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091118求杨辉三角中的项

lzj520 — Wed, 18 Nov 2009 08:11:00 GMT

public static int recursion(int l,int n){
  int r=0;
  if(l==1|n==1|l==n){
   return 1;
  }else{
  return recursion(l-1,n-1)+recursion(l-1,n);
  }
}

public static void main(String[] args){
System.out.println(recursion(7,4));
}

lzj520 2009-11-18 16:11 发表评论

090310 Exercise 1.11. recursive process and iterative

lzj520 — Tue, 10 Mar 2009 00:51:00 GMT

Exercise 1.11.  A function f is defined by the rule that f(n) = n if n<3 and f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n> 3. Write a procedure that computes f by means of a recursive process. Write a procedure that computes f by means of an iterative process.

recursive:

(define (fn n)
(cond ((>= n 3) (+ (+ (fn (- n 1)) (* 2 (fn (- n 2)))) (* 3 (fn (- n 3)))))
        ((< n 3) n)
   ))

iterative:

(define (re n)
(if (< n 3)
      n
      (iter 2 1 0 n)
      ))
(define (iter a b c n)
(if(= n 3)
   (ca a b c)
   (iter (ca a b c) a b (- n 1))
   )
)
(define (ca a b c)
(+ a (* 2 b) (* 3 c) )
)

lzj520 2009-03-10 08:51 发表评论

090306 Exercise 1.8 cube-root procedures

lzj520 — Fri, 06 Mar 2009 08:22:00 GMT

Exercise 1.8. Newton's method for cube roots is based on the fact that if y is an approximation to
the cube root of x, then a better approximation is given by the value
(x/y2+2y)/3
Use this formula to implement a cube-root procedure analogous to the square-root procedure. (In
section 1.3.4 we will see how to implement Newton's method in general as an abstraction of these
square-root and cube-root procedures.)

(define (cube x)
(* x x x))
(define (square x)
(* x x ))
(define (result x y)
(/ (+ (/ x (square y)) (* 2 y)) 3))
(define (improve x guess)
(result   x guess))
(define (good-enough? x guess)
(< (abs (- (* guess guess guess ) x))0.001))
(define (sqrt-iter x guess)
(if (good-enough? x guess)
      guess
      (sqrt-iter x (improve x guess)
                 )))

lzj520 2009-03-06 16:22 发表评论

090306 Exercise 1.6 Square Roots by Newton's Method

lzj520 — Fri, 06 Mar 2009 07:19:00 GMT

Exercise 1.6. Alyssa P. Hacker doesn't see why if needs to be provided as a special form. ``Why
can't I just define it as an ordinary procedure in terms of cond?'' she asks. Alyssa's friend Eva Lu
Ator claims this can indeed be done, and she defines a new version of if:
(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
        (else else-clause)))
Eva demonstrates the program for Alyssa:
(new-if (= 2 3) 0 5)
5
(new-if (= 1 1) 0 5)
0
Delighted, Alyssa uses new-if to rewrite the square-root program:
32(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x)
                     x)))
What happens when Alyssa attempts to use this to compute square roots? Explain.

(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
      (else-clause)))
(define (average x y)
(/ (+ x y) 2))
(define (improve guess x)
(average guess (/ x guess)))
(define (good-enough? guess x)
(< (abs (- (square guess) x))0.001))
(define (square x)
(* x x))
(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x)
                     x)))

sqrt-iter (improve guess x)作�ؓ参数来传递给new-if�Q�在执行new-if的时候，��L��会执行sqrt-iter (improve guess x)�Q�造成了死循环�?br />

lzj520 2009-03-06 15:19 发表评论

090305 Exercise 1.3 returns the sum of the squares of the two larger numbers

lzj520 — Thu, 05 Mar 2009 11:56:00 GMT

Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.

(define (compare x y) (- x y))
(define (sumsquares x y)(+(* x x)(* y y)))
(define (returnlarge a b c)
(cond ((and (>= (compare a b) 0) (>= (compare c b) 0)) (sumsquares a c))
        ((and (>= (compare a c) 0) (>= (compare b c) 0)) (sumsquares a b))
        ((and (>= (compare c a) 0) (>= (compare b a) 0)) (sumsquares b c))
         )
)
(returnlarge 3 3 2)

>18

lzj520 2009-03-05 19:56 发表评论