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
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|Example
#include using namespace std;//left and right strings will be filled up, di will hold the difference between left and rightvoid 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输出
100001100010101011110011100100101101101110110101110110111111登录后复制
以上就是长度为n的所有可能的二进制数，两半部分的和相等？的详细内容，更多请关注【创想鸟】其它相关文章！
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