长度为n的所有可能的二进制数，两半部分的和相等？

Here we will see all possible binary numbers of n bit (n is given by the user) where the sum of each half is same. For example, if the number is 10001 here 10 and 01 are same because their sum is same, and they are in the different halves. Here we will generate all numbers of that type.

Algorithm

genAllBinEqualSumHalf(n, left, right, diff)

left and right are initially empty, diff is holding difference between left and right

艺帆集团公司企业网站源码1.7.5

艺帆集团公司企业网站源码基于艺帆企业cms制作，全站div+css 制作；它包含了单页设置、单页分类设置、新闻、产品、下载、在线招聘、在线留言、幻灯管理、友情链接管理和数据库备份等功能。 DIV+CSS布局优势一.精简代码，减少重构难度。网站使用DIV+CSS布局使代码很是精简，相信大多朋友也都略有所闻，css文件可以在网站的任意一个页面进行调用，而若是使用table表格修改部分页面却是显得很麻烦

下载

Begin
   if n is 0, then
      if diff is 0, then
         print left + right
      end if
      return
   end if
   if n is 1, then
      if diff is 0, then
         print left + 0 + right
         print left + 1 + right
      end if
      return
   end if
   if 2* |diff| <= n, then
      if left is not blank, then
         genAllBinEqualSumHalf(n-2, left + 0, right + 0, diff)
         genAllBinEqualSumHalf(n-2, left + 0, right + 1, diff-1)
      end if
      genAllBinEqualSumHalf(n-2, left + 1, right + 0, diff + 1)
      genAllBinEqualSumHalf(n-2, left + 1, right + 1, diff)
   end if
End

Example

#include 
using namespace std;
//left and right strings will be filled up, di will hold the difference between left and right
void genAllBinEqualSumHalf(int n, string left="", string right="", int di=0) {
   if (n == 0) { //when the n is 0
      if (di == 0) //if diff is 0, then concatenate left and right
         cout << left + right << " ";
      return;
   }
   if (n == 1) {//if 1 bit number is their
      if (di == 0) { //when difference is 0, generate two numbers one with 0 after left, another with 1 after left, then add right
         cout << left + "0" + right << " ";
         cout << left + "1" + right << " ";
      }
      return;
   }
   if ((2 * abs(di) <= n)) {
      if (left != ""){ //numbers will not start with 0
         genAllBinEqualSumHalf(n-2, left+"0", right+"0", di);
         //add 0 after left and right
         genAllBinEqualSumHalf(n-2, left+"0", right+"1", di-1);
         //add 0 after left, and 1 after right, so difference is 1 less
      }
      genAllBinEqualSumHalf(n-2, left+"1", right+"0", di+1); //add 1 after left, and 0 after right, so difference is 1 greater
      genAllBinEqualSumHalf(n-2, left+"1", right+"1", di); //add 1 after left and right
   }
}
main() {
   int n = 5;
   genAllBinEqualSumHalf(n);
}

输出

100001
100010
101011
110011
100100
101101
101110
110101
110110
111111