adroid机子的apps问题 - 未名空间MITBBS历史存档

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adroid机子的apps问题

adroid机子的apps问题# PDA - 掌中宝

s*02012-10-07 07:10

1 楼

第一道题是什么意思啊？高等数学的英语题还能看懂，初等数学的题反而看不懂了：
1. The odds against event A occurring are 3:1 and the odds in favor of event
B occurring are 3:5. What is the probability that at least one of these
independent events occurs? Express your answer as a common fraction.
第二道题看起来有点难，三十年不做初等数学题了，技巧全忘光了：
2. In how many ways can 18 be written as the sum of four distinct positive
integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
ways.

f*e2012-10-07 07:10

2 楼

【以下文字转载自 Travel 讨论区】
发信人: fage (fage), 信区: Travel
标题: 请推荐加州中文旅行社
发信站: BBS 未名空间站 (Tue Apr 27 15:21:09 2010, 美东)
想送父母去加州，LA，LV和SD旅游，请推荐加州中文旅行社。
谢谢！！

k*r2012-10-07 07:10

3 楼

现在有个HP的touchpad,刚刷了CM9,发现很多有用的免费的apps.
如果我买adroid的手机，也能搞到这些apps么？
问题就是，adroid的apps是不是统合在一个大市场里，不同的adroid机型都可以access？
谢谢！

B*12012-10-07 07:10

4 楼

1. The odds against event A occurring are 3:1 and the odds in favor of event
B occurring are 3:5. What is the probability that at least one of these
independent events occurs? Express your answer as a common fraction.
At least one occurs = odds A + odds B - odds both A and B
=1/4 + 3/8 -(1/4)*(3/8)
= (8+12-3)/32
= 17/32
since A and B are independent

a92012-10-07 07:10

5 楼

一般来讲是的，但是有些app是适配机型的。

access？

【在 k****r 的大作中提到】

: 现在有个HP的touchpad,刚刷了CM9,发现很多有用的免费的apps.
: 如果我买adroid的手机，也能搞到这些apps么？
: 问题就是，adroid的apps是不是统合在一个大市场里，不同的adroid机型都可以access？
: 谢谢！

B*12012-10-07 07:10

6 楼

2. In how many ways can 18 be written as the sum of four distinct positive
integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
ways.
1 2 3 12
1 2 4 11
. . . .
1 2 7 8 (5 ways)
1 3 4 10
1 3 5 9
1 3 6 8 (3 ways)
1 4 5 8
1 4 6 7 (2 ways)
2 3 4 9
2 3 5 8
2 3 6 7 (3 ways)
2 4 5 7 (1 way)
3 4 5 6 (1 way)
15 * 4! = 360

k*r2012-10-07 07:10

7 楼

嗯，谢谢！
adroid 2.3(Gingerbread)比起4.1 (icecream sandwich)能运行的app要少很多，是么？

【在 a9 的大作中提到】

: 一般来讲是的，但是有些app是适配机型的。
:
: access？

s*02012-10-07 07:10

8 楼

我是反过来做的，先取大的数：
9 6 2 1
9 5 3 1
9 4 3 2 (3 ways)
8 7 2 1
8 6 3 1
8 5 4 1
8 5 3 2 (4 ways)
7 6 4 1
7 6 3 2
7 5 4 2 (3 ways)
6 5 4 3 (1 ways)
total: (3 + 4 + 3 + 1）4！ = 11 X 4！=264
感觉自己的方法太笨了，没有更好的技巧吗？

【在 B******1 的大作中提到】

: 2. In how many ways can 18 be written as the sum of four distinct positive
: integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
: ways.
: 1 2 3 12
: 1 2 4 11
: . . . .
: 1 2 7 8 (5 ways)
: 1 3 4 10
: 1 3 5 9
: 1 3 6 8 (3 ways)

a92012-10-07 07:10

9 楼

应该不多。2.3毕竟现在是市场上占有率最高的android版本。

么？

【在 k****r 的大作中提到】

: 嗯，谢谢！
: adroid 2.3(Gingerbread)比起4.1 (icecream sandwich)能运行的app要少很多，是么？

x*12012-10-07 07:10

10 楼

2. In how many ways can 18 be written as the sum of four distinct positive
integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
ways.
excluding 13, 14, 15, 16, 17, 18, all other number (1 to 12) could be used
as one of four distinct positive integers, so the total will be:
total: (18-6)*3!=72

G*n2012-10-07 07:10

11 楼

啊？ 2.3是占有率最多的？
现在新机型都用4.0以上了吧？那以后2.3不是要被抛弃。

【在 a9 的大作中提到】

: 应该不多。2.3毕竟现在是市场上占有率最高的android版本。
:
: 么？

B*12012-10-07 07:10

12 楼

You did not try 12, 11, or 10 as the biggest possible number.

