给定长度为N的二进制字符串S ，任务是检查二进制字符串的十进制表示是否可被9整除。
例子：
Input: S = 1010001
Output:Yes
Explanation: The decimal representation of the binary string S is 81, which is divisible by 9. Therefore, the required output is Yes.
Input: S = 1010011
Output: No
Explanation: The decimal representation of the binary string S is 83, which is not divisible by 9. Therefore, the required output is No.
天真的方法：解决此问题的最简单方法是将二进制数字转换为十进制数字，然后检查十进制数字是否可被9整除。如果发现为真，则打印“真”。否则，打印False。
时间复杂度： O(N)
辅助空间： O(1)
高效的方法：如果二进制字符串的长度大于64，则二进制字符串的十进制表示形式将导致溢出。因此，减少溢出问题的想法是将二进制字符串转换成八进制表示并检查二进制字符串的八进制表示是9或不能整除。请按照以下步骤解决问题：
- 将二进制字符串转换为八进制表示形式。
- 初始化一个变量，说Oct_9存储的9八进制表示。
- 找出数字的总和，例如在二进制字符串的八进制表示形式中偶数位置处出现了evenSum 。
- 找出数字的总和，例如在二进制字符串的八进制表示形式中的奇数位置处存在奇数求和。
- 检查abs(oddSum – EvenSum)％Oct_9 == 0 。如果发现是真的，则打印“是” 。
- 否则，打印No。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to convert the binary string
// into octal representation
string ConvertequivalentBase8(string S)
{
// Stores binary representation of
// the decimal value [0 - 7]
map mp;
// Stores the decimal values
// of binary strings [0 - 7]
mp["000"] = '0';
mp["001"] = '1';
mp["010"] = '2';
mp["011"] = '3';
mp["100"] = '4';
mp["101"] = '5';
mp["110"] = '6';
mp["111"] = '7';
// Stores length of S
int N = S.length();
if (N % 3 == 2) {
// Update S
S = "0" + S;
}
else if (N % 3 == 1) {
// Update S
S = "00" + S;
}
// Update N
N = S.length();
// Stores octal representation
// of the binary string
string oct;
// Traverse the binary string
for (int i = 0; i < N; i += 3) {
// Stores 3 consecutive characters
// of the binary string
string temp = S.substr(i, 3);
// Append octal representation
// of temp
oct.push_back(mp[temp]);
}
return oct;
}
// Function to check if binary string
// is divisible by 9 or not
string binString_div_9(string S, int N)
{
// Stores octal representation
// of S
string oct;
oct = ConvertequivalentBase8(S);
// Stores sum of elements present
// at odd positions of oct
int oddSum = 0;
// Stores sum of elements present
// at odd positions of oct
int evenSum = 0;
// Stores length of oct
int M = oct.length();
// Traverse the string oct
for (int i = 0; i < M; i += 2) {
// Update oddSum
oddSum += int(oct[i] - '0');
}
// Traverse the string oct
for (int i = 1; i < M; i += 2) {
// Update evenSum
evenSum += int(oct[i] - '0');
}
// Stores cotal representation
// of 9
int Oct_9 = 11;
// If absolute value of (oddSum
// - evenSum) is divisible by Oct_9
if (abs(oddSum - evenSum) % Oct_9
== 0) {
return "Yes";
}
return "No";
}
// Driver Code
int main()
{
string S = "1010001";
int N = S.length();
cout << binString_div_9(S, N);
} Java
// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to convert the binary string
// into octal representation
static String ConvertequivalentBase8(String S)
{
// Stores binary representation of
// the decimal value [0 - 7]
HashMap mp = new HashMap();
// Stores the decimal values
// of binary Strings [0 - 7]
mp.put("000", '0');
mp.put("001", '1');
mp.put("010", '2');
mp.put("011", '3');
mp.put("100", '4');
mp.put("101", '5');
mp.put("110", '6');
mp.put("111", '7');
// Stores length of S
int N = S.length();
if (N % 3 == 2)
{
// Update S
S = "0" + S;
}
else if (N % 3 == 1)
{
// Update S
S = "00" + S;
}
// Update N
N = S.length();
// Stores octal representation
// of the binary String
String oct = "";
// Traverse the binary String
for(int i = 0; i < N; i += 3)
{
// Stores 3 consecutive characters
// of the binary String
String temp = S.substring(i, i + 3);
// Append octal representation
// of temp
oct += mp.get(temp);
}
return oct;
}
// Function to check if binary String
// is divisible by 9 or not
static String binString_div_9(String S, int N)
{
// Stores octal representation
// of S
String oct = "";
oct = ConvertequivalentBase8(S);
// Stores sum of elements present
// at odd positions of oct
int oddSum = 0;
// Stores sum of elements present
// at odd positions of oct
int evenSum = 0;
// Stores length of oct
int M = oct.length();
// Traverse the String oct
for(int i = 0; i < M; i += 2)
// Update oddSum
oddSum += (oct.charAt(i) - '0');
// Traverse the String oct
for(int i = 1; i < M; i += 2)
{
// Update evenSum
