给定两个二进制字符串，执行运算直到 B > 0 并打印结果
原文:https://www . geesforgeks . org/给定-两个二进制字符串-执行-操作-直到-B- 0-并打印结果/
给定两个长度为 N 和 M 的二进制字符串 A 和 B (最多 10 个 5 个)。任务是重复下面的过程，找到答案。
Initialize ans = 0
while (B > 0)
ans += A & B (bitwise AND)
B = B / 2
print ans
注:答案可以很大所以打印答案% 100000007。 例:
Input: A = "1001", B = "10101"
Output: 11
1001 & 10101 = 1, ans = 1, B = 1010
1001 & 1010 = 8, ans = 9, B = 101
1001 & 101 = 1, ans = 10, B = 10
1001 & 10 = 0, ans = 10, B = 1
1001 & 1 = 1, ans = 11, B = 0
Input: A = "1010", B = "1101"
Output: 12
方法:由于在所有迭代中只有 B 受到影响，将一个二进制数除以 2 意味着将其右移 1 位，因此可以观察到 A 中的一个位将仅受到位于左侧的 B 中的设置位的影响，即比当前位(包括当前位)更重要。例如A =“1001”*B =“10101”*， A 中的最低有效位将仅受 B 中的设置位的影响，即总共 3 位和 A 中的最高有效位将仅受 B 中的单个设置位的影响，即 B 中的最高有效位，因为在执行逐位“与”时，所有其他设置位在循环的任何迭代中都不会对其产生影响 所以最终的结果会是20 3+23 1 = 3+8 = 11。 以下是上述办法的实施情况。
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mod (int)(1e9 + 7)
// Function to return the required result
ll BitOperations(string a, int n, string b, int m)
{
// Reverse the strings
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
// Count the number of set bits in b
int c = 0;
for (int i = 0; i < m; i++)
if (b[i] == '1')
c++;
// To store the powers of 2
ll power[n];
power[0] = 1;
// power[i] = pow(2, i) % mod
for (int i = 1; i < n; i++)
power[i] = (power[i - 1] * 2) % mod;
// To store the final answer
ll ans = 0;
for (int i = 0; i < n; i++) {
if (a[i] == '1') {
// Add power[i] to the ans after
// multiplying it with the number
// of set bits in b
ans += c * power[i];
if (ans >= mod)
ans %= mod;
}
// Divide by 2 means right shift b>>1
// if b has 1 at right most side than
// number of set bits will get decreased
if (b[i] == '1')
c--;
// If no more set bits in b i.e. b = 0
if (c == 0)
break;
}
// Return the required answer
return ans;
}
// Driver code
int main()
{
string a = "1001", b = "10101";
int n = a.length(), m = b.length();
cout << BitOperations(a, n, b, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
static int mod = (int)(1e9 + 7);
// Function to return the required result
static int BitOperations(String a,
int n, String b, int m)
{
// Reverse the strings
char[] ch1 = a.toCharArray();
reverse( ch1 );
a = new String( ch1 );
char[] ch2 = b.toCharArray();
reverse( ch2 );
b = new String( ch2 );
// Count the number of set bits in b
int c = 0;
for (int i = 0; i < m; i++)
if (b.charAt(i) == '1')
c++;
// To store the powers of 2
int[] power = new int[n];
power[0] = 1;
// power[i] = pow(2, i) % mod
for (int i = 1; i < n; i++)
power[i] = (power[i - 1] * 2) % mod;
// To store the final answer
int ans = 0;
for (int i = 0; i < n; i++)
{
if (a.charAt(i) == '1')
{
// Add power[i] to the ans after
// multiplying it with the number
// of set bits in b
ans += c * power[i];
if (ans >= mod)
ans %= mod;
}
// Divide by 2 means right shift b>>1
// if b has 1 at right most side than
// number of set bits will get decreased
if (b.charAt(i) == '1')
c--;
// If no more set bits in b i.e. b = 0
if (c == 0)
break;
}
// Return the required answer
return ans;
}
static void reverse(char a[])
{
int i, k,n=a.length;
char t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
// Driver code
public static void main(String[] args)
{
String a = "1001", b = "10101";
int n = a.length(), m = b.length();
System.out.println(BitOperations(a, n, b, m));
}
}
// This code contributed by Rajput-Ji
Python 3
# Python 3 implementation of the approach
mod = 1000000007
# Function to return the required result
def BitOperations(a, n, b, m):
# Reverse the strings
a = a[::-1]
b = b[::-1]
# Count the number of set
# bits in b
c = 0
for i in range(m):
if (b[i] == '1'):
c += 1
# To store the powers of 2
power = [None] * n
power[0] = 1
# power[i] = pow(2, i) % mod
for i in range(1, n):
power[i] = (power[i - 1] * 2) % mod
# To store the final answer
ans = 0
for i in range(0, n):
if (a[i] == '1'):
# Add power[i] to the ans after
# multiplying it with the number
# of set bits in b
ans += c * power[i]
if (ans >= mod):
ans %= mod
