使用按位运算符
生成 N 的前 K 倍
原文:https://www . geesforgeks . org/generate-first-k-倍数-n-use-bitwise-operator/
给定一个整数 N ，任务是使用按位运算符打印 N 的前 K 倍。
示例:
输入: N = 16，K = 7 T3】输出:T5】16 * 1 = 16 16 * 2 = 32 16 * 3 = 48 16 * 4 = 64 16 * 5 = 80 16 * 6 = 96 16 * 7 = 112 输入: N = 7， K = 10 输出: 7 * 1 = 7 7 * 2 = 14 7 * 3 = 21 7 * 4 = 28 7 * 5 = 35 7 * 6 = 42 7 * 7 = 49 7 * 8 = 56 7 * 9 = 63 7 * 10 = 70
进场:
按照以下步骤解决问题:
- 迭代到 K 。
- 每次迭代，打印 N 的当前值。
- 然后，计算 N 的每一个 i 第组位的2IT3】之和。将此总和加到 N 上，重复上述步骤。
以下是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the first K
// multiples of N
void Kmultiples(int n, int k)
{
int a = n;
for (int i = 1; i <= k; i++) {
// Print the value of N*i
cout << n << " * " << i << " = "
<< a << endl;
int j = 0;
// Iterate each bit of N and add
// pow(2, pos), where pos is the
// index of each set bit
while (n >= (1 << j)) {
// Check if current bit at
// pos j is fixed or not
a += n & (1 << j);
// For next set bit
j++;
}
}
}
// Driver Code
int main()
{
int N = 16, K = 7;
Kmultiples(N, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to print the first K
// multiples of N
static void Kmultiples(int n, int k)
{
int a = n;
for (int i = 1; i <= k; i++)
{
// Print the value of N*i
System.out.print(n+ " * " + i+ " = "
+ a +"\n");
int j = 0;
// Iterate each bit of N and add
// Math.pow(2, pos), where pos is the
// index of each set bit
while (n >= (1 << j))
{
// Check if current bit at
// pos j is fixed or not
a += n & (1 << j);
// For next set bit
j++;
}
}
}
// Driver Code
public static void main(String[] args)
{
int N = 16, K = 7;
Kmultiples(N, K);
}
}
// This code is contributed by Rohit_ranjan
Python 3
# Python3 program to implement
# the above approach
# Function to print the first K
# multiples of N
def Kmultiples(n, k):
a = n
for i in range(1, k + 1):
# Print the value of N*i
print("{} * {} = {}".format(n, i, a))
j = 0
# Iterate each bit of N and add
# pow(2, pos), where pos is the
# index of each set bit
while(n >= (1 << j)):
# Check if current bit at
# pos j is fixed or not
a += n & (1 << j)
# For next set bit
j += 1
# Driver Code
N = 16
K = 7
Kmultiples(N, K)
# This code is contributed by Shivam Singh
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print the first K
// multiples of N
static void Kmultiples(int n, int k)
{
int a = n;
for(int i = 1; i <= k; i++)
{
// Print the value of N*i
Console.Write(n + " * " + i +
" = " + a + "\n");
int j = 0;
// Iterate each bit of N and add
// Math.Pow(2, pos), where pos is
// the index of each set bit
while (n >= (1 << j))
{
// Check if current bit at
// pos j is fixed or not
a += n & (1 << j);
// For next set bit
j++;
}
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 16, K = 7;
Kmultiples(N, K);
}
}
// This code is contributed by Amit Katiyar
java 描述语言
<script>
// javascript program to implement
// the above approach
// Function to print the first K
// multiples of N
function Kmultiples(n , k)
{
var a = n;
for (i = 1; i <= k; i++)
{
// Print the value of N*i
document.write(n+ " * " + i+ " = "
+ a +"<br>");
var j = 0;
// Iterate each bit of N and add
// Math.pow(2, pos), where pos is the
// index of each set bit
while (n >= (1 << j))
{
// Check if current bit at
// pos j is fixed or not
a += n & (1 << j);
// For next set bit
j++;
}
}
}
// Driver Code
var N = 16, K = 7;
Kmultiples(N, K);
// This code contributed by Princi Singh
</script>
Output:
16 * 1 = 16
16 * 2 = 32
16 * 3 = 48
16 * 4 = 64
16 * 5 = 80
16 * 6 = 96
16 * 7 = 112
时间复杂度:O(Klog2N) 辅助空间: O(1)
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