evenSum += (oct.charAt(i) - '0');
}
// Stores octal representation
// of 9
int Oct_9 = 11;
// If absolute value of (oddSum
// - evenSum) is divisible by Oct_9
if (Math.abs(oddSum - evenSum) % Oct_9 == 0)
{
return "Yes";
}
return "No";
}
// Driver Code
public static void main(String[] args)
{
String S = "1010001";
int N = S.length();
System.out.println(binString_div_9(S, N));
}
}
// This code is contributed by grand_master Python3
# Python3 program to implement
# the above approach
# Function to convert the binary
# string into octal representation
def ConvertequivalentBase8(S):
# Stores binary representation of
# the decimal value [0 - 7]
mp = {}
# Stores the decimal values
# of binary strings [0 - 7]
mp["000"] = '0'
mp["001"] = '1'
mp["010"] = '2'
mp["011"] = '3'
mp["100"] = '4'
mp["101"] = '5'
mp["110"] = '6'
mp["111"] = '7'
# Stores length of S
N = len(S)
if (N % 3 == 2):
# Update S
S = "0" + S
elif (N % 3 == 1):
# Update S
S = "00" + S
# Update N
N = len(S)
# Stores octal representation
# of the binary string
octal = ""
# Traverse the binary string
for i in range(0, N, 3):
# Stores 3 consecutive characters
# of the binary string
temp = S[i: i + 3]
# Append octal representation
# of temp
if temp in mp:
octal += (mp[temp])
return octal
# Function to check if binary string
# is divisible by 9 or not
def binString_div_9(S, N):
# Stores octal representation
# of S
octal = ConvertequivalentBase8(S)
# Stores sum of elements present
# at odd positions of oct
oddSum = 0
# Stores sum of elements present
# at odd positions of oct
evenSum = 0
# Stores length of oct
M = len(octal)
# Traverse the string oct
for i in range(0, M, 2):
# Update oddSum
oddSum += ord(octal[i]) - ord('0')
# Traverse the string oct
for i in range(1, M, 2):
# Update evenSum
evenSum += ord(octal[i]) - ord('0')
# Stores cotal representation
# of 9
Oct_9 = 11
# If absolute value of (oddSum
# - evenSum) is divisible by Oct_9
if (abs(oddSum - evenSum) % Oct_9 == 0):
return "Yes"
return "No"
# Driver Code
if __name__ == "__main__":
S = "1010001"
N = len(S)
print(binString_div_9(S, N))
# This code is contributed by chitranayalC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to convert the binary string
// into octal representation
static String ConvertequivalentBase8(String S)
{
// Stores binary representation of
// the decimal value [0 - 7]
Dictionary mp = new Dictionary();
// Stores the decimal values
// of binary Strings [0 - 7]
mp.Add("000", '0');
mp.Add("001", '1');
mp.Add("010", '2');
mp.Add("011", '3');
mp.Add("100", '4');
mp.Add("101", '5');
mp.Add("110", '6');
mp.Add("111", '7');
// Stores length of S
int N = S.Length;
if (N % 3 == 2)
{
// Update S
S = "0" + S;
}
else if (N % 3 == 1)
{
// Update S
S = "00" + S;
}
// Update N
N = S.Length;
// Stores octal representation
// of the binary String
String oct = "";
// Traverse the binary String
for(int i = 0; i < N; i += 3)
{
// Stores 3 consecutive characters
// of the binary String
String temp = S.Substring(0, N);
// Append octal representation
// of temp
if (mp.ContainsKey(temp))
oct += mp[temp];
}
return oct;
}
// Function to check if binary String
// is divisible by 9 or not
static String binString_div_9(String S, int N)
{
// Stores octal representation
// of S
String oct = "";
oct = ConvertequivalentBase8(S);
// Stores sum of elements present
// at odd positions of oct
int oddSum = 0;
// Stores sum of elements present
// at odd positions of oct
int evenSum = 0;
// Stores length of oct
int M = oct.Length;
// Traverse the String oct
for(int i = 0; i < M; i += 2)
// Update oddSum
oddSum += (oct[i] - '0');
// Traverse the String oct
for(int i = 1; i < M; i += 2)
{
// Update evenSum
evenSum += (oct[i] - '0');
}
// Stores octal representation
// of 9
int Oct_9 = 11;
// If absolute value of (oddSum
// - evenSum) is divisible by Oct_9
if (Math.Abs(oddSum - evenSum) % Oct_9 == 0)
{
return "Yes";
}
return "No";
}
// Driver Code
public static void Main(String[] args)
{
String S = "1010001";
int N = S.Length;
Console.WriteLine(binString_div_9(S, N));
}
}
// This code is contributed by shikhasingrajput 输出：
Yes时间复杂度： O(N)
辅助空间： O(N)