打印前 n 个自然数的正方形
原文:
给定一个自然数“n”，不使用、/和-打印前 n 个自然数的平方。 例:*
input: n = 5
output: 0 1 4 9 16
input: n = 6
output: 0 1 4 9 16 25
强烈建议尽量减少浏览器，先自己试试这个。 方法一:思路是利用之前的平方值计算下一个平方。考虑以下 x 的平方和(x-1)之间的关系。我们知道(x-1)的平方是(x-1)2–2 * x 1。我们可以把 x 2 写成
x2 = (x-1)2 2*x - 1
x2 = (x-1)2 x (x - 1)
在编写迭代程序时，我们可以跟踪 x 的先前值，并将 x 的当前值和先前值加到 square 的当前值上。这样我们甚至不用“-”运算符。 以下是上述方法的实施:
c
// c program to print squares of first 'n' natural numbers
// without using *, / and -
#include
using namespace std;
void printsquares(int n)
{
// initialize 'square' and previous value of 'x'
int square = 0, prev_x = 0;
// calculate and print squares
for (int x = 0; x < n; x )
{
// update value of square using previous value
square = (square x prev_x);
// print square and update prev for next iteration
cout << square << " ";
prev_x = x;
}
}
// driver program to test above function
int main()
{
int n = 5;
printsquares(n);
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print squares
// of first 'n' natural numbers
// without using *, / and
import java.io.*;
class gfg
{
static void printsquares(int n)
{
// initialize 'square' and
// previous value of 'x'
int square = 0, prev_x = 0;
// calculate and
// print squares
for (int x = 0; x < n; x )
{
// update value of square
// using previous value
square = (square x prev_x);
// print square and update
// prev for next iteration
system.out.print( square " ");
prev_x = x;
}
}
// driver code
public static void main (string[] args)
{
int n = 5;
printsquares(n);
}
}
// this code is contributed
// by akt_mit
python 3
# python 3 program to print squares of first
# 'n' natural numbers without using *, / and -
def printsquares(n):
# initialize 'square' and previous
# value of 'x'
square = 0; prev_x = 0;
# calculate and print squares
for x in range(0, n):
# update value of square using
# previous value
square = (square x prev_x)
# print square and update prev
# for next iteration
print(square, end = " ")
prev_x = x
# driver code
n = 5;
printsquares(n);
# this code is contributed
# by akanksha rai
c
// c# program to print squares
// of first 'n' natural numbers
// without using *, / and
using system;
public class gfg{
static void printsquares(int n)
{
// initialize 'square' and
// previous value of 'x'
int square = 0, prev_x = 0;
// calculate and
// print squares
for (int x = 0; x < n; x )
{
// update value of square
// using previous value
square = (square x prev_x);
// print square and update
// prev for next iteration
console.write( square " ");
prev_x = x;
}
}
// driver code
static public void main (){
int n = 5;
printsquares(n);
}
}
// this code is contributed
// by ajit
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
0 1 4 9 16
时间复杂度:0(n)
辅助空间:0(1)
方法二:前 n 个奇数之和是从 1 到 n 的自然数的平方，比如 1，1 3，1 3 5，1 3 5 7，1 3 5 7 9，…。 以下是基于上述概念的程序。感谢 aadithya umashanker 和 raviteja 提出这种方法。
c
// c program to print squares of first 'n' natural numbers
// without using *, / and -
#include
using namespace std;
void printsquares(int n)
{
// initialize 'square' and first odd number
int square = 0, odd = 1;
// calculate and print squares
for (int x = 0; x < n; x )
{
// print square
cout << square << " ";
// update 'square' and 'odd'
square = square odd;
odd = odd 2;
}
}
// driver program to test above function
int main()
{
int n = 5;
printsquares(n);
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print
// squares of first 'n'
// natural numbers without
// using *, / and -
import java.io.*;
class gfg
{
static void printsquares(int n)
{
// initialize 'square'
// and first odd number
int square = 0, odd = 1;
// calculate and
// print squares
for (int x = 0; x < n; x )
{
// print square
system.out.print(square
" " );
// update 'square'
// and 'odd'
square = square odd;
odd = odd 2;
}
}
// driver code
public static void main (string[] args)
{
int n = 5;
printsquares(n);
}
}
// this code is contributed
// by ajit
python 3
# python3 program to print squares
# of first 'n' natural numbers
# without using *, / and -
def printsquares(n):
# initialize 'square' and
# first odd number
square = 0
odd = 1
# calculate and print squares
for x in range(0 , n):
# print square
print(square, end= " ")
# update 'square' and 'odd'
square = square odd
odd = odd 2
# driver code
n = 5;
printsquares(n)
# this code is contributed
# by rajput-ji
c
// c# program to print squares of first 'n'
// natural numbers without using *, / and -
using system;
class gfg
{
static void printsquares(int n)
{
// initialize 'square'
// and first odd number
int square = 0, odd = 1;
// calculate and
// print squares
for (int x = 0; x < n; x )
{
// print square
console.write(square " " );
// update 'square'
// and 'odd'
square = square odd;
odd = odd 2;
}
}
// driver code
public static void main ()
{
int n = 5;
printsquares(n);
}
}
// this code is contributed
// by inder_verma..
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
0 1 4 9 16
时间复杂度:0(n)
辅助空间:0(1)
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