Problem

Thanh wants to paint a wonderful mural on a wall that is N sections long. Each section of the wall has a beauty score, which indicates how beautiful it will look if it is painted. Unfortunately, the wall is starting to crumble due to a recent flood, so he will need to work fast!

At the beginning of each day, Thanh will paint one of the sections of the wall. On the first day, he is free to paint any section he likes. On each subsequent day, he must paint a new section that is next to a section he has already painted, since he does not want to split up the mural.

At the end of each day, one section of the wall will be destroyed. It is always a section of wall that is adjacent to only one other section and is unpainted (Thanh is using a waterproof paint, so painted sections can't be destroyed).

The total beauty of Thanh's mural will be equal to the sum of the beauty scores of the sections he has painted. Thanh would like to guarantee that, no matter how the wall is destroyed, he can still achieve a total beauty of at least B. What's the maximum value of B for which he can make this guarantee?
Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing an integer N. Then, another line follows containing a string of N digits from 0 to 9. The i-th digit represents the beauty score of the i-th section of the wall.
Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum beauty score that Thanh can guarantee that he can achieve, as described above.
Limits

1 ≤ T ≤ 100.
Small dataset

2 ≤ N ≤ 100.
Large dataset

For exactly 1 case, N = 5 × 106; for the other T - 1 cases, 2 ≤ N ≤ 100.
Sample

Input
4
4
1332
4
9583
3
616
10
1029384756

output:
Case #1: 6
Case #2: 14
Case #3: 7
Case #4: 31

In the first sample case, Thanh can get a total beauty of 6, no matter how the wall is destroyed. On the first day, he can paint either section of wall with beauty score 3. At the end of the day, either the 1st section or the 4th section will be destroyed, but it does not matter which one. On the second day, he can paint the other section with beauty score 3.

In the second sample case, Thanh can get a total beauty of 14, by painting the leftmost section of wall (with beauty score 9). The only section of wall that can be destroyed is the rightmost one, since the leftmost one is painted. On the second day, he can paint the second leftmost section with beauty score 5. Then the last unpainted section of wall on the right is destroyed. Note that on the second day, Thanh cannot choose to paint the third section of wall (with beauty score 8), since it is not adjacent to any other painted sections.

In the third sample case, Thanh can get a total beauty of 7. He begins by painting the section in the middle (with beauty score 1). Whichever section is destroyed at the end of the day, he can paint the remaining wall at the start of the second day.

t = int(input())
for case in range(t):
    n = int(input())
    s = input()
    l = []
    for i in range(len(s)):
        l.append(int(s[i]))
    # print(l)
    l1 = (n+1)//2
    count = sum(l[:l1])
    temp = count
    for i in range(l1,len(l)):
        temp = temp + l[i] - l[i-l1]
        count = max(count,temp)
    print("Case #",end='')
    print(case+1,end='')
    print(": ",end='')
    print(count)

講道理，這道題很簡單。程式碼也是三道題裡最短的。讀懂題目有些難度，大意是有一睹牆，畫家從任意一個起點開始修復這面牆，獲得這面牆的價值，然後洪水會從最左邊或者最右邊侵蝕一睹牆，已經被侵蝕的牆則不可以再被修復。然後畫家向左或是向右繼續修復然後洪水亦復如是。直到所有的牆都被修復或被侵蝕為止。求畫家可以獲得的最大價值。
如果用模擬法，一步步考慮地話就進坑了。觀察到，畫家可以修復的牆的長度是可以直接確定的(n+1)//2，而修復的牆總是連續的，所以也就是求(n+1)//2長度的最大值。
這裡的一個優化就是，使用字首和來減少計算量。
small和large都可以通過。