{"trustable":true,"sections":[{"title":"","value":{"format":"HTML","content":"\u003cdiv align\u003d\"left\"\u003eConsider a monochrome computer monitor with N x M pixels resolution. It can display K symbols. Symbol is a non-empty set of black pixels on the screen. You are to write a program, that computes minimal possible amount of pixels, that are enough to recognize a symbol. \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003e\u003cb\u003eInput\u003c/b\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003eFirst line of input contains integers N, M, K (1 \u0026lt;\u003d N, M \u0026lt;\u003d 10, 2 \u0026lt;\u003d K \u0026lt;\u003d 6). K blocks follow, separated by empty lines. Each block consists of N rows of M characters. It\u0027s a symbols representation on the N x M screen. `1\u0027 denotes black pixel, `0\u0027 denotes white pixel. You may assume that symbols are unique. \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003e\u003cb\u003eOutput\u003c/b\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003ePrint on the first line of output the minimal possible amount of pixels that enough to recognize symbol. I.e. you have to find set of pixels that has different color in each pair of symbols. If there are several sets with equal amount of pixels, output any of them. Then print N lines M characters each. `1\u0027 denotes that corresponding pixel is in your set. `0\u0027 denotes that corresponding pixel is not in your set. \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003e\u003cb\u003eSample test(s)\u003c/b\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003eInput\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cfont face\u003d\"Courier New\"\u003e\u003c/font\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cpre\u003e\u003c/pre\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e2 4 3 \u003cbr\u003e0000 \u003cbr\u003e0001 \u003cbr\u003e \u003cbr\u003e1000 \u003cbr\u003e0001 \u003cbr\u003e \u003cbr\u003e1111 \u003cbr\u003e1001 \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003eOutput\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cfont face\u003d\"Courier New\"\u003e\u003c/font\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cpre\u003e\u003c/pre\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e2 \u003cbr\u003e1000 \u003cbr\u003e1000 \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"left\"\u003e\u003cbr\u003e\u003cb\u003eNote\u003c/b\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003eIn this example output set consists of two most left pixels. They are both white in first symbol; black and white in second symbol; both black in third symbol. So, it\u0027s possible to recognize each symbol using this set of pixels. \u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"right\"\u003e \u003c/div\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003cdiv align\u003d\"right\"\u003e \u003c/div\u003e\u003c/div\u003e\u003cdiv align\u003d\"left\"\u003e\u003chr\u003e\u003c/div\u003e\u003ctable align\u003d\"left\" cellspacing\u003d\"7\"\u003e\u003ctbody\u003e\u003ctr\u003e\u003ctd\u003eAuthor:\u003c/td\u003e\u003ctd\u003eAndrew V. Lazarev \u003c/td\u003e\u003c/tr\u003e\u003ctr\u003e\u003ctd\u003eResource:\u003c/td\u003e\u003ctd\u003eSaratov SU Contest: Golden Fall 2004 \u003c/td\u003e\u003c/tr\u003e\u003ctr\u003e\u003ctd\u003eDate:\u003c/td\u003e\u003ctd\u003eOctober 2, 2004 \u003c/td\u003e\u003c/tr\u003e\u003c/tbody\u003e\u003c/table\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003cbr\u003e\u003c/div\u003e \u003c/div\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e\r\n\u003c/div\u003e"}}]}