# Divide by 2 means right shift b>>1
# if b has 1 at right most side than
# number of set bits will get decreased
if (b[i] == '1'):
c -= 1
# If no more set bits in b i.e. b = 0
if (c == 0):
break
# Return the required answer
return ans
# Driver code
if __name__ == '__main__':
a = "1001"
b = "10101"
n = len(a)
m = len(b)
print(BitOperations(a, n, b, m))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the approach
using System;
using System.Collections;
class GFG
{
static int mod = (int)(1e9 + 7);
// Function to return the required result
static int BitOperations(string a,
int n, string b, int m)
{
// Reverse the strings
char[] ch1 = a.ToCharArray();
Array.Reverse( ch1 );
a = new string( ch1 );
char[] ch2 = b.ToCharArray();
Array.Reverse( ch2 );
b = new string( ch2 );
// Count the number of set bits in b
int c = 0;
for (int i = 0; i < m; i++)
if (b[i] == '1')
c++;
// To store the powers of 2
int[] power = new int[n];
power[0] = 1;
// power[i] = pow(2, i) % mod
for (int i = 1; i < n; i++)
power[i] = (power[i - 1] * 2) % mod;
// To store the final answer
int ans = 0;
for (int i = 0; i < n; i++)
{
if (a[i] == '1')
{
// Add power[i] to the ans after
// multiplying it with the number
// of set bits in b
ans += c * power[i];
if (ans >= mod)
ans %= mod;
}
// Divide by 2 means right shift b>>1
// if b has 1 at right most side than
// number of set bits will get decreased
if (b[i] == '1')
c--;
// If no more set bits in b i.e. b = 0
if (c == 0)
break;
}
// Return the required answer
return ans;
}
// Driver code
static void Main()
{
string a = "1001", b = "10101";
int n = a.Length, m = b.Length;
Console.WriteLine(BitOperations(a, n, b, m));
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
$GLOBALS['mod'] = (1e9 + 7);
// Function to return the required result
function BitOperations($a, $n, $b, $m)
{
// Reverse the strings
$a = strrev($a);
$b = strrev($b);
// Count the number of set bits in b
$c = 0;
for ($i = 0; $i < $m; $i++)
if ($b[$i] == '1')
$c++;
# To store the powers of 2
$power = array() ;
$power[0] = 1;
// power[i] = pow(2, i) % mod
for ($i = 1; $i < $n; $i++)
$power[$i] = ($power[$i - 1] * 2) %
$GLOBALS['mod'];
// To store the final answer
$ans = 0;
for ($i = 0; $i < $n; $i++)
{
if ($a[$i] == '1')
{
// Add power[i] to the ans after
// multiplying it with the number
// of set bits in b
$ans += $c * $power[$i];
if ($ans >= $GLOBALS['mod'])
$ans %= $GLOBALS['mod'];
}
// Divide by 2 means right shift b>>1
// if b has 1 at right most side than
// number of set bits will get decreased
if ($b[$i] == '1')
$c--;
// If no more set bits in b i.e. b = 0
if ($c == 0)
break;
}
// Return the required answer
return $ans;
}
// Driver code
$a = "1001";
$b = "10101";
$n = strlen($a);
$m = strlen($b);
echo BitOperations($a, $n, $b, $m);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// JavaScript implementation of the approach
let mod = (1e9 + 7);
// Function to return the required result
function BitOperations(a, n, b, m)
{
// Reverse the strings
let ch1 = a.split('');
ch1.reverse();
a = ch1.join("");
let ch2 = b.split('');
ch2.reverse();
b = ch2.join("");
// Count the number of set bits in b
let c = 0;
for (let i = 0; i < m; i++)
if (b[i] == '1')
c++;
// To store the powers of 2
let power = new Array(n);
power[0] = 1;
// power[i] = pow(2, i) % mod
for (let i = 1; i < n; i++)
power[i] = (power[i - 1] * 2) % mod;
// To store the final answer
let ans = 0;
for (let i = 0; i < n; i++)
{
if (a[i] == '1')
{
// Add power[i] to the ans after
// multiplying it with the number
// of set bits in b
ans += c * power[i];
if (ans >= mod)
ans %= mod;
}
// Divide by 2 means right shift b>>1
// if b has 1 at right most side than
// number of set bits will get decreased
if (b[i] == '1')
c--;
// If no more set bits in b i.e. b = 0
if (c == 0)
break;
}
// Return the required answer
return ans;
}
let a = "1001", b = "10101";
let n = a.length, m = b.length;
document.write(BitOperations(a, n, b, m));
</script>
Output:
11
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