E1.31don't care about others ,hope be to fluncy希望越来越顺手

gyumni | Aug. 26, 2019, 11:56 p.m.

sicp学上之后，就不在意别人的态度了

(define (factorial n)
(if (= n 1)
1
(* n (factorial (- n 1)))))

product counter · product

counter counter + 1

(define (factorial n)
(fact-iter 1 1 n))

(define (fact-iter product counter max-count)
(if (> counter max-count)
product
(fact-iter (* counter product)
(+ counter 1)
max-count)))

factorial a b count-a count-b

if count-a count-b even?

(b+2)b/s(a)

if count-a count-b odd?

b/a or b*s(b+2)/a*s(a+2)

if count-a  even count-b odd?

b/s(a) or b*s(b+2)*（b+4）/s(a)

if count-a odd count-b even?

b*(b+2)/a or s(a)(a+2)要不要give up and down08/28/2019/07:08:56/range of a b

@1 2*4  3*3           b*(b+2)  s(a)  product = up/down

up =

if count-b is odd?

write a function = if 1 1

3 2

5 3

if n is dont'need count same time for one time

counter-b

2                   x 1   product 2      counter-b = count-b + 1/2

2*4*4            x2   2*s(2+2)

2*4*4*6*6    x3    2*s(2+2)s(2+2+2)

s(b)

define  name  b count-b

define   callup x product counter-b

if (= x counter-b)

product

callup x+1 x*s(product+2)  counter-b

callup 1 2 count-b + 1/2

if count-b is even?

counter-b

2*4                   x 2*(2+2)   product 2*4      counter-b = count-b /2

2*4*4*6            x2   2*s(2+2)*(2+2+2)

2*4*4*6*6    x3    2*s(2+2)s(2+2+2)(this is term)

define  name  b count-b

define   callup x product counter-b

if (= x counter-b)

product

callup x+1 x*(product+2)  counter-b

callup 1 2* count-b/2

b

define <name> <a> <b><term><next>

define

define  <name>  <b> <count-b>

define  <callup> <x> <product> <counter-b>

if (= x counter-b)

product

callup <x+1> <x*(product+2)>  <counter-b>

callup <1> <2*> <count-b/2>

 define        define            if (= x counter-b)                 product                 callup       callup <1> <2*> define              (I need more and more stop now I don't fear anymore)     define            if (= x counter-b)                 product                 callup       callup <1> <2*4>

down

2*4*4 3*3*5      b*s(b+2) s(a)*(a+2) product = up/down

if a is even?

3*3                  product=3*3 counter-a =1

3*3*5*5           product=3*3*5*5   2

3*3*5*5*7*     product=3*3*5*5*7*7 3

define  name  a count-a

define   callup x product counter-a

if (= x counter-a)

product

callup x+1 x*s(product+2)  counter-a

callup 1 3*3 count-a /2

if a is odd?

3                  product=3 counter-a =1

3*3*5           product=3*3*5   2

3*3*5*5*7     product=3*3*5*5*7 3

define  name  a count-a

define   callup x product counter-a

if (= x counter-a)

product

callup x+1 x*s(product+2)  counter-a

callup 1 3*3 count-a /2

@2 2      3

if a is even?

3*3                  product=3*3 counter-a =1

3*3*5*5           product=3*3*5*5   2

3*3*5*5*7*     product=3*3*5*5*7*7 3

define  a-name  b term next   count-b

define   callup x product counter-a

if (= x counter-a)

product

callup x+1 x*s(product+2)[product*trem (product + 2)]  counter-a

callup 1 3*3 count-a /2

four become one procedure，first do simply

define next term a b

if is odd a? term a  [up/down]

if is even a ? term a

if is odd b? term b

if is even b? term b

then ,second

define a procedure that can take any odd or even a and b,then given th up/down(oh ,that isn.t  [up/down]okay)

define a b (this equal to a-count but not a-counter)

define(how to do sum of that?)

may be (name term next )

@3  2    3*3

@4 2*4  3 they have the common function of      2 *(4*4)*(6*6)*(8*8)/

(3*3)*(5*5)*(7*7)

A*B/(A+1)(A+1)=( /4)

A*B=(A+1)*(A+1)*( /4)

A= (A+1)*(A+1)*( /4)/B

give a clue of eveything(iteratively)

define  b-odd  b count-b

define   callup x product counter-b

if (= x counter-b)

product

callup x+1 x*s(product+2)  counter-b

callup 1 2 count-b + 1/2

define  b-even  b count-b

define   callup x product counter-b

if (= x counter-b)

product

callup x+1 x*(product+2)  counter-b

callup 1 2*4 count-b/2

define  a-even  a count-a

define   callup x product counter-a

if (= x counter-a)

product

callup x+1 x*s(product+2)  counter-a

callup 1 3*3 count-a /2

define  a-odd  a count-a

define   callup x product counter-a

if (= x counter-a)

