个因数,例如当n=12时,6将不会被打印出来。
循环运行至数字的平方根(即i ≤ sqrt(n)),并打印出i和n/i(两者都是n的倍数)。
{
int n;
int i=2;
scanf("%d",&n);
while(i <= sqrt(n))
{
if(n%i==0) {
printf("%d,",i);
if (i != (n / i)) {
printf("%d,",n/i);
}
}
i++;
}
getch();
}
注意:
i*i == n
。#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
unsigned int FindDivisors(unsigned long long divisors[], unsigned long long N) {
unsigned int lastdiv = 0;
divisors[lastdiv++] = 1;
unsigned long long powerfactor = 1;
unsigned long long number = N;
while ((number & 1) == 0) {
powerfactor <<= 1;
divisors[lastdiv++] = powerfactor;
number >>= 1;
}
unsigned long long factor = 3; unsigned long long upto = lastdiv;
powerfactor = 1;
while (factor * factor <= number) {
if (number % factor == 0) {
powerfactor *= factor;
for (unsigned int i = 0; i < upto; i++)
divisors[lastdiv++] = divisors[i] * powerfactor;
number /= factor;
}
else {
factor += 2; upto = lastdiv;
powerfactor = 1;
}
}
if (number > 1) {
if (number != factor) {
upto = lastdiv;
powerfactor = 1;
}
powerfactor *= number;
for (unsigned int i = 0; i < upto; i++)
divisors[lastdiv++] = divisors[i] * powerfactor;
}
return lastdiv;
}
int cmp(const void *a, const void *b) {
if( *(long long*)a-*(long long*)b < 0 ) return -1;
if( *(long long*)a-*(long long*)b > 0 ) return 1;
return 0;
}
int main(int argc, char *argv[]) {
unsigned long long N = 2;
unsigned int Ndigit = 1;
if (argc > 1) {
N = strtoull(argv[1], NULL, 10);
Ndigit = strlen(argv[1]);
}
unsigned int maxdiv[] = {1, 4, 12, 32, 64, 128, 240, 448, 768, 1344,
2304, 4032, 6720, 10752, 17280, 26880, 41472, 64512, 103680};
unsigned long long divisors[maxdiv[Ndigit]];
unsigned int size = FindDivisors(divisors, N);
printf("Number of divisors = %u\n", size);
qsort(divisors, size, sizeof(unsigned long long), cmp);
for (unsigned int i = 0; i < size; i++)
printf("%llu ", divisors[i]);
printf("\n");
return 0;
}
public static long GetDivisors(long number)
{
long divisors = 0;
long boundary = (long)Math.Sqrt(number);
for (int i = 1; i <= boundary; i++)
{
if (number % i == 0)
{
divisors++;
if(i != (number / i))
{
if (i * i != number)
{
divisors++;
}
}
}
}
return divisors;
}
在其中一个答案中呈现的代码存在一个难以一眼看出的错误。如果sqrt(n)是一个有效的除数;但n不是一个完全平方数,则会省略两个结果。
例如,尝试n = 15
,看看会发生什么;sqrt(15) = 3
,因此while循环的最后一个值为2。执行下一个语句if (i * i == n)
将被执行为if(3 * 3 == 15)
。因此,3未列为除数,也错过了5。
以下内容将正确处理正整数的一般情况。
{
int n;
int i=2;
scanf("%d",&n);
while(i <= sqrt(n))
{
if(n%i==0) {
printf("%d,",i);
if (i != (n / i)) {
printf("%d,",n/i);
}
}
i++;
}
getch();
}
当给定的数字为奇数时,我们甚至可以跳过偶数。 接受代码的轻微改进 :)
这是查找给定数字因子的Java代码。
import java.util.Scanner;
public class Factors {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t=scanner.nextInt();
while(t-- > 0) {
int n = scanner.nextInt();
if(n % 2 == 0) {
for(int i = 1; i <= Math.sqrt(n); i++) {
if(n % i == 0) {
System.out.println(i + ", ");
if(i != n/i) {
System.out.println(n/i + ", ");
}
}
}
}
else {
for(int i = 1; i <= Math.sqrt(n); i=i+2) {
if(n % i == 0) {
System.out.println(i + ", ");
if(i != n/i) {
System.out.println(n/i + ", ");
}
}
}
}
}
}
}
int count = 2;
//long childsum = 0;
long _originalvalue = sum;
dividend = "1";
for (int i = 2; i < sum; i++)
{
if (_originalvalue % i == 0)
{
sum = _originalvalue / i;
//sum = childsum;
dividend = dividend + "," + i+","+sum;
if (sum == i)
{
count++;
}
else
{
count = count + 2;
}
}
}
return count;
printf("%d,", i);printf("%d,", n/i);
改为printf("%d,", i); if(i != n/i){printf("%d,", n/i);}
。对于一个完全平方数,它不会打印两次根数。 - MYMNeoi < sqrt(n)
来避免多次检查n/i != i
,这意味着n/i != i
,然后像单独的边缘情况一样在while循环外检查i=sqrt(n)
。 - chepneri <= sqrt(n)
会将 i 提升为 double 类型。你可以使用i <= (int)sqrt(n)
来避免这种情况。将if (i != (n / i))
改为if (i*i != n)
可以提高代码的运行速度。 - rcgldr