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

也许企业和fl是两个等级

发现式子不重要delete

(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  <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  <name> <a> <next> <b><term>            (I need more and more stop now I don't fear anymore)

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

          if (= x counter-b)

                product

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

    callup <1> <2*4> <count-b/2>

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

About Us

brown1730and...

走一样的经历,然后知(感觉这才是程序2019-08-29/1:40:01 PM)道每一个思绪

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)

 

怎么办

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

是不是还要有这种情况,a和b差很多

是要iteratively

a b 

那就 a-counter is odd

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 

这如果是个定理,那是不是,乘以pi/4  or  所以 (a-count = b-c) or (a-count>b-c) or (a-count<b-c)

(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

 

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