長度為n的所有可能的二進制數，兩半部的和相等？

這裡我們將看到所有可能的n位元二進位數（n由使用者給出），其中每一半的和相同。例如，如果數字是 10001，這裡 10 和 01 是相同的，因為它們的總和相同，並且它們位於不同的一半。這裡我們將產生該類型的所有數字。

演算法

genAllBinEqualSumHalf(n, left, right, diff)

left和right最初為空，diff保存差異左右之間

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

範例

#include <bits/stdc++.h>
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);
}

輸出

100001
100010
101011
110011
100100
101101
101110
110101
110110
111111

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