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Автор AbhayChandna, история, 5 лет назад, По-английски

https://mirror.codeforces.com/contest/1353/problem/E

On input 
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0

Output is 
1
2
5
4
0
0

In the second testcase, how is the output 2?

if we swithc all off all lamps, 4 moves if we switch on lamps on index 0 3 6 , moves 3 (switch off {1,2} switch on{6})

if we switch on lamps on index 1 4 7 , moves 4 (switch off {0,2,3} switch on{7})

if we switch on lamps on index 2 5 8 , moves 4 (switch off {0,1,3} switch on{8})

So , how is the minimum required moves 2?

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5 лет назад, # |
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Auto comment: topic has been updated by AbhayChandna (previous revision, new revision, compare).

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5 лет назад, # |
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In question it is given that "The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k"- this means all the adjacent lamps switched ON in the answer should have distance k between them. This doesnot means that all the adjacent lamps at distance k should be switched ON. Answer for 2nd case is to turn off lamp at index 1 & 2(100100000). Switched ON index are 0 and 3 and distance between them is 3.

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we need to spend the minimum number of moves to make this string of kind contiguous block of zeros, contiguous block of ones and again contiguous block of zeros

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5 лет назад, # |
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It is not mandatory that all the lamps at a distance of k should be turned on. Instead it means that if there are two "on lamps" they must be at a distance of k from each other. Like for 111100000, turning off lamps at indices 1,2 gives 100100000.(making it as 100100100 is not needed,since it takes 3 moves).