Tenka1 Programmer Contest/Tenka1 Programmer Beginner Contest 2018

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1 Tenka1 Programmer Contest/Tenka1 Programmer Beginner Contest 2018 DEGwer 2018/10/27 For International Readers: English editorial starts on page 5. A: Measure #include <s t d i o. h> #include <s t r i n g > #include <iostream > using namespace std ; i n t main ( ) s t r i n g s ; c i n >> s ; i f ( s. s i z e ( ) == 2) cout << s << endl ; e l s e cout << s [ 2 ] << s [ 1 ] << s [ 0 ] << endl ; B: Exchange #include <s t d i o. h> using namespace std ; i n t main ( ) i n t a, b, k ; s c a n f ( %d%d%d, &a, &b, &k ) ; f o r ( i n t i = 0 ; i < k ; i ++) 1

2 i f ( i % 2 == 0) b += a / 2, a /= 2 ; e l s e a += b / 2, b /= 2 ; p r i n t f ( %d %d\n, a, b ) ; C: Align p 1,..., p N p i 1 < p i < p i+1 p i 1 > p i > p i+1 i p i p 1 p 2 p 3... p 1 p 2 p 3... (p 1 p 2 ) + (p 3 p 2 ) + (p 3 p 4 ) + (p 5 p 4 ) +... i p i ( ) p i D: Crossing k 2 1 1, 2,..., N 2 ( ) k 1 N = k(k 1)/2 1, 2,..., k(k 1)/2 1 E: Equilateral a, b, c, d b + d = a + c + d = a + b + c a + c = b = d b = d (a + c = b = d ) 45 O(max(N, M) 3 ) 2

3 1 D 2 E F: Circular A ( ) 1 1 ( K ) K A i i 3

4 i i i i ( ) i i i ( K ) i i K i i i i K ( K ) i K i i i K i i i 4

5 Tenka1 Programmer Contest/Tenka1 Programmer Beginner Contest 2018 Editorial DEGwer 2018/10/27 A: Measure #include <s t d i o. h> #include <s t r i n g > #include <iostream > using namespace std ; i n t main ( ) s t r i n g s ; c i n >> s ; i f ( s. s i z e ( ) == 2) cout << s << endl ; e l s e cout << s [ 2 ] << s [ 1 ] << s [ 0 ] << endl ; 5

6 B: Exchange #include <s t d i o. h> using namespace std ; i n t main ( ) i n t a, b, k ; s c a n f ( %d%d%d, &a, &b, &k ) ; f o r ( i n t i = 0 ; i < k ; i++) i f ( i % 2 == 0)b += a / 2, a /= 2 ; e l s e a += b / 2, b /= 2 ; p r i n t f ( %d %d\n, a, b ) ; 6

7 C: Align Suppose that after rearranging the sequence, we get p 1,..., p N. We can assume that there is no i that satisfies p i 1 < p i < p i+1 or p i 1 > p i > p i+1. If such i exists, we can move p i to the end of the sequence, and the sum of absolute differences never decreases. (If this operation results in new increasing (or decreasing) three consecutive elements at the end of the sequence, swap the last two elements again.) Thus, there are two cases: p 1 p 2 p 3... p 1 p 2 p 3... (Strictly speaking, there are some other cases like p 1 < p 2 = p 3 > p 4, but we can eliminate them with the same observation as above). Since these two cases are symmetric, let s describe the first case. We want to maximize the value (p 1 p 2 )+(p 3 p 2 )+(p 3 p 4 )+(p 5 p 4 )+... Here, we can forget about the constraint p 1 p 2 p 3... because when this condition doens t hold the value never becomes the maximum. For example, when n = 5, (p 1 p 2 ) + (p 3 p 2 ) + (p 3 p 4 ) + (p 5 p 4 ) = (+1) p 1 + ( 2) p 2 + (+2) p 3 + ( 2) p 4 + (+1) p 5. To achieve the maximum, we should assign the greatest elements to the position with the greatest coefficients. So, in the decreasing order of elements, we should assign them to positions 3, 1, 5, 2, 4 in this order. For general n, we can compute coefficients in a similar way, sort them, and assign the greatest elements to the position with the greatest coefficients. 7

