# GCJ 2015 – Qualification Problem C – Dijkstra (迪科斯彻) – 解题报告

## 第三题：迪科斯彻 / Dijkstra

### 顶礼膜拜 Dijkstra 祖师 Orz.. Orz.. Orz.. ### 解题报告：

##### 结论：

ok，有了上述的定理二，我们求整串乘积就可以直接将重复数 `K` 裁减为 `K % 4`，这样就可以在 `O(L)` 的时间空间复杂度里面验证定理一。

``````S = single * (X % 4 + (4 if X >= 4 else 0))
``````

##### 例程：
``````# 对于四元数的数据结构表示，这里使用一个 python 二元组
# 第一位是符号，第二位是字符，然后用判断逻辑实现了乘法
def multiply(a, b):
s1, c1 = a
s2, c2 = b
s, c = s1 * s2, '1'
if c1 == '1':
c = c2
elif c2 == '1':
c = c1
elif c1 == c2:
s *= -1
c = '1'
elif c1 == 'i' and c2 == 'j':
c = 'k'
elif c1 == 'i' and c2 == 'k':
s *= -1
c = 'j'
elif c1 == 'j' and c2 == 'k':
c = 'i'
elif c1 == 'j' and c2 == 'i':
s *= -1
c = 'k'
elif c1 == 'k' and c2 == 'i':
c = 'j'
elif c1 == 'k' and c2 == 'j':
s *= -1
c = 'i'
return s, c

def solve():

for t in range(int(input())):

L, X = map(int, input().split())
single = str(input())

# 先按照定理一判断整体乘积：
val = (1, '1')
for c in single * (X % 4):
val = multiply(val, (1, c))

result = val == (-1, '1')

if result:

# 生成简化串
S = single * (X % 4 + (4 if X >= 4 else 0))

# 记录查找位置
p1 = p2 = -1

# 从前往后
n1 = (1, '1')
for i in range(len(S)):
n1 = multiply(n1, (1, S[i]))
if n1 == (1, 'i'):
p1 = i
break

# 从后往前
n2 = (1, '1')
for i in range(len(S)-1, -1, -1):
n2 = multiply((1, S[i]), n2)
if n2 == (1, 'k'):
p2 = i
break

# 必须都找到，而且 p1 在 p2 前面，但是记得加上忽略掉的长度
result = (p1 > -1 and p2 > -1) and p1 < p2 + (X * L - len(S))

print('Case #%d:' % (t + 1), ('YES' if result else 'NO'))

solve()
``````

### 原题内容：

##### Problem

The Dutch computer scientist Edsger Dijkstra made many important contributions to the field, including the shortest path finding algorithm that bears his name. This problem is not about that algorithm.

You were marked down one point on an algorithms exam for misspelling “Dijkstra” — between D and stra, you wrote some number of characters, each of which was either i, j, or k. You are prepared to argue to get your point back using quaternions, an actual number system (extended from complex numbers) with the following multiplicative structure: To multiply one quaternion by another, look at the row for the first quaternion and the column for the second quaternion. For example, to multiply i by j, look in the row for i and the column for j to find that the answer is k. To multiply j by i, look in the row for j and the column for i to find that the answer is -k.

As you can see from the above examples, the quaternions are not commutative — that is, there are some a and b for which a * b != b * a. However they are associative — for any a, b, and c, it’s true that a * (b * c) = (a * b) * c.

Negative signs before quaternions work as they normally do — for any quaternions a and b, it’s true that -a * -b = a * b, and -a * b = a * -b = -(a * b).

You want to argue that your misspelling was equivalent to the correct spelling ijk by showing that you can split your string of is, js, and ks in two places, forming three substrings, such that the leftmost substring reduces (under quaternion multiplication) to i, the middle substring reduces to j, and the right substring reduces to k. (For example, jij would be interpreted as j * i * j; j * i is -k, and -k * j is i, so jij reduces to i.) If this is possible, you will get your point back. Can you find a way to do it?

##### Input

The first line of the input gives the number of test cases, T. T test cases follow. Each consists of one line with two space-separated integers L and X, followed by another line with L characters, all of which are i, j, or k. Note that the string never contains negative signs, 1s, or any other characters. The string that you are to evaluate is the given string of L characters repeated X times. For instance, for L = 4, X = 3, and the given string kiij, your input string would be kiijkiijkiij.

##### 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 either YES or NO, depending on whether the string can be broken into three parts that reduce to i, j, and k, in that order, as described above.

##### Limits

1 ≤ T ≤ 100. 1 ≤ L ≤ 10000.

###### Small dataset

1 ≤ X ≤ 10000. 1 ≤ L * X ≤ 10000.

###### Large dataset

1 ≤ X ≤ 1012. 1 ≤ L * X ≤ 1016.

##### Sample
``````Input

5
2 1
ik
3 1
ijk
3 1
kji
2 6
ji
1 10000
i

Output

Case #1: NO
Case #2: YES
Case #3: NO
Case #4: YES
Case #5: NO
``````

In Case #1, the string is too short to be split into three substrings.

In Case #2, just split the string into i, j, and k.

In Case #3, the only way to split the string into three parts is k, j, i, and this does not satisfy the conditions.

In Case #4, the string is jijijijijiji. It can be split into jij (which reduces to i), iji (which reduces to j), and jijiji (which reduces to k).

In Case #5, no matter how you choose your substrings, none of them can ever reduce to a j or a k.