在这里我们将看到 Baum Sweet 序列。该序列是一个二进制序列。如果数字n有奇数个连续的0，则第n位将为0，否则第n位将为1。
我们有一个自然数n。我们的任务是找到 Baum Sweet 序列的第 n 项。所以我们必须检查它是否有奇数长度的连续零块。
如果数字是 4，则该项将为 1，因为 4 是 100。所以它有两个（偶数）个0。
算法
BaumSweetSeqTerm (G, s) -
begin
define bit sequence seq of size n
baum := 1
len := number of bits in binary of n
for i in range 0 to len, do
j := i + 1
count := 1
if seq[i] = 0, then
for j in range i + 1 to len, do
if seq[j] = 0, then
increase count
else
break
end if
done
if count is odd, then
baum := 0
end if
end if
done
return baum
end示例
#includeusing namespace std; int BaumSweetSeqTerm(int n) { bitset<32> sequence(n); //store bit-wise representation int len = 32 - __builtin_clz(n); //builtin_clz() function gives number of zeroes present before the first 1 int baum = 1; // nth term of baum sequence for (int i = 0; i < len;) { int j = i + 1; if (sequence[i] == 0) { int count = 1; for (j = i + 1; j < len; j++) { if (sequence[j] == 0) // counts consecutive zeroes count++; else break; } if (count % 2 == 1) //check odd or even baum = 0; } i = j; } return baum; } int main() { int n = 4; cout << BaumSweetSeqTerm(n); }
输出
1