8 1 Problem D D: Crossing Let k be the number of subsets we choose. From the condition in the statement, for each pair (i, j) such that 1 i < j k, there must be exactly one element that is contained in both S i and S j. Also, all elements must appear this way, because all elements are contained in exactly two sets. Thus, once we fix the value k, we can essentially uniquely determine subsets. N must be the number of pairs (i, j), thus N = k(k 1)/2. We first find k that satisfies N = k(k 1)/2 (If such k doesn t exist, there s no answer). Then, we can uniquely determine the subsets except for labelling. For example, when k = 4, There is an element contained in S 1 and S 2. Let s call it A. There is an element contained in S 1 and S 3. Let s call it B. There is an element contained in S 1 and S 4. Let s call it C. There is an element contained in S 2 and S 3. Let s call it D. There is an element contained in S 2 and S 4. Let s call it E. There is an element contained in S 3 and S 4. Let s call it F. Then we get S 1 = A, B, C, S 2 = A, D, E, and so on (and somehow assign the letters to integers between 1 and N). The following picture illustrates the case k = 6. 8

9 2 Problem E E: Equilateral There are two types of arrangements of three points on a plane: two points form a diagonal of a rectangle and the other point is inside the rectangle (like the picture on the left), or not (like the picture on the right). It s easy to see that left-type arrangement never satisfies the condition in the statement, so let s consider the right-type arrangement, and define the lengths a, b, c, d as in the picture. There lengths must satisfy b + d = a + c + d = a + b + c. Thus, we get b = d. This means that two points must be on the same diagonal (whose slope is 45 degrees). Let s fix these two points. For example, suppose that these points are (x, y) and (x + d, y + d). Then, the candidates of the coordinates of the other point are: (x d + i, y + d + i) for 0 i d. (x + d + i, y d + i) for 0 i d. With proper pre-computation (similar to prefix sums), we can count the number of existing points among these candidate coordinates in O(1). Since there are O(max(N, M) 3 ) ways to fix two points on the same diagonal, we can solve the entire problem in O(max(N, M) 3 ). Note that it s important to avoid double-countings. One possible way to do this is, in each phase, for the third point, only consider these coordinates: (x d + i, y + d + i) for 0 i < d. Then after each phase rotate the entire board by 90 degrees. After four phases, each valid triplet will be counted once. 9

10 F: Circular If all elements in given A are equal, the problem is easy, so let us assume otherwise. First, it can be seen that the region where each number occurs in A must form a contiguous segment (on the circle). Additionally, it can also be seen that the operation was performed K times, where K is the number of times 1 occurs in a row, minus 1. If there is a number that does not occur in a contiguous segment, or occurs K + 1 or more times in a row, the answer is 0. Otherwise, let us go over the numbers in A in ascending order, and for each number, determine its position in the original permutation. Let us say a number i to be existing when the given sequence A contains i. Assume we are looking at existing number i. If the number occurs just before the segment where i occurs is greater than i, the original position of i is uniquely determined (as the beginning of that segment). If the number occurs just after the segment where i occurs is greater than i, the original position of i is also uniquely determined (as K positions before the end of that segment). Note that if both numbers around the segment are greater than i and the number of times i occurs in a row is not K + 1, the answer is 0. Let us consider the remaining case, that is, both numbers around the segment where i occurs are less than i. Because of how the original positions of numbers are determined, there is no number whose original position is inside the segment where i occurs, or K or less positions before the beginning of that segment (particularly, K or less positions before the end of that segment). Thus, any position that is between 0 and K positions before the beginning of the segment where i occurs can be specified as the original position of i. Regarding the original position of non-existing i, any position that is inside the segment where numbers less than i occurs, and not among the last K positions of such segment, and not already specified as the original position of some other number, can be specified as the original position of i. Therefore, if both numbers around the segment where i occurs are less than i, or i does not exist, the number of ways to specify the original position of i does not depend on how the original positions of previous numbers are specified. We can just multiply the answer by some constant in these cases, and the problem is solved. 10