将 m 插入 n，使得 m 从位 j 开始，到位 I 结束
原文:https://www . geesforgeks . org/insert-m-n-m-starts-bit-j-ends-bit/
我们有两个数字 n 和 m，以及两个位的位置 I 和 j。从 j 到 I 开始将 m 的位插入 n。我们可以假设位 j 到 I 有足够的空间来容纳所有 m。也就是说，如果 m = 10011，您可以假设 j 和 I 之间至少有 5 位。例如，您不会有 j = 3 和 i = 2，因为 m 不能完全容纳在位 3 和位 2 之间。 例:
Input : n = 1024
m = 19
i = 2
j = 6;
Output : n = 1100
Binary representations of input numbers
m in binary is (10011)2
n in binary is (10000000000)2
Binary representations of output number
(10000000000)2
Input : n = 5
m = 3
i = 1
j = 2
Output : 7
算法:
1\. Clear the bits j through i in n
2\. Shift m so that it lines up with bits j through i
3\. Return Bitwise AND of m and n.
最棘手的部分是第一步。我们如何清除 n 中的位？我们可以用面具来做。除了位 j 到 I 中的 0，此掩码将全部为 1。我们通过首先创建掩码的左半部分，然后创建右半部分来创建此掩码。 以下是上述方法的实施。
C++
// C++ program for implementation of updateBits()
#include <bits/stdc++.h>
using namespace std;
// Function to updateBits M insert to N.
int updateBits(int n, int m, int i, int j)
{
/* Create a mask to clear bits i through j
in n. EXAMPLE: i = 2, j = 4\. Result
should be 11100011\. For simplicity, we'll
use just 8 bits for the example. */
int allOnes = ~0; // will equal sequence of all ls
// ls before position j, then 0s. left = 11100000
int left= allOnes << (j + 1);
// l's after position i. right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put min there */
int n_cleared = n & mask; // Clear bits j through i.
int m_shifted = m << i; // Move m into correct position.
return (n_cleared | m_shifted); // OR them, and we're done!
}
// Driver Code
int main()
{
int n = 1024; // in Binary N= 10000000000
int m = 19; // in Binary M= 10011
int i = 2, j = 6;
cout << updateBits(n,m,i,j);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for implementation of updateBits()
class UpdateBits
{
// Function to updateBits M insert to N.
static int updateBits(int n, int m, int i, int j)
{
/* Create a mask to clear bits i through j
in n. EXAMPLE: i = 2, j = 4\. Result
should be 11100011\. For simplicity, we'll
use just 8 bits for the example. */
int allOnes = ~0; // will equal sequence of all ls
// ls before position j, then 0s. left = 11100000
int left= allOnes << (j + 1);
// l's after position i. right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put min there */
// Clear bits j through i.
int n_cleared = n & mask;
// Move m into correct position.
int m_shifted = m << i;
// OR them, and we're done!
return (n_cleared | m_shifted);
}
public static void main (String[] args)
{
// in Binary N= 10000000000
int n = 1024;
// in Binary M= 10011
int m = 19;
int i = 2, j = 6;
System.out.println(updateBits(n,m,i,j));
}
}
Python 3
# Python3 program for implementation
# of updateBits()
# Function to updateBits M insert to N.
def updateBits(n, m, i, j):
# Create a mask to clear bits i through
# j in n. EXAMPLE: i = 2, j = 4\. Result
# should be 11100011\. For simplicity,
# we'll use just 8 bits for the example.
# will equal sequence of all ls
allOnes = ~0
# ls before position j,
# then 0s. left = 11100000
left = allOnes << (j + 1)
# l's after position i. right = 00000011
right = ((1 << i) - 1)
# All ls, except for 0s between
# i and j. mask 11100011
mask = left | right
# Clear bits j through i then put min there
n_cleared = n & mask
# Move m into correct position.
m_shifted = m << i
return (n_cleared | m_shifted)
# Driver Code
n = 1024 # in Binary N = 10000000000
m = 19 # in Binary M = 10011
i = 2; j = 6
print(updateBits(n, m, i, j))
# This code is contributed by Anant Agarwal.
C
// C# program for implementation of
// updateBits()
using System;
class GFG {
// Function to updateBits M
// insert to N.
static int updateBits(int n, int m,
int i, int j)
{
/* Create a mask to clear bits i
through j in n. EXAMPLE: i = 2,
j = 4\. Result should be 11100011.
For simplicity, we'll use just 8
bits for the example. */
// will equal sequence of all ls
int allOnes = ~0;
// ls before position j, then 0s.
// left = 11100000
int left= allOnes << (j + 1);
// l's after position i.
// right = 00000011
int right = ((1 << i) - 1);
// All ls, except for 0s between i
// and j. mask 11100011
int mask = left | right;
/* Clear bits j through i then put
min there */
// Clear bits j through i.
int n_cleared = n & mask;
// Move m into correct position.
int m_shifted = m << i;
// OR them, and we're done!
return (n_cleared | m_shifted);
}
public static void Main()
{
// in Binary N= 10000000000
int n = 1024;
// in Binary M= 10011
int m = 19;
int i = 2, j = 6;
Console.WriteLine(updateBits(n, m, i, j));
}
}
//This code is contributed by Anant Agarwal.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for implementation
// of updateBits()
// Function to updateBits
// M insert to N.
function updateBits($n, $m, $i, $j)
{
// Create a mask to clear
// bits i through j in n.
// EXAMPLE: i = 2, j = 4.
// Result should be 11100011.
// For simplicity, we'll use
// just 8 bits for the example.
// will equal sequence of all ls
$allOnes = ~0;
// ls before position j, then
// 0s. left = 11100000
$left= $allOnes << ($j + 1);
// l's after position i.
// right = 00000011
$right = ((1 << $i) - 1);
// All ls, except for 0s between
// i and j. mask 11100011
$mask = $left | $right;
// Clear bits j through i
// then put min there
// Clear bits j through i.
$n_cleared = $n & $mask;
// Move m into correct position.
$m_shifted = $m << $i;
// OR them, and we're done!
return ($n_cleared | $m_shifted);
}
// Driver Code
// in Binary N= 10000000000
$n = 1024;
// in Binary M= 10011
$m = 19;
$i = 2;
$j = 6;
echo updateBits($n, $m, $i, $j);
// This code is contributed by Ajit
?>
java 描述语言
<script>
// JavaScript program for implementation of updateBits()
// Function to updateBits M insert to N.
function updateBits(n,m,i,j)
{
/* Create a mask to clear bits i through j
in n. EXAMPLE: i = 2, j = 4\. Result
should be 11100011\. For simplicity, we'll
use just 8 bits for the example. */
let allOnes = ~0; // will equal sequence of all ls
// ls before position j, then 0s. left = 11100000
let left= allOnes << (j + 1);
// l's after position i. right = 00000011
let right = ((1 << i) - 1);
// All ls, except for 0s between i and j. mask 11100011
let mask = left | right;
/* Clear bits j through i then put min there */
// Clear bits j through i.
let n_cleared = n & mask;
// Move m into correct position.
let m_shifted = m << i;
// OR them, and we're done!
return (n_cleared | m_shifted);
}
// in Binary N= 10000000000
let n = 1024;
// in Binary M= 10011
let m = 19;
let i = 2, j = 6;
document.write(updateBits(n,m,i,j));
// This code is contributed by sravan
</script>
输出:
1100 // in Binary (10001001100)2
本文由 Somesh Awasthi 先生供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果你发现任何不正确的地方，或者你想分享更多关于上面讨论的话题的信息，请写评论。
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