IOS9没法关闭background刷新？ - 未名空间MITBBS历史存档

国际科技财经博客移民网络热点娱乐民生时事公众号

Redian新闻

>未名空间

>Apple - 家有苹果

IOS9没法关闭background刷新？

IOS9没法关闭background刷新？# Apple - 家有苹果

j*72015-09-19 07:09

1 楼

Given two square matrices(initial and final) consisting of elements either
1 or 0. Using minimum number of toggles change the initial to final matrix.
You can toggle either a single row or column at a time. If ith row is
toggled all 1's become 0 and vice versa in that row.
What will be the correct algorithm for this?
For example
|0 0 1|
|1 1 1|
|1 0 1|
to
|1 1 1|
|1 1 0|
|1 0 0| would require
1st row and last column toggle.
Thanks

d*y2015-09-19 07:09

2 楼

包子或或者paypal。
谢谢。

a*g2015-09-19 07:09

3 楼

b*e2015-09-19 07:09

4 楼

通知关掉了
微博app设置里面推送通知也都关掉了
在电池里面，2小时内还有30多分钟的background使用，奇怪

g*s2015-09-19 07:09

5 楼

a* search? eval::=final - trial and always pick up the move closest to
all-0 matrix.
however it also needs some mathematical thinking about the solvability.

either
matrix.

【在 j****7 的大作中提到】

: Given two square matrices(initial and final) consisting of elements either
: 1 or 0. Using minimum number of toggles change the initial to final matrix.
: You can toggle either a single row or column at a time. If ith row is
: toggled all 1's become 0 and vice versa in that row.
: What will be the correct algorithm for this?
: For example
: |0 0 1|
: |1 1 1|
: |1 0 1|
: to

s*12015-09-19 07:09

6 楼

Co ask.

b*d2015-09-19 07:09

7 楼

两根备用jj长在头顶了

F*T2015-09-19 07:09

8 楼

Low Power Mode

z*e2015-09-19 07:09

9 楼

my understanding is it's the same as the "word ladder" problem: giving A and
B and some mutation rules, find out the min mutations from A=>B.
so a straightforward solution is to use BFS to find the shortest path from A
=>B.
Unfortunately, the bfs is a "general" solution, but not the best. wonder if
there is better solution for this particular question.

.

【在 j****7 的大作中提到】

s*12015-09-19 07:09

10 楼

Co ask.

z*n2015-09-19 07:09

11 楼

下边那个小孩儿是打酱油的？

b*e2015-09-19 07:09

12 楼

Yes, there is a simple solution here, which is no more than to solve a
set of linear equations.
Taking the example in OP as an example:
1. computer the xor, we get:
1 1 0
0 0 1
0 1 1
2. linearize it, top to bottom, left to right, let R be:
1 1 0 0 0 1 0 1 1
3. List all the transformations you can do, let T be:
1 1 1 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0
0 0 0 0 0 0 1 1 1
1 0 0 1 0 0 1 0 0
0 1 0 0 1 0 0 1 0
0 0 1 0 0 1 0 0 1
4. Let T' be the transit of T, and R' be the transit of R (so R's is a
column). Now the problem becomes as simple as to find solution for:
T' * X = R'
where multiplication and addition are all modulo 2.
The key is to note that, you will only need to flip a row/col at most
once,
and the order in which you flip it doesn't matter.