具体数学第三章作业解答

老师的具体数学作业要电子版了，那就把我自己的解答放在这里。

10.

[egin{array}{l} left lceil frac{2x+1} {2} ight ceil-left lceil frac{2x+1} {4} ight ceil+left lfloor frac{2x+1} {4} ight floor \ =left lceil frac{2x+1} {2} ight ceil-(left lceil frac{2x+1} {4} ight ceil-left lfloor frac{2x+1} {4} ight floor)\ =left lceil x+frac{1} {2} ight ceil-[frac{2x+1}{4}不是整数] end{array} ]

若(frac{2x+1}{4})是整数，则：

[egin{array}{l}2x+1=4N quad (N为整数)\ 2x=4N-1\ x=2N-frac{1}{2} end{array} ]

• 若({x} eqfrac{1} {2})，则原式=(left lceil x+frac{1}{2} ight ceil-1=left lfloor x ight floor)
• 若(x=frac{1} {2})，则原式=(left lceil x+frac{1}{2} ight ceil=x+frac{1}{2}=left lceil x ight ceil)

12.

[egin{array}{l}left lceil frac{n}{m} ight ceil=left lfloor frac{n+m-1}{m} ight floor\ left lfloor frac{n+m-1}{m} ight floor=left lfloor frac{n}{m} +1-frac{1}{m} ight floor=left lfloor frac{n-1}{m} ight floor+1 end{array} ]

则证明:(left lceil frac{n}{m} ight ceil-left lfloor frac{n-1}{m} ight floor=1)即可

易知:(0<frac{n}{m}-frac{n-1}{m}leq1)（当且仅当m=1时，等式成立）

• 当m=1时，(left lceil n ight ceil-left lfloor n-1 ight floor=n-n+1=1成立)

• 当(m eq1)时，

  • 若(frac{n}{m})为整数，则(frac{n-1}{m}<frac{n-1}{m}且frac{n-1}{m}不为整数)

    则(left lceil frac{n}{m} ight ceil-left lfloor frac{n-1}{m} ight floor=frac{n}{m}-left lfloor frac{n-1}{m} ight floor=1)

  • 若(frac{n-1}{m})为整数，则(frac{n-1}{m}<frac{n-1}{m}且frac{n}{m}不为整数)

    则(left lceil frac{n}{m} ight ceil-left lfloor frac{n-1}{m} ight floor=leftlfloorfrac{n}{m} ight floor-frac{n-1}{m}=1)

  • 若(frac{n-1}{m}和frac{n}{m})均非整数，则n mod m<1 ,(n-1) mod m<1且(left lfloor frac{n}{m} ight floor=left lfloor frac{n-1}{m} ight floor), 则(left lceil frac{n}{m} ight ceil-left lfloor frac{n-1}{m} ight floor=1)

证毕

23.

设第n个元素为(x_n)且为第m组, 则(x_n=m)

此时:

[egin{array}{l} frac{1}{2}m(m-1)<nleqfrac{1}{2}m(m+1)\ m^2-m<2nleq m^2+m\ m^2-m+frac{1}{4}<2n<m^2+m+frac{1}{4} quad m,n均为正整数，左侧小于2n，在加上frac{1}{4}，大小关系不改变\ (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\ 则m=left lfloor sqrt{2n}+frac{1}{2} ight floor\ 即x_n=left lfloor sqrt{2n}+frac{1}{2} ight floor\ end{array} ]

约瑟夫环

n个人，每隔q个人去掉1人，最终剩下的人的编号?

n个人，初始编号为1, 2, ..., n

重新编号，第1个人：n+1，第2个人：n+2，直至第q个人：去掉，第q+1个人：n+q

假设当前去掉的人的编号为kq，此时去掉了k个人，接下来的人的编号为n+k(q-1)+1

也即：原来kq+d -> 现在n+k(q-1)+d

最后去掉的人编号为nq

令N=n+k(q-1)+d

上一次编号为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} ight floor)

上一次编号为：

(left lfloor frac{N-n-1}{q-1} ight floor+N-n)

令D=qn+1-N替代N

则

[egin{array}{l}D = qn + 1 - N \ = qn + 1 - left( {leftlfloor {frac{ {(qn + 1 - D) - n - 1}}{ {q - 1}}} ight floor + qn + 1 - D - n} ight)\ = n + D - leftlfloor {frac{ {(q - 1)n - D}}{ {q - 1}}} ight floor \ = D - leftlfloor {frac{ { - D}}{ {q - 1}}} ight floor \ = D + leftlceil {frac{D}{ {q - 1}}} ight ceil \ = leftlceil {frac{q}{ {q - 1}}D} ight ceil end{array} ]