product

callup x+1 x*s(product+2)  counter-a

callup 1 3 count-a+1 /2

define  a-start a next term b(会完善程序了)the thoughes to get the code better

define   callup x product counter-a

if (= x counter-a)

product

callup next x  x*term(product+2)  counter-a

callup 1 3*3 a /2

this is a clue of all clues

and ,if I can know

how to get input a out the product,like b is the samee way,and the four of  one,they are has the product a (need to be odd or even ,diff is with the counter-a) and next is same of plus one, the input procedure could be diffient ,may be t's ,how can you put diffient a and b ,out the diffient though the different term of produ  and count

define t term a b product-default   count-default

define  a-start a next term b(会完善程序了)the thoughes to get the code better

define   callup x product counter-a

if (= x counter-a)

product

callup next x  x*term(product+2)  counter-a

callup 1 3*3 a /2

define aoutput term1/ term2

(a-count = b-c) /4

if  (a-count > b-c)

b/a=( /4)/a-b+b/b

(a-count < b-c)

b/a=( /4)*(b-a)+a/a

if is term2 a?( t x*square a    3 count-a+1 /2 )  [up/down]

if is term1 a ?(t  x*square a    3*3 count-a /2)

if is term2 b?(t x*square b   2  count-b+1/2)

if is term1 b? (t x* b 2*4  count-b /2)

if is odd a? x*square a    3 count-a+1 /2   [up/down]

if is even a ? x*square a    3*3 count-a /2

if is odd b? x*square b   2  count-b+1/2

if is even b? x* b 2*4  count-b /2

sometimes feeling be fooled

(a-count = b-c) /4

if  (a-count > b-c)

A*B/(A+1)(A+1)=( /4)

A*B=(A+1)*(A+1)*( /4)

A= (A+1)*(A+1)*( /4)/B

A/(A+1)*(A+1)= ( /4)/B

b/a=( /4)/a-b+b/b

(a-count < b-c) A*B/(A+1)=( /4) *(A+1) b/a=( /4)*(b-a)+a/a

define eo? (a b)

cond and even? a even? b

(x*square a    3*3 count-a /2

x* b 2*4  count-b /2)

cond and odd? a even? b

(x*square a    3 count-a+1 /2

x* b 2*4  count-b /2)

else and even? a odd? b

(x*square a    3*3 count-a /2

x*square b   2  count-b+1/2

"term1=even?

term2=odd?"

define aoutput term1 term2

if is term2 a? x*square a    3 count-a+1 /2   [up/down]

if is term1 a ? x*square a    3*3 count-a /2

if is term2 b? x*square b   2  count-b+1/2

if is term1 b? x* b 2*4  count-b /2

 想在纸上写

> (aoutput 3)

37575

> (aoutput 4)

97575

> (aoutputb 4)

8192192

> (aoutputb 6)

8192192192

> (aoutputb 8)

8192192192192

> (aoutputb 3)

23232

> (aoutputb 5)

2323232

 give v or next(+ v 2) over

08/29/2019/2:33:32 PM

> (st 6 6)

66661024/1225

> (aoutput 6)

611025

> (aoutputb b)

recursive over 4:28:46 PM/08/29/2019（same day）

> (st 3 3)

332 32 33 45 32/45

brown1730and...

 51  a vast number of similar sequences as over a given range is a and b 这个式子上面的和偶，factorial 上面的奇偶，n*n-1

1 2 product = 1*2

2 3 p =pr*3

2 4 pro-even = 2*4

(2*4) 4 pro-even = 4 * product-even

(2*4*4) 6 pro-even = 6 * product-even

(2*4*4*6) 6 pro-even =6 *product-even

if b - a = 1吧

3  3 pro-odd = 3*3

(3*3)  5 pro-odd = 5*pro-odd

(3*3*5)  5 pro-odd = 5*pro-odd

(3*3*5*5)  7 pro-odd = 7

double 吧

2 *(4*4)*(6*6)*(8*8)/

(3*3)*(5*5)*(7*7)

2 * s(n+2)/s(n+1)

get a n is juduct n is order if n =  if b = if a-b = 1

(/ 2*(n+2)

s(b+1))

if a-b = -1

a b

a:3/3*3*5

b-counter is even

b:2*4

a-count is even

a:3*3

b-count is odd

b:2/2*4*4

(a-count = b-c) 2*4*4/3*3*5  or(2/3) 2*4/3*3 or (order  2*4*4*6/3*3*5*5)

(a-count>b-c)2*4*4*6/3*3*5

= /4

2*4/3=3"before is *3/3"*( /4)(总觉得重要的英文一定会ka，但是我想好的答案是不会)

A*(A+2)/（B）(B)= /4 B=A+1

A*B/(A+1)=( /4) *(A+1)

（a-count<b-count)2*4/3*3

=( /4

2*4/3*3=( /4)

A*B/(A+1)(A+1)=( /4)

A*B=(A+1)*(A+1)*( /4)

A= (A+1)*(A+1)*( /4)/B