给定数字N ，找到所有长度为2 * N的二进制序列，以使前N个位的总和与后N个位的总和相同。
例子：
Input: N = 2
Output:
0000
0101
0110
1001
1010
1111
Input: N = 1
Output:
00
11
注意：可在此处找到针对此问题的递归方法。
方法：
一种简单的方法来运行从0到2 2 * N的循环并转换为二进制形式，并检查上半部分的总和是否等于下半部分的总和。
如果上述条件成立，则打印该号码，否则检查下一个号码。
下面是上述方法的实现：
C++
// C++ implementation
#include
#include
#include
using namespace std;
// Function to convert the
// number into binary and
// store the number into
// an array
void convertToBinary(int num, int a[], int n)
{
int pointer = n - 1;
while (num > 0) {
a[pointer] = num % 2;
num = num / 2;
pointer--;
}
}
// Function to check if the
// sum of the digits till
// the mid of the array and
// the sum of the digits
// from mid till n is the
// same, if they are same
// then print that binary
void checkforsum(int a[], int n)
{
int sum1 = 0;
int sum2 = 0;
int mid = n / 2;
// Calculating the sum from
// 0 till mid and store
// in sum1
for (int i = 0; i < mid; i++)
sum1 = sum1 + a[i];
// Calculating the sum
// from mid till n and
// store in sum2
for (int j = mid; j < n; j++)
sum2 = sum2 + a[j];
// If sum1 is same as
// sum2 print the binary
if (sum1 == sum2) {
for (int i = 0; i < n; i++)
cout << a[i];
cout << "\n";
}
}
// Function to print sequence
void print_seq(int m)
{
int n = (2 * m);
// Creating the array
int a[n];
// Initialize the array
// with 0 to store the
// binary nnumbers
for (int j = 0; j < n; j++) {
a[j] = 0;
}
for (int i = 0; i < (int)pow(2, n); i++) {
// Converting the number
// into binary first
convertToBinary(i, a, n);
// Checking if the sum of
// the first half of the
// array is same as the
// sum of the next half
checkforsum(a, n);
}
}
// Driver Code
int main()
{
int m = 2;
print_seq(m);
return 0;
} Java
// Java code implementation for above approach
class GFG
{
// Function to convert the
// number into binary and
// store the number into
// an array
static void convertToBinary(int num,
int a[], int n)
{
int pointer = n - 1;
while (num > 0)
{
a[pointer] = num % 2;
num = num / 2;
pointer--;
}
}
// Function to check if the
// sum of the digits till
// the mid of the array and
// the sum of the digits
// from mid till n is the
// same, if they are same
// then print that binary
static void checkforsum(int a[], int n)
{
int sum1 = 0;
int sum2 = 0;
int mid = n / 2;
// Calculating the sum from
// 0 till mid and store
// in sum1
for (int i = 0; i < mid; i++)
sum1 = sum1 + a[i];
// Calculating the sum
// from mid till n and
// store in sum2
for (int j = mid; j < n; j++)
sum2 = sum2 + a[j];
// If sum1 is same as
// sum2 print the binary
if (sum1 == sum2)
{
for (int i = 0; i < n; i++)
System.out.print(a[i]);
System.out.println();
}
}
// Function to print sequence
static void print_seq(int m)
{
int n = (2 * m);
// Creating the array
int a[] = new int[n];
// Initialize the array
// with 0 to store the
// binary nnumbers
for (int j = 0; j < n; j++)
{
a[j] = 0;
}
for (int i = 0; i < (int)Math.pow(2, n); i++)
{
// Converting the number
// into binary first
convertToBinary(i, a, n);
// Checking if the sum of
// the first half of the
// array is same as the
// sum of the next half
checkforsum(a, n);
}
}
// Driver Code
public static void main (String[] args)
{
int m = 2;
print_seq(m);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of above approach
# Function to convert the number into binary
# and store the number into an array
def convertToBinary(num, a, n):
pointer = n - 1
while (num > 0):
a[pointer] = num % 2
num = num // 2
pointer -= 1
# Function to check if the sum of the digits till
# the mid of the array and the sum of the digits
# from mid till n is the same, if they are same
# then print that binary
def checkforsum(a, n):
sum1 = 0
sum2 = 0
mid = n // 2
# Calculating the sum from 0 till mid
# and store in sum1
for i in range(mid):
sum1 = sum1 + a[i]
# Calculating the sum from mid till n
# and store in sum2
for j in range(mid, n):
sum2 = sum2 + a[j]
# If sum1 is same as sum2 print the binary
if (sum1 == sum2):
for i in range(n):
print(a[i], end = "")
print()
# Function to prsequence
def print_seq(m):
n = (2 * m)
# Creating the array
a = [0 for i in range(n)]
for i in range(pow(2, n)):
# Converting the number
# into binary first
convertToBinary(i, a, n)
# Checking if the sum of the first half
# of the array is same as the sum of
# the next half
checkforsum(a, n)
# Driver Code
m = 2
print_seq(m)
# This code is contributed by mohit kumarC#
// C# code implementation for above approach
using System;
class GFG
{
// Function to convert the
// number into binary and
// store the number into
// an array
static void convertToBinary(int num,
int []a, int n)
{
int pointer = n - 1;
while (num > 0)
{
a[pointer] = num % 2;
num = num / 2;
pointer--;
}
}
// Function to check if the
// sum of the digits till
// the mid of the array and
// the sum of the digits
// from mid till n is the
// same, if they are same
// then print that binary
static void checkforsum(int []a, int n)
{
int sum1 = 0;
int sum2 = 0;
int mid = n / 2;
// Calculating the sum from
// 0 till mid and store
// in sum1
for (int i = 0; i < mid; i++)
sum1 = sum1 + a[i];
// Calculating the sum
// from mid till n and
// store in sum2
for (int j = mid; j < n; j++)
sum2 = sum2 + a[j];
// If sum1 is same as
// sum2 print the binary
if (sum1 == sum2)
{
for (int i = 0; i < n; i++)
Console.Write(a[i]);
Console.WriteLine();
}
}
// Function to print sequence
static void print_seq(int m)
{
int n = (2 * m);
// Creating the array
int []a = new int[n];
// Initialize the array
// with 0 to store the
// binary nnumbers
for (int j = 0; j < n; j++)
{
a[j] = 0;
}
for (int i = 0; i < (int)Math.Pow(2, n); i++)
{
// Converting the number
// into binary first
convertToBinary(i, a, n);
// Checking if the sum of
// the first half of the
// array is same as the
// sum of the next half
checkforsum(a, n);
}
}
// Driver Code
public static void Main (String[] args)
{
int m = 2;
print_seq(m);
}
}
// This code is contributed by PrinciRaj1992输出：
0000
0101
0110
1001
1010
1111