





?/p>
2
?/p>
?
?
?/p>
2
?/p>
?/p>
?
?/p>
2
?
?
?
第二?/p>
1
.(
1
?/p>
35
?/p>
?/p>
100011
?/p>
[
35]
?/p>
10100011
[
35]
?/p>
11011100
[
35]
?/p>
11011101
?/p>
2
?/p>
[127]
?/p>
?/p>
01111111
[127]
?/p>
?/p>
01111111
[127]
?/p>
?/p>
01111111
?/p>
3
?/p>
127
?/p>
?/p>
1111111
?/p>
[
127]
?/p>
11111111
[
127]
?/p>
10000001
[
127]
?/p>
10000000
?/p>
4
?/p>
1
?/p>
?/p>
00000001
?/p>
[
1]
?/p>
10000001
[
1]
?/p>
11111111
[
1]
?/p>
11111110
2
?/p>
[x]
?/p>
= a
0
. a
1
a
2
?/p>
a
6
解法一?/p>
?/p>
1
?/p>
?/p>
a
0
= 0,
?/p>
x > 0,
也满?/p>
x > -0.5
此时
a
1
?/p>
a
6
可任?/p>
?
2
?/p>
?/p>
a
0
= 1,
?/p>
x <= 0,
要满?/p>
x > -0.5,
需
a
1
= 1
?/p>
a
0
= 1, a
1
= 1, a
2
?/p>
a
6
有一个不?/p>
0
解法二?/p>
-0.5 = -0.1
(2)
= -0.100000 = 1, 100000
?/p>
1
?/p>
?/p>
x >= 0,
?/p>
a0 = 0, a
1
?/p>
a
6
任意
即可?/p>
?/p>
2
?/p>
[x]
?/p>
= x = a
0
. a
1
a
2
?/p>
a
6
?/p>
2
?/p>
?/p>
x < 0,
?/p>
x > -0.5
只需
-x < 0.5, -x > 0
[x]
?/p>
= -x, [0.5]
?/p>
= 01000000
?/p>
[-x]
?/p>
< 01000000
a
0
*
a
1
*
a
2
a
6
?/p>
1
?/p>
01000000
a
0
*
a
1
*
a
2
a
6
?/p>
00111111