原文:
给定一个数字 n ,任务是找到不同的 n 数字,这样他们的产品就是一个。 举例:
输入: n = 3 输出: 1,8,27 解释: 输出数的乘积= 1 * 8 * 27 = 216,是 6 的完美立方(6 3 = 216) 输入: n = 2 输出: 1 8 解释:t20
方法:pg电子试玩链接的解决方案基于 的事实
第一个‘n’个完美立方数的乘积永远是一个完美立方。
因此,前 n 个自然数的将被打印为输出。 例如:
for n = 1 => [1]
product is 1
and cube root of 1 is also 1
for n = 2 => [1, 8]
product is 8
and cube root of 8 is 2
for n = 3 => [1, 8, 27]
product is 216
and cube root of 216 is 6
and so on
以下是上述方法的实现:
c
// c program to find n numbers such that
// their product is a perfect cube
#include
using namespace std;
// function to find the n numbers such
//that their product is a perfect cube
void findnumbers(int n)
{
int i = 1;
// loop to traverse each
//number from 1 to n
while (i <= n) {
// print the cube of i
//as the ith term of the output
cout << (i * i * i)
<< " ";
i ;
}
}
// driver code
int main()
{
int n = 4;
// function call
findnumbers(n);
}
java 语言(一种计算机语言,尤用于创建网站)
// java program to find n numbers such that
// their product is a perfect cube
import java.util.*;
class gfg
{
// function to find the n numbers such
//that their product is a perfect cube
static void findnumbers(int n)
{
int i = 1;
// loop to traverse each
//number from 1 to n
while (i <= n) {
// print the cube of i
//as the ith term of the output
system.out.print( (i * i * i)
" ");
i ;
}
}
// driver code
public static void main (string []args)
{
int n = 4;
// function call
findnumbers(n);
}
}
// this code is contributed by chitranayal
python 3
# python3 program to find n numbers such that
# their product is a perfect cube
# function to find the n numbers such
# that their product is a perfect cube
def findnumbers(n):
i = 1
# loop to traverse each
# number from 1 to n
while (i <= n):
# print the cube of i
# as the ith term of the output
print((i * i * i), end=" ")
i = 1
# driver code
if __name__ == '__main__':
n = 4
# function call
findnumbers(n)
# this code is contributed by mohit kumar 29
c
// c# program to find n numbers such that
// their product is a perfect cube
using system;
class gfg
{
// function to find the n numbers such
//that their product is a perfect cube
static void findnumbers(int n)
{
int i = 1;
// loop to traverse each
//number from 1 to n
while (i <= n) {
// print the cube of i
//as the ith term of the output
console.write( (i * i * i)
" ");
i ;
}
}
// driver code
public static void main (string []args)
{
int n = 4;
// function call
findnumbers(n);
}
}
// this code is contributed by yash_r
java 描述语言
output:
1 8 27 64
业绩分析:
- 时间复杂度:和上面的方法一样,我们正在寻找 n 个数的完美立方,因此需要 o(n) 时间。
- 辅助空间复杂度:和上面的方法一样,没有使用额外的空间;因此辅助空间的复杂性将是 o(1) 。
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