【在 s****0 的大作中提到】

: 我是反过来做的，先取大的数：
: 9 6 2 1
: 9 5 3 1
: 9 4 3 2 (3 ways)
: 8 7 2 1
: 8 6 3 1
: 8 5 4 1
: 8 5 3 2 (4 ways)
: 7 6 4 1
: 7 6 3 2

g*g2012-10-07 07:10

13 楼

新机是4.0/4.1，不过很多老机都没有官方升级，除非自己刷/升

【在 G***n 的大作中提到】

: 啊？ 2.3是占有率最多的？
: 现在新机型都用4.0以上了吧？那以后2.3不是要被抛弃。

d*a2012-10-07 07:10

14 楼

2. In how many ways can 18 be written as the sum of four distinct positive
integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
ways.
解法：
#include
#define N 18
int main()
{
int i, j, k, l;
int m = 0;
for (i = 1; i < N; i++)
for (j = 1; j < N; j++)
for (k = 1; k < N; k++)
for (l = 1; l < N; l++) {
if (i+j+k+l == N &&
i != j && i != k && i != l &&
j != k && j != l && k != l)
m++;
}
printf("m = %d\n", m);
return 0;
}
$ gcc -o comb-add -Wall comb-add.c
$ ./comb-add
m = 360
Beida的答案是对的。

a*m2012-10-07 07:10

15 楼

多数厂家到今年下半年才开始广泛预装和升级4.0。

【在 G***n 的大作中提到】

: 啊？ 2.3是占有率最多的？
: 现在新机型都用4.0以上了吧？那以后2.3不是要被抛弃。

B*12012-10-07 07:10

16 楼

Man, your fancy program language is very powerful, and more convincing than
my handwritings!

【在 d***a 的大作中提到】

: 2. In how many ways can 18 be written as the sum of four distinct positive
: integers? Note: 1 + 3 + 5 + 9 and 5 + 1 + 3 + 9 are counted as different
: ways.
: 解法：
: #include
: #define N 18
: int main()
: {
: int i, j, k, l;
: int m = 0;

d*a2012-10-07 07:10

17 楼

Too much programming hurts mathematical thinking, even though programming
languages were first created by mathematicians. :)

than

【在 B******1 的大作中提到】

: Man, your fancy program language is very powerful, and more convincing than
: my handwritings!

x*12012-10-07 07:10

18 楼

太牛了!

【在 d***a 的大作中提到】

s*02012-10-07 07:10

19 楼

我自己人为加了个约束，以为只能用个位数，结果范围被缩小了。

【在 B******1 的大作中提到】

: You did not try 12, 11, or 10 as the biggest possible number.

B*12012-10-07 07:10

20 楼

Make sense.

【在 s****0 的大作中提到】

: 我自己人为加了个约束，以为只能用个位数，结果范围被缩小了。

s*02012-10-07 07:10

21 楼

这个是一个Excel VB code，可以在Excel里面把表列出来，360组。
Sub look_up_18()
Sheet1.Select
Range("A1").Select
mmm = 0
NN = 18
For I1 = 1 To NN
For I2 = 1 To NN
For I3 = 1 To NN
For I4 = 1 To NN

Sum = I1 + I2 + I3 + I4
prod = (I1 - I2) * (I1 - I3) * (I1 - I4) * (I2 - I3) * (I2 - I4) *
(I3 - I4)

If (Sum = NN) And (prod <> 0) Then
mmm = mmm + 1
ActiveCell.Offset(mmm, 1) = mmm
ActiveCell.Offset(mmm, 2) = I1
ActiveCell.Offset(mmm, 3) = I2
ActiveCell.Offset(mmm, 4) = I3
ActiveCell.Offset(mmm, 5) = I4
ActiveCell.Offset(mmm, 7) = Sum

End If

Next I4
Next I3
Next I2
Next I1

End Sub

【在 B******1 的大作中提到】

: Make sense.

s*d2012-10-07 07:10

22 楼

1 - 1/4*5/8

event

【在 B******1 的大作中提到】

: 1. The odds against event A occurring are 3:1 and the odds in favor of event
: B occurring are 3:5. What is the probability that at least one of these
: independent events occurs? Express your answer as a common fraction.
: At least one occurs = odds A + odds B - odds both A and B
: =1/4 + 3/8 -(1/4)*(3/8)
: = (8+12-3)/32
: = 17/32
: since A and B are independent

B*12012-10-07 07:10

23 楼

Should be:
1- 3/4 * 5/8

【在 s**********d 的大作中提到】

: 1 - 1/4*5/8
:
: event

s*02012-10-07 07:10

24 楼

能把第一道题翻译成中文吗？odds against 和 odds in favor of 是专门的用词？
感觉跟国内有些小学生看不懂中文题目一个意思。

【在 s**********d 的大作中提到】

: 1 - 1/4*5/8
:
: event

s*02012-10-07 07:10

25 楼

试了一下，把10到20之间的数都找一遍：
Sub look_up_18()
Sheet1.Select
Range("A1").Select
mmm = 0
For NN = 9 To 20
NNN = 0
For I1 = 1 To NN
For I2 = I1 + 1 To NN
For I3 = I2 + 1 To NN
For I4 = I3 + 1 To NN

Sum = I1 + I2 + I3 + I4
prod = (I1 - I2) * (I1 - I3) * (I1 - I4) * (I2 - I3) * (I2 - I4) *
(I3 - I4)

If (Sum = NN) And (prod <> 0) Then
mmm = mmm + 1
NNN = NNN + 1
ActiveCell.Offset(mmm, 3) = NNN
ActiveCell.Offset(mmm, 4) = I1
ActiveCell.Offset(mmm, 5) = I2
ActiveCell.Offset(mmm, 6) = I3
ActiveCell.Offset(mmm, 7) = I4
ActiveCell.Offset(mmm, 9) = Sum

End If

Next I4
Next I3
Next I2
Next I1
Next NN
End Sub

结果：
四数之和组合个数
10 1
11 1
12 2
13 3
14 5
15 6
16 9
17 11
18 15
19 18
20 23