具體數學第三章作業解答
老師的具體數學作業要電子版了,那就把我自己的解答放在這裏。
10.
\[ \begin{array}{l} \left \lceil \frac{2x+1} {2} \right \rceil-\left \lceil \frac{2x+1} {4} \right \rceil+\left \lfloor \frac{2x+1} {4} \right \rfloor \=\left \lceil \frac{2x+1} {2} \right \rceil-(\left \lceil \frac{2x+1} {4} \right \rceil-\left \lfloor \frac{2x+1} {4} \right \rfloor)\=\left \lceil x+\frac{1} {2} \right \rceil-[\frac{2x+1}{4}不是整數] \end{array} \]
若\(\frac{2x+1}{4}\)是整數,則:
\[
\begin{array}{l}2x+1=4N \quad (N為整數)\2x=4N-1\x=2N-\frac{1}{2} \end{array}
\]
- 若\({x}\neq\frac{1} {2}\),則原式=\(\left \lceil x+\frac{1}{2} \right \rceil-1=\left \lfloor x \right \rfloor\)
- 若\(x=\frac{1} {2}\),則原式=\(\left \lceil x+\frac{1}{2} \right \rceil=x+\frac{1}{2}=\left \lceil x \right\rceil\)
12.
\[ \begin{array}{l}\left \lceil \frac{n}{m} \right \rceil=\left \lfloor \frac{n+m-1}{m} \right \rfloor\\left \lfloor \frac{n+m-1}{m} \right \rfloor=\left \lfloor \frac{n}{m} +1-\frac{1}{m}\right \rfloor=\left \lfloor \frac{n-1}{m} \right \rfloor+1 \end{array} \]
則證明:\(\left \lceil \frac{n}{m} \right \rceil-\left \lfloor \frac{n-1}{m} \right \rfloor=1\)
易知:\(0<\frac{n}{m}-\frac{n-1}{m}\leq1\)(當且僅當m=1時,等式成立)
當m=1時,\(\left \lceil n \right \rceil-\left \lfloor n-1 \right \rfloor=n-n+1=1成立\)
當\(m\neq1\)時,
若\(\frac{n}{m}\)為整數,則\(\frac{n-1}{m}<\frac{n-1}{m}且\frac{n-1}{m}不為整數\)
則\(\left \lceil \frac{n}{m} \right \rceil-\left \lfloor \frac{n-1}{m} \right \rfloor=\frac{n}{m}-\left \lfloor \frac{n-1}{m}\right \rfloor=1\)
若\(\frac{n-1}{m}\)為整數,則\(\frac{n-1}{m}<\frac{n-1}{m}且\frac{n}{m}不為整數\)
則\(\left \lceil \frac{n}{m} \right \rceil-\left \lfloor \frac{n-1}{m} \right \rfloor=\left\lfloor\frac{n}{m}\right\rfloor-\frac{n-1}{m}=1\)
若\(\frac{n-1}{m}和\frac{n}{m}\)均非整數,則n mod m<1 ,(n-1) mod m<1且\(\left \lfloor \frac{n}{m} \right \rfloor=\left \lfloor \frac{n-1}{m} \right \rfloor\), 則\(\left \lceil \frac{n}{m} \right \rceil-\left \lfloor \frac{n-1}{m} \right \rfloor=1\)
證畢
23.
設第n個元素為\(x_n\)且為第m組, 則\(x_n=m\)
此時:
\[
\begin{array}{l} \frac{1}{2}m(m-1)<n\leq\frac{1}{2}m(m+1)\m^2-m<2n\leq m^2+m\m^2-m+\frac{1}{4}<2n<m^2+m+\frac{1}{4} \quad m,n均為正整數,左側小於2n,在加上\frac{1}{4},大小關系不改變\(m-\frac{1}{2})^2<2n<(m+\frac{1}{2})^2\m-\frac{1}{2}<\sqrt{2n}<m+\frac{1}{2}\m<\sqrt{2n}+\frac{1}{2}<m+1\則m=\left \lfloor \sqrt{2n}+\frac{1}{2} \right \rfloor\即x_n=\left \lfloor \sqrt{2n}+\frac{1}{2} \right \rfloor\\ \end{array}
\]
約瑟夫環
n個人,每隔q個人去掉1人,最終剩下的人的編號?
n個人,初始編號為1, 2, ..., n
重新編號,第1個人:n+1,第2個人:n+2,直至第q個人:去掉,第q+1個人:n+q
假設當前去掉的人的編號為kq,此時去掉了k個人,接下來的人的編號為n+k(q-1)+1
也即:原來kq+d -> 現在n+k(q-1)+d
最後去掉的人編號為nq
令N=n+k(q-1)+d
上一次編號為kq+d=kq+N-n-k(q-1)=k+N-n
\(k=\frac{N-n-d}{q-1}=\left \lfloor \frac{N-n-1}{q-1} \right \rfloor\)
上一次編號為:
\(\left \lfloor \frac{N-n-1}{q-1} \right \rfloor+N-n\)
令D=qn+1-N替代N
則
\[ \begin{array}{l}D = qn + 1 - N \\ = qn + 1 - \left( {\left\lfloor {\frac{ {(qn + 1 - D) - n - 1}}{ {q - 1}}} \right\rfloor + qn + 1 - D - n} \right)\\ = n + D - \left\lfloor {\frac{ {(q - 1)n - D}}{ {q - 1}}} \right\rfloor \\ = D - \left\lfloor {\frac{ { - D}}{ {q - 1}}} \right\rfloor \\ = D + \left\lceil {\frac{D}{ {q - 1}}} \right\rceil \\ = \left\lceil {\frac{q}{ {q - 1}}D} \right\rceil \end{array} \]
具體數學第三章作業解答