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<p> |
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Pierre Peintre is slaving away over a new abstract painting entitled <i>Rain Over New York</i>. |
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This will be a simple yet powerful piece, omitting incidental details such as busy city dwellers shielding themselves with umbrellas, |
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and instead focusing on the fundamental atmosphere of a rainy metropolitan day. |
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It will be painted on a canvas which is subdivided into a grid of 1cm x 1cm cells, with <strong>N</strong> rows and <strong>M</strong> columns. |
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Each cell in this grid will be filled in with a solid color, either black, white, grey, or blue. |
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</p> |
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<p> |
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The lower portion of <i>Rain Over New York</i> will depict the skyline of New York City. |
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In each column <em>i</em>, the bottom-most <strong>H<sub>i</sub></strong> cells will be painted grey to represent an austere skyscraper. |
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</p> |
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<p> |
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Somewhere above the buildings, Pierre will place a single, innocent raincloud. In particular, the cloud can be any rectangle of white cells on the canvas, as long as none of them are supposed to be grey. |
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</p> |
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<p> |
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Below the cloud, there must be a gentle rainfall, of course. Every cell which has a white cell somewhere directly above it and a grey cell somewhere directly below it, and which isn't supposed to be white or grey itself, should be painted blue. |
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Note that there may be no such cells, if the cloud is immediately above the skyline. |
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</p> |
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<p> |
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All of the remaining cells in the painting will be painted black, providing a serene nighttime backdrop for the scene. |
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</p> |
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<p> |
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Pierre knows that every painting he can produce like this will sell for an enormous sum of money, but only if it's unique. |
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As such, he'll paint as many different paintings as he can by varying the position and dimensions of the raincloud depicted in them. |
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Two paintings are considered distinct if at least one cell on the canvas is a different color in one painting than it is in the other. |
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</p> |
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<p> |
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As an example, below is an illustration of 1 of the 246 possible paintings for the fourth sample case: |
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<img src="{{PHOTO_ID:374610330166776}}" /> |
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<p> |
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Thanks to the incredible sum of money which Pierre is sure to make from these works, he'll be able to purchase all of the paint that he'll need. |
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He always buys his paint in cans of a fixed size, each of which contains just enough to cover a surface of 1,000,000,007 cm<sup>2</sup>, |
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and for each color, he'll buy just enough such cans in order to be able to complete all possible distinct variations of his painting, once each. |
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Always one to plan ahead, Pierre would like to figure out exactly how much paint of each color he'll have left over when he's done. |
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</p> |
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<p> |
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The sequence <strong>H<sub>1..M</sub></strong> can be constructed by concatenating <strong>K</strong> temporary |
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sequences of values <strong>S<sub>1..K</sub></strong>, the <em>i</em>th of which has a length of <strong>L<sub>i</sub></strong>. |
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It's guaranteed that the sum of these sequences' lengths is equal to <strong>M</strong>. |
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For each sequence <strong>S<sub>i</sub></strong>, you're given <strong>S<sub>i,1</sub></strong>, |
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and <strong>S<sub>i,2..L<sub>i</sub></sub></strong> may then be calculated as follows, |
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using given constants <strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>: |
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<strong>S<sub>i,j</sub></strong> = ((<strong>A<sub>i</sub></strong> * <strong>S<sub>i,j-1</sub></strong> + <strong>B<sub>i</sub></strong>) % (<strong>N</strong> - 1)) + 1 |
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</p> |
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<h3>Input</h3> |
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<p> |
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Input begins with an integer <strong>T</strong>, the number of different base skylines Pierre wants to use. |
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For each skyline, there is first a line containing the three space-separated integers, <strong>N</strong>, <strong>M</strong>, and <strong>K</strong>. |
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Then <strong>K</strong> lines follow, the <em>i</em>th of which contains the four space-separated integers |
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<strong>L<sub>i</sub></strong>, <strong>S<sub>i,1</sub></strong>, <strong>A<sub>i</sub></strong>, and <strong>B<sub>i</sub></strong>. |
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</p> |
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<h3>Output</h3> |
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<p> |
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For the <em>i</em>th skyline, print a line containing "Case #<strong>i</strong>: " followed by four space-separated integers, |
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the total amount of black, white, grey, and blue paint which Pierre will have left over, respectively (in cm<sup>2</sup>), after completing all possible variations of his painting. |
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</p> |
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<h3>Constraints</h3> |
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<p> |
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1 ≤ <strong>T</strong> ≤ 100 <br /> |
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2 ≤ <strong>N</strong> ≤ 1,000,000,000 <br /> |
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1 ≤ <strong>M</strong> ≤ 200,000 <br /> |
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1 ≤ <strong>K</strong> ≤ 100 <br /> |
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1 ≤ <strong>L<sub>i</sub></strong> ≤ M <br /> |
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1 ≤ <strong>H<sub>i</sub></strong>, <strong>S<sub>i,j</sub></strong> ≤ <strong>N</strong> - 1 <br /> |
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0 ≤ <strong>A<sub>i</sub></strong>, <strong>B<sub>i</sub></strong> < <strong>N</strong> - 1 <br /> |
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</p> |
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<h3>Explanation of Sample</h3> |
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<p> |
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In the first case, there's only one possible painting, with the top cell painted white, and the remaining two cells painted grey. Pierre will buy 1 can each of white and grey paint, and have 1,000,000,006 and 1,000,000,005 cm<sup>2</sup> left over of those colors, respectively. |
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</p> |
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<p> |
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In the second case, there are 6 possible paintings: three with the cloud covering one cell, two with the cloud covering two cells, and one with the cloud covering three cells. Therefore, Pierre will use 10 cm<sup>2</sup> of white paint in total. |
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</p> |
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<p> |
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The fourth case corresponds to the picture shown above. |
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