【Javascript】兩視窗資料傳遞的方法

這幾天遇到一個需求，必須另開視窗產生完超商代收的 barcode 資訊，
再回傳相關資訊給原本視窗，接下來再 POST 所有資訊到下一頁，
在這邊記錄一下問題的大致做法。

主流程： ```html <button type="button" onclick="genBarcode();>超商代收</button> ```

另開視窗頁面： ```html ```

【Python】Codility in Python : Lesson 13 - Fibonacci Numbers【Ladder】

markdown Fibonacci Numbers 第一題：【Ladder】
Count the number of different ways of climbing to the top of a ladder.

---

You have to climb up a ladder. The ladder has exactly N rungs, numbered from 1 to N.
With each step, you can ascend by one or two rungs. More precisely:

with your first step you can stand on rung 1 or 2,
if you are on rung K, you can move to rungs K + 1 or K + 2,
finally you have to stand on rung N.
Your task is to count the number of different ways of climbing to the top of the ladder.

For example, given N = 4, you have five different ways of climbing, ascending by:

　　1, 1, 1 and 1 rung,
　　1, 1 and 2 rungs,
　　1, 2 and 1 rung,
　　2, 1 and 1 rungs, and
　　2 and 2 rungs.

Given N = 5, you have eight different ways of climbing, ascending by:

　　1, 1, 1, 1 and 1 rung,
　　1, 1, 1 and 2 rungs,
　　1, 1, 2 and 1 rung,
　　1, 2, 1 and 1 rung,
　　1, 2 and 2 rungs,
　　2, 1, 1 and 1 rungs,
　　2, 1 and 2 rungs, and
　　2, 2 and 1 rung.

The number of different ways can be very large,
so it is sufficient to return the result modulo 2P, for a given integer P.

Write a function:

　　def solution(A, B)

that, given two non-empty zero-indexed arrays A and B of L integers,
returns an array consisting of L integers specifying the consecutive answers;
position I should contain the number of different ways of climbing the ladder with A[I] rungs modulo 2B[I].

For example, given L = 5 and:

　　A[0] = 4 B[0] = 3
　　A[1] = 4 B[1] = 2
　　A[2] = 5 B[2] = 4
　　A[3] = 5 B[3] = 3
　　A[4] = 1 B[4] = 1

the function should return the sequence [5, 1, 8, 0, 1], as explained above.

Assume that:
　　L is an integer within the range [1..30,000];
　　each element of array A is an integer within the range [1..L];
　　each element of array B is an integer within the range [1..30].

---

方法一：參考 Code Says 的做法
Correctness：100%、Performance：100%
```python def solution(A, B): limit = len(A) result = [0] * len(A) B = [(1 << item) - 1 for item in B] # 取得費氏數列 fib = [0] * (limit + 2) fib[1] = 1 for i in range(2, limit + 2): fib[i] = fib[i - 1] + fib[i - 2] for i in range(limit): result[i] = fib[A[i] + 1] & B[i] return result ```

完整練習題 source code 請參閱：github

【Python】Codility in Python : Lesson 12 - Euclidean Algorithm【ChocolatesByNumbers】

Euclidean Algorithm 第一題：【ChocolatesByNumbers】
There are N chocolates in a circle. Count the number of chocolates you will eat.

---

Two positive integers N and M are given.
Integer N represents the number of chocolates arranged in a circle, numbered from 0 to N − 1.

You start to eat the chocolates. After eating a chocolate you leave only a wrapper.

You begin with eating chocolate number 0.
Then you omit the next M − 1 chocolates or wrappers on the circle, and eat the following one.

More precisely, if you ate chocolate number X,
then you will next eat the chocolate with number (X + M) modulo N (remainder of division).

You stop eating when you encounter an empty wrapper.

For example, given integers N = 10 and M = 4.
You will eat the following chocolates: 0, 4, 8, 2, 6.

The goal is to count the number of chocolates that you will eat, following the above rules.

Write a function:

　　def solution(N, M)

that, given two positive integers N and M, returns the number of chocolates that you will eat.

For example, given integers N = 10 and M = 4. the function should return 5, as explained above.

Assume that:
　　N and M are integers within the range [1..1,000,000,000].

---

方法一：參考 Maurice Dassen 大大的做法
Correctness：100%、Performance：100%
```python def solution(N, M): x = 0 while x != 1: x, y = N, M while y != 0: x, y = y, x % y N, M = N / x, M / x return N ``` 完整練習題 source code 請參閱：github

【PHP】解決 SQLSTATE[HY000]: General error: 2031 No data supplied for parameters in prepared statement.

記錄本日工作時遇到的詭異問題，PHP 在撈 DB 資料時出現了下列錯誤訊息：
SQLSTATE[HY000]: General error: 2031 No data supplied for parameters in prepared statement.

原因是如果參數有兩個名稱重複時，就會造成這個錯誤
```sql select * from student where sid = :id or tid = :id ```

將參數改為不同名稱就能解決這個問題
```sql select * from student where sid = :sid or tid = :tid ```