以递减顺序打印给定整数的质因数
原文:
给定一个整数 n ，任务是使用。
示例:
输入: n = 34 输出: 17 2 说明: 数字 34 的质因数是 2 和 17。
输入:n = 8 t3】输出: 2
方式:思路是使用存储nt7】的所有中的所有值。按照以下步骤解决问题:
- 初始化一个，比如 st 。
- 同时 n！= 1 。从 i = 2、开始，对于 i 的每个值，运行一个循环，直到 n % i == 0 和 st 并将 n 更新为 n/i.
- 最后， st 。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to print prime factors
// of n in decreasing order
void primefactors(int n)
{
// stores prime factors of n
// in decreasing order
stack st;
int i = 2;
while (n != 1) {
if (n % i == 0) {
// insert i into stack
st.push(i);
while (n % i == 0) {
// update n
n = n / i;
}
}
// update i
i ;
}
// print value of stack st
while (!st.empty()) {
printf("%d ", st.top());
st.pop();
}
}
// driver code
int main()
{
int n = 8;
// function call
primefactors(n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.util.*;
class gfg{
// function to print prime factors
// of n in decreasing order
static void primefactors(int n)
{
// stores prime factors of n
// in decreasing order
stack st = new stack<>();
int i = 2;
while (n != 1)
{
if (n % i == 0)
{
// insert i into stack
st.push(i);
while (n % i == 0)
{
// update n
n = n / i;
}
}
// update i
i ;
}
// print value of stack st
while (!st.isempty())
{
system.out.println(st.peek());
st.pop();
}
}
// driver code
public static void main (string[] args)
{
int n = 8;
// function call
primefactors(n);;
}
}
// this code is contributed by susmitakundugoaldanga
python 3
# python3 program for the above approach
# function to print prime factors
# of n in decreasing order
def primefactors(n):
# stores prime factors of n
# in decreasing order
st = []
i = 2
while (n != 1):
if (n % i == 0):
# insert i into stack
st.append(i)
while (n % i == 0):
# update n
n = n // i
# update i
i = 1
# print value of stack st
while (len(st) != 0):
print(st[-1])
st.pop()
# driver code
if __name__ == "__main__":
n = 8
# function call
primefactors(n)
# this code is contributed by chitranayal.
c
// c# program for the above approach
using system;
using system.collections.generic;
class gfg{
// function to print prime factors
// of n in decreasing order
static void primefactors(int n)
{
// stores prime factors of n
// in decreasing order
stack st = new stack();
int i = 2;
while (n != 1)
{
if (n % i == 0)
{
// insert i into stack
st.push(i);
while (n % i == 0)
{
// update n
n = n / i;
}
}
// update i
i ;
}
// print value of stack st
while (st.count != 0)
{
console.write(st.peek());
st.pop();
}
}
// driver code
public static void main ()
{
int n = 8;
// function call
primefactors(n);;
}
}
// this code is contributed by code_hunt
java 描述语言
output:
2
时间复杂度: o(sqrt(n)) 辅助空间: o(